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CHAPTER 6<br />

Introduction to Vectors<br />

Review of Prerequisite Skills, p. 273<br />

"3<br />

1<br />

"2<br />

1. a. c. e.<br />

2<br />

2<br />

2<br />

"3<br />

b. 2"3 d. f. 1<br />

2<br />

2. Find BC using the Pythagorean theorem,<br />

AC 2 5 AB 2 1 BC 2 .<br />

BC 2 5 AC 2 2 AB 2<br />

5 10 2 2 6 2<br />

5 64<br />

BC 5 8<br />

Next, use the ratio tan A 5 opposite<br />

tan A 5 BC<br />

adjacent .<br />

AB<br />

5 8 6<br />

5 4 3<br />

3. a. To solve DABC, find measures of the sides and<br />

angles whose values are not given: AB, /B, and<br />

/C. Find AB using the Pythagorean theorem,<br />

BC 2 5 AB 2 1 AC 2 .<br />

AB 2 5 BC 2 2 AC 2<br />

5 (37.0) 2 2 (22.0) 2<br />

5 885<br />

AB 5 "885<br />

8 29.7<br />

Find /B using the ratio sin B 5<br />

opposite<br />

sin B 5 AC<br />

hypotenuse .<br />

BC<br />

5 22.0<br />

37.0<br />

/B 8 36.5°<br />

/C 5 90° 2 /B<br />

/C 5 90° 2 36.5°<br />

/C 8 53.5°<br />

b. Find measures of the angles whose values are not<br />

given. Find /A using the cosine law,<br />

cos A 5 b2 1 c 2 2 a 2<br />

.<br />

2bc<br />

5 52 1 8 2 2 10 2<br />

2(5)(8)<br />

5 211<br />

80<br />

/A 8 97.9°<br />

Find /B using the sine law.<br />

sin B<br />

5 sin A<br />

b a<br />

sin B sin (97.9°)<br />

5<br />

5 10<br />

sin B 8 0.5<br />

/B 8 29.7°<br />

Find /C using the sine law.<br />

sin C<br />

5 sin A<br />

c a<br />

sin C sin (97.9°)<br />

5<br />

8 10<br />

sin C 8 0.8<br />

/C 8 52.4°<br />

4. Since the sum of the internal angles of a triangle<br />

equals 180°, determine the measure of /Z using<br />

/X 5 60° and /Y 5 70°.<br />

/Z 5 180°2 (/X 1 /Y)<br />

5 180°2 (60° 1 70°)<br />

5 50°<br />

Find XY and YZ using the sine law.<br />

XY<br />

sin Y 5 XY<br />

sin Z<br />

XY<br />

sin 70° 5 6<br />

sin 50°<br />

XZ 8 7.36<br />

YZ<br />

sin X 5 XY<br />

sin Z<br />

YZ<br />

sin 60° 5 6<br />

sin 50°<br />

YZ 8 6.78<br />

5. Find each angle using the cosine law.<br />

cos R 5 RS2 1 RT 2 2 ST 2<br />

2(RS)(RT)<br />

5 42 1 7 2 2 5 2<br />

2(4)(7)<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-1


5 5 7<br />

/R 8 44°<br />

cos S 5 RS2 1 ST 2 2 RT 2<br />

2(RS)(ST)<br />

5 42 1 5 2 2 7 2<br />

2(4)(5)<br />

52 1 5<br />

/S 8 102°<br />

cos T 5 RT 2 1 ST 2 2 RS 2<br />

2(RT)(ST)<br />

5 72 1 5 2 2 4 2<br />

2(7)(5)<br />

5 29<br />

35<br />

/T 8 34°<br />

6.<br />

A<br />

3.5 km<br />

T<br />

70°<br />

8.<br />

C<br />

100 km/h<br />

48°<br />

A<br />

T<br />

80 km/h<br />

Find AC and AT using the speed of each vehicle and<br />

the elapsed time (in hours) until it was located,<br />

distance 5 speed 3 time.<br />

AC 5 100 km>h 3 1 4 h<br />

5 25 km<br />

AT 5 80 km>h 3 1 3 h<br />

5 26 2 3 km<br />

Find CT using the cosine law.<br />

CT 2 5 AC 2 1 AT 2 2 2(AC)(AT) cos A<br />

5 (25 km) 2 1 a26 2 2<br />

3 kmb<br />

6 km<br />

2 2(25 km) a26 2 kmb cos 48°<br />

3<br />

8 443.94 km 2<br />

CT 8 21.1 km<br />

9.<br />

A<br />

Find AB (the distance between the airplanes) using<br />

the cosine law.<br />

AB 2 5 AT 2 1 BT 2 2 2(AT)(BT)cos T<br />

5 (3.5 km) 2 1 (6 km) 2<br />

2 2(3.5 km)(6 km) cos 70°<br />

8 33.89 km 2<br />

AB 8 5.82 km<br />

7. P<br />

2 km 7 km<br />

Q 142°<br />

R<br />

Find QR using the cosine law.<br />

QR 2 5 PQ 2 1 PR 2 2 2(PQ)(PR) cos P<br />

5 (2 km) 2 1 (7 km) 2<br />

2 2(2 km)(7 km) cos 142°<br />

8 75.06 km 2<br />

QR 8 8.66 km<br />

B<br />

5 cm<br />

5 cm<br />

B<br />

C<br />

C<br />

The pentagon can be divided into 10 congruent<br />

right triangles with height AC and base BC.<br />

10 3 /A 5 360°<br />

/A 5 36°<br />

Find AC and BC using trigonometric ratios.<br />

AC 5 AB 3 cos A<br />

5 5 cos 36°<br />

8 4.0 cm<br />

BC 5 AB 3 sin A<br />

5 5 sin 36°<br />

8 2.9 cm<br />

The area of the pentagon is the sum of the areas of<br />

the 10 right triangles. Use the area of ^ABC to<br />

determine the area of the pentagon.<br />

6-2 <strong>Chapter</strong> 6: Introduction to Vectors


Area pentagon 5 10 3 1 2 (BC)(AC) 6-3<br />

5 10 3 1 (2.9 cm)(4.0 cm)<br />

2<br />

5 59.4 cm 2<br />

6.1 An Introduction to Vectors,<br />

pp. 279–281<br />

1. a. False. Two vectors with the same magnitude<br />

can have different directions, so they are not equal.<br />

b. True. Equal vectors have the same direction and<br />

the same magnitude.<br />

c. False. Equal or opposite vectors must be parallel<br />

and have the same magnitude. If two parallel vectors<br />

have different magnitude, they cannot be equal or<br />

opposite.<br />

d. False. Equal or opposite vectors must be parallel<br />

and have the same magnitude. Two vectors with the<br />

same magnitude can have directions that are not<br />

parallel, so they are not equal or opposite.<br />

2. Vectors must have a magnitude and direction. For<br />

some scalars, it is clear what is meant by just the<br />

number. Other scalars are related to the magnitude<br />

of a vector.<br />

• Height is a scalar. Height is the distance (see below)<br />

from one end to the other end. No direction is given.<br />

• Temperature is a scalar. Negative temperatures are<br />

below freezing, but this is not a direction.<br />

• Weight is a vector. It is the force (see below) of<br />

gravity acting on your mass.<br />

• Mass is a scalar. There is no direction given.<br />

• Area is a scalar. It is the amount space inside a<br />

two-dimensional object. It does not have<br />

direction.<br />

• Volume is a scalar. It is the amount of space inside<br />

a three-dimensional object. No direction is given.<br />

• Distance is a scalar. The distance between two<br />

points does not have direction.<br />

• Displacement is a vector. Its magnitude is related<br />

to the scalar distance, but it gives a direction.<br />

• Speed is a scalar. It is the rate of change of<br />

distance (a scalar) with respect to time, but does<br />

not give a direction.<br />

• Force is a vector. It is a push or pull in a certain<br />

direction.<br />

• Velocity is a vector. It is the rate of change of<br />

displacement (a vector) with respect to time. Its<br />

magnitude is related to the scalar speed.<br />

3. Answers may vary. For example: Friction resists<br />

the motion between two surfaces in contact by<br />

acting in the opposite direction of motion.<br />

• A rolling ball stops due to friction which resists<br />

the direction of motion.<br />

• A swinging pendulum stops due to friction<br />

resisting the swinging pendulum.<br />

4. Answers may vary. For example:<br />

a. AD ><br />

b. AD > 5 BC > ; AB ><br />

ED > 52CB > 5<br />

52EB > AB > DC > ; AE ><br />

c. AC > & DB > DA > 52CD > 5 EC ><br />

;<br />

; AE > 52BC > AE > ; DE > 5<br />

;<br />

; 52CE > EB ><br />

;<br />

& EB > ; EC ><br />

& DE > ; AB ><br />

& CB ><br />

5.<br />

B<br />

H<br />

D<br />

E<br />

A<br />

C<br />

G<br />

a. AB ><br />

b. AB > 5 CD ><br />

c. @AB > 52EF ><br />

@ but<br />

d. GH > 5 @ EF > @<br />

e. AB > 5 2AB > AB > 2 EF ><br />

522JI ><br />

7. a. 100 km><br />

h, south<br />

b. 50 km><br />

h, west<br />

c. 100 km><br />

h, northeast<br />

d. 25 km><br />

h, northwest<br />

e. 60 km><br />

h, east<br />

F<br />

6. a. b. c. d. e.<br />

I<br />

J<br />

Calculus and Vectors <strong>Solutions</strong> Manual


8. a. 400 km><br />

h, due south<br />

b. 70 km><br />

h, southwesterly<br />

c. 30 km><br />

h southeasterly<br />

d. 25 km><br />

h, due east<br />

9. a. i. False. They have equal magnitude, but<br />

opposite direction.<br />

ii. True. They have equal magnitude.<br />

iii. True. The base has sides of equal length, so the<br />

vectors have equal magnitude.<br />

iv. True. They have equal magnitude and direction.<br />

b. E<br />

H<br />

To calculate @BD > @ , @BE > @ and @BH > @ , find the lengths<br />

of their corresponding line segments BD, BE and<br />

BH using the Pythagorean theorem.<br />

BD 2 5 AB 2 1 AD 2<br />

5 3 2 1 3 2<br />

BD 5 "18<br />

BE 2 5 AB 2 1 AE 2<br />

5 3 2 1 8 2<br />

BE 5 "73<br />

BH 2 5 BD 2 1 DH 2<br />

5 ("18) 2 1 8 2<br />

BH 5 "82<br />

10. a. The tangent vector describes James’s velocity<br />

at that moment. At point A his speed is 15 km><br />

h<br />

and he is heading north. The tangent vector shows<br />

his velocity is 15 km><br />

h, north.<br />

b. The length of the vector represents the magnitude<br />

of James’s velocity at that point. James’s speed is<br />

the same as the magnitude of James’s velocity.<br />

c. The magnitude of James’s velocity (his speed) is<br />

constant, but the direction of his velocity changes at<br />

every point.<br />

d. Point C<br />

e. This point is halfway between D and A, which is<br />

of the way around the circle. Since he is running<br />

7<br />

8<br />

A<br />

B<br />

F<br />

D<br />

G<br />

C<br />

15 km><br />

h and the track is 1 km in circumference, he<br />

can run around the track 15 times in one hour. That<br />

7<br />

means each lap takes him 4 minutes. 8 of 4 minutes<br />

is 3.5 minutes.<br />

3<br />

f. When he has travelled 8 of a lap, James will be<br />

halfway between B and C and will be heading<br />

southwest.<br />

11. a. Find the magnitude of AB ><br />

using the distance<br />

formula.<br />

@AB > @ 5 "(x A 2 x B ) 2 1 (y B 2 y A ) 2<br />

5 "(24 1 1) 2 1 (3 2 2) 2<br />

5 or 3.16<br />

b. CD > "10<br />

5 AB > . AB ><br />

moves from A(24, 2) to<br />

B(21, 3) or (x B , y B ) 5 (x A 1 3, y A 1 1). Use this<br />

to find point D.<br />

( x D , y D ) 5 (x C 1 3, y C 1 1)<br />

5 (26 1 3, 0 1 1)<br />

c. EF > 5 (23,<br />

5 AB > 1)<br />

. Find point E using<br />

( x A , y A ) 5 (x B 2 3, y B 2 1).<br />

( x E , y E ) 5 (x F 2 3, y F 2 1)<br />

5 (3 2 3, 22 2 1)<br />

d. GH > 5 (0, 23)<br />

52AB > , and moves in the opposite<br />

direction as AB > .<br />

( x H , y H ) 5 (x G 2 3, y G 2 1).<br />

( x H , y H ) 5 (x G 2 3, y G 2 1)<br />

5 (3 2 3, 1 2 1)<br />

5 (0, 0)<br />

6.2 Vector Addition, pp. 290–292<br />

1. a.<br />

b.<br />

c.<br />

x – y<br />

y – x<br />

x<br />

–x<br />

x + y<br />

x<br />

y<br />

–y<br />

y<br />

6-4 <strong>Chapter</strong> 6: Introduction to Vectors


d.<br />

–x<br />

–y –x<br />

2. a. BA > A<br />

C<br />

b. 0 > A<br />

–y<br />

B<br />

c.<br />

d.<br />

a<br />

a<br />

–b<br />

–c<br />

a – b – c<br />

–b<br />

c<br />

c. CB ><br />

C<br />

A<br />

B<br />

4. a.<br />

a – b + c<br />

b<br />

d. CA ><br />

3. a.<br />

C<br />

C<br />

c<br />

A<br />

B<br />

B<br />

b.<br />

a<br />

a<br />

(b + c)<br />

a + (b + c)<br />

b<br />

(a + b)<br />

c<br />

c<br />

b.<br />

b<br />

a + b + c<br />

–c<br />

a<br />

b<br />

(a + b) + c<br />

c. The resultant vectors are the same. The order in<br />

which you add vectors does not matter.<br />

Aa > 1 b > B 1 c > 5 a > 1 Ab > 1 c > B<br />

5. a. PS ><br />

R –RQ Q<br />

RS<br />

PR PQ<br />

S<br />

PS<br />

b. 0 > P<br />

Q<br />

a + b – c<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

a<br />

S<br />

RQ<br />

–RS<br />

R –PQ<br />

6. x > 1 y > PS<br />

5 MR > P<br />

z > 1 t > 5 MS > 1 RS ><br />

5 ST ><br />

5 SQ > 1 TQ ><br />

so<br />

( x > 1 y > ) 1 (z > 1 t > ) 5 MS ><br />

5 MQ > 1 SQ ><br />

6-5


7. a.<br />

b. y > 2x ><br />

c. x > 1<br />

d. 2x > y ><br />

e. x > 1<br />

1<br />

f. 2x > y > y ><br />

1<br />

g. 2x > 2 y > z ><br />

h. 2x > 1 y ><br />

2z > 1 z ><br />

8. a.<br />

y<br />

b. See the figure in part a. for the drawn vectors.<br />

0 y > 2 and<br />

so 0 y > 2 x > 0 2 5 0 x > 2 y > 2<br />

0 2x > x > 0 2 5<br />

0 5 0 x > 0 y > 0 2 1 0 x > 0 2 2 20 y > 002x > 0 cos (u)<br />

0,<br />

0<br />

9. a. Maria’s velocity is 11 km><br />

h downstream.<br />

b.<br />

4 km/h<br />

c.<br />

y –x<br />

7 km/h<br />

4 km/h<br />

A<br />

7 km/h<br />

Maria’s speed is 3 km><br />

h.<br />

10. a.<br />

x<br />

u<br />

11 km/h<br />

3 km/h<br />

f 1 + f 2<br />

u<br />

f 1<br />

ship<br />

f 2<br />

b. The vectors form a triangle with side lengths<br />

S S S S S S<br />

@ f1 @ , @ f2 @ and @ f1 1 f2 @ . Find @ f 1 1 f2 @ using the<br />

cosine law.<br />

S S<br />

@ f 1 1 f2 @ 2 S<br />

5 @ f 1<br />

@ 2 S<br />

1 @ f 2<br />

@ 2 S S<br />

2 2 @ f @@f2<br />

1<br />

@ cos (u)<br />

S S S<br />

@ f 1 1 f2 0 5 $ @ f 1<br />

@ 2 S<br />

1 @ f 2<br />

@ 2 S S<br />

2 2 @ f @@f2<br />

1<br />

@ cos (u)<br />

B<br />

y<br />

y<br />

x – y<br />

D<br />

u<br />

x<br />

C<br />

11.<br />

a + w<br />

Find using the Pythagorean theorem.<br />

0 a > 0 a ><br />

1 w > 1 w > 0<br />

0 2 5 0 a > 0 2 1 0 w > 2<br />

0<br />

5 (150 km>h) 2 1 (80 km>h) 2<br />

0 a > 1 w > 5 28900<br />

0 2 5 170<br />

Find the direction of a > 1 w > using the ratio<br />

tan(u) 5 0 w> 0<br />

0 a > 0<br />

u5tan 21<br />

150 km>h<br />

a > 1<br />

12. and form a right triangle. Find<br />

0 x > x > w > 8 N 28.1° W<br />

y > 5 170<br />

using the Pythagorean theorem.<br />

0 x > 1 y > x > km>h,<br />

1 y > N 28.1° W<br />

, ,<br />

1 y > 0<br />

0 2 5 0 x > 0 2 1 0 y > 2<br />

0<br />

5 7 2 1 24 2<br />

0 x > 1 y > 5 625<br />

0 5 25<br />

Find the angle between x > and x > 1 y > using the ratio<br />

tan (u) 5 0 y> 0<br />

0 x > 0<br />

21<br />

24<br />

u5tan<br />

7<br />

8 73.7°<br />

13. Find @AB > 1 AC > @ using the cosine law and the<br />

supplement to the angle between and<br />

@AB > 1<br />

5 @AB > AC > AB ><br />

AC > .<br />

2<br />

@<br />

@ 2 1 @AC > @ 2 2 2 @AB > @ @AC > @ cos (30°)<br />

5 1 2 1 1 2 "3<br />

2 2(1)(1)<br />

2<br />

5 2<br />

@AB > 2 "3<br />

1 AC > @ 8 0.52<br />

14. D<br />

A<br />

w<br />

u<br />

a<br />

80 km>h<br />

E<br />

B<br />

The diagonals of a parallelogram bisect each other.<br />

So EA > 52EC ><br />

and<br />

Therefore, EA > ED ><br />

1 EB > 52EB ><br />

1 EC > .<br />

1 ED > 5 0 > .<br />

C<br />

6-6 <strong>Chapter</strong> 6: Introduction to Vectors


15. P<br />

T<br />

b<br />

R<br />

b S<br />

Multiple applications of the Triangle Law for<br />

adding vectors show that<br />

RM > 1 b > 5 a > 1 TP ><br />

(since both are equal to the<br />

undrawn vector and that<br />

RM > 1 a > 5 b > TM ><br />

1 SQ > ),<br />

(since both are equal to the<br />

undrawn vector RQ > )<br />

Adding these two equations gives<br />

2 RM > 1 a > 1 b > 5 a > 1> b > 1 TP > ><br />

1 SQ ><br />

16. a > 1 b >2 RM> 5 TP 1 SQ<br />

and a > 2 b ><br />

represent the diagonals of a<br />

parallelogram with sides a > and b > .<br />

b<br />

a<br />

M<br />

a – b<br />

Since @a > a<br />

1 b > @ 5 @a > 2 b > @ and the only parallelogram<br />

with equal diagonals is a rectangle, the parallelogram<br />

must also be a rectangle.<br />

17. P<br />

G<br />

a<br />

a + b<br />

Q<br />

M<br />

R<br />

Let point M be defined as shown. Two applications<br />

of the Triangle Law for adding vectors show that<br />

GQ ><br />

GR > 1 QM ><br />

1 RM > 1 MG ><br />

1 MG > 5 0 ><br />

5 0 ><br />

Adding these two equations gives<br />

GQ > 1 QM > 1 2 MG > 1 GR > 1 RM > 5 0 ><br />

From the given information,<br />

2 MG > and<br />

QM > 5 GP ><br />

1 RM > 5 0 ><br />

(since they are opposing vectors of<br />

equal length), so<br />

GQ > 1 GP > 1 GR > 5 0 > , as desired.<br />

Q<br />

6.3 Multiplication of a Vector by a<br />

Scalar, pp. 298–301<br />

1. A vector cannot equal a scalar.<br />

2. a.<br />

3 cm<br />

b.<br />

c.<br />

d.<br />

3. E25°N describes a direction that is 25° toward the<br />

north of due east ( 90° east of north), in other words<br />

90° 2 25° 5 65° toward the east of due north. N65°E<br />

and “a bearing of 65° ” both describe a direction that<br />

is 65° toward the east of due north. So, each is<br />

describing the same direction in a different way.<br />

4. Answers may vary. For example:<br />

a.<br />

2 cm<br />

v ><br />

6 cm<br />

9 cm<br />

2v ><br />

b. 1<br />

c.<br />

2 v><br />

d. e. 1<br />

0 v > v ><br />

0<br />

22v > 2 2 3 v><br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-7


5. a.<br />

y<br />

b. c.<br />

3b<br />

–3b<br />

y<br />

y<br />

x + 3y<br />

x<br />

b.<br />

–y<br />

x<br />

–y<br />

x – 3y<br />

c.<br />

–2x + y<br />

–x<br />

y –x<br />

d.<br />

–x<br />

–x<br />

–2x – y<br />

–y<br />

6. Answers may vary. For example:<br />

a<br />

b<br />

–y<br />

d.<br />

e.<br />

7. a.<br />

m(2a > ) 1 n a 3 2 a> b 5 0 ><br />

m(4a > ) 1 n (3a > ) 5 0 ><br />

m 5 3 and n 524 satisfy the equation, as does any<br />

multiple of the pair (3, 24). There are infinitely<br />

many values possible.<br />

b.<br />

da > 1 ea 3 2 a> b 1 f(2a > ) 5 0 ><br />

2da > 1 3ea > 1 4f a > 5 0 ><br />

d 5 2, e 5 0, and f 521 satisfy the equation, as<br />

does any multiple of the triple (2, 0, 21). There are<br />

infinitely many values possible.<br />

8. or<br />

a > and b ><br />

are collinear, so where k is a nonzero<br />

scalar. Since 0 a > 0 5 @b > a > 5 kb > ,<br />

@ , k can only be 21 or 1.<br />

9.<br />

b<br />

a<br />

2a – 3b<br />

c > 5 2a > , b > 5 3<br />

mc > 1 nb > 2 a><br />

5 0 ><br />

c > 5 2a > , b > 5 3<br />

da > 1 eb > 1 fc > 5 0 > 2 a><br />

a<br />

a<br />

4a<br />

a<br />

2a + 3b<br />

a<br />

–b<br />

–b<br />

–b<br />

b<br />

b<br />

b<br />

a.<br />

2a<br />

–2b<br />

Yes<br />

6-8 <strong>Chapter</strong> 6: Introduction to Vectors


10. Two vectors are collinear if and only if they can<br />

be related by a scalar multiple. In this case<br />

a. collinear<br />

b. not collinear<br />

c. not collinear<br />

d. collinear<br />

1<br />

11. a. is a vector with length 1 unit in the same<br />

direction as x > .<br />

b. 2 1<br />

0 x > 0 x > is a vector with length 1 unit in the<br />

opposite direction of x > .<br />

12.<br />

a<br />

a<br />

b<br />

b<br />

b<br />

1<br />

2<br />

sin u 8 (1)<br />

2.91<br />

u 8 9.9° from x > towards y ><br />

16. b > 5 1<br />

0 a > a > 0<br />

@ b > 1<br />

@ 5 `<br />

0 a > a >` 0<br />

@b > @ 5 1<br />

b > @b > 0 a > 0 a > 0<br />

0<br />

@ 5 1<br />

is a positive multiple of so it points in the<br />

same direction as a > a > ,<br />

and has magnitude 1. It is a<br />

unit vector in the same direction as a > .<br />

17.<br />

A<br />

0 x > 0 x > a > 2 kb > 6-9<br />

m 5 2 3<br />

13. a. 2 2 3 a><br />

1<br />

b.<br />

3 a> 1<br />

c.<br />

3 0 a> 0<br />

2<br />

d.<br />

3 0 a> 0<br />

4<br />

e.<br />

3 a><br />

14. and make an angle of , so you may find<br />

0 2x > x ><br />

using the Pythagorean theorem.<br />

0 2x > 1 y > y ><br />

90°<br />

1 y > 0<br />

0 2 5 0 2x > 0 2 1 0 y > 2<br />

0<br />

0 2x > 1 y > 5 2 2 1 1 2<br />

0 5 "5 or 2.24<br />

Find the direction of 2x > 1 y > using the ratio<br />

tan (u) 5 0 y> 0<br />

0 2x > 0<br />

u5tan 1 21 2<br />

8 26.6° from towards 2<br />

15. Find 0 2x > 1 y > x > x > 1 y ><br />

0 using the cosine law, and the<br />

supplement to the angle between and<br />

0 2x > 1 y > 0 2 5 0 2x > 0 2 1 0 y > 0 2 2 20 2x > x ><br />

00y > y > .<br />

0 cos (150°)<br />

2"3<br />

5 2 2 1 1 2 2 2(2)(1)<br />

0 2x > 1 y > 2<br />

0 8 2.91<br />

Find the direction of 2x > 1 y > using the sine law.<br />

sin u<br />

0 y > sin (150°)<br />

5<br />

0 0 2x > 1 y > 0<br />

B<br />

D<br />

C<br />

AD ><br />

AD > 5 c ><br />

2 AD > 5 b > 1 CD ><br />

5<br />

But CD > c > 1 BD ><br />

1 b ><br />

So 2 AD > 1 BD > 1 CD > 1 BD ><br />

5 c > 5<br />

1 b > 0.<br />

, or AD > 5 1<br />

18. PM > and<br />

so MN > 5 a ><br />

5 PN > PN ><br />

PQ > 5<br />

5 and<br />

so QR > 2a > b > 2<br />

2 a > PM > 5 b > 2 c> 1 1 2 b> .<br />

5 PR > PR > 5<br />

5 2b > 2 PQ > 2b ><br />

2 2a ><br />

Notice that<br />

2MN > 5 2b > 2<br />

5 QR > 2a ><br />

We can conclude that is parallel to and<br />

@QR > @ 5 2 @MN > QR ><br />

MN ><br />

@ .<br />

19. A B<br />

C<br />

c<br />

AD<br />

E<br />

b<br />

D<br />

Calculus and Vectors <strong>Solutions</strong> Manual


Answers may vary. For example:<br />

a. u > and<br />

b. u > 5 AB ><br />

and<br />

c. u > 5 AD > v ><br />

and<br />

d. u > 5 AC > v > 5 CD ><br />

5 ED > v > 5 AE ><br />

and v > 5 DB ><br />

5 AD ><br />

20. a. Since the magnitude of is three times the<br />

magnitude of y > x ><br />

and because the given sum is<br />

must be in the opposite direction of ny > 0, mx ><br />

and<br />

0 n 0 5 30 m 0.<br />

b. Whether x > and y > are collinear or not, m 5 0 and<br />

n 5 0 will make the given equation true.<br />

21. a.<br />

b. BE > CD > 5<br />

5 2b > b > 2<br />

2<br />

5 2(b > 2a > a ><br />

2<br />

5 2CD > a > )<br />

The two are therefore parallel (collinear) and<br />

@BE > @ 5 2 @CD > @<br />

22. A<br />

B<br />

E<br />

D<br />

C<br />

Applying the triangle law for adding vectors shows<br />

that<br />

AC > 5 AD > 1 DC ><br />

The given information states that<br />

AB > 5 2 3 DC><br />

3<br />

2 AB> 5 DC ><br />

By the properties of trapezoids, this gives<br />

3<br />

2 AE > 5 EC > , and since<br />

AC > 5 AE > 1 EC ><br />

, the original equation gives<br />

AE > 1 3 2 AE> 5 AD > 1 3 2 AB><br />

5<br />

2 AE> 5 AD > 1 3 2 AB><br />

6.4 Properties of Vectors, pp. 306–307<br />

AE > 5 2 5 AD> 1 3 5 AB><br />

1. a. 0<br />

b. 1<br />

c. 0 ><br />

d. 1<br />

2. a > 1 b > 5 b > 1 a ><br />

b<br />

3.<br />

4. Answers may vary. For example:<br />

a<br />

a + b<br />

k(a + b)<br />

k a<br />

k b<br />

5. PQ > 5 RQ > 1<br />

5 (RQ > SR > 1<br />

5 (SR > 1 SR > TS > 1<br />

5 SQ > 1 RQ > ) 1 (TS > PT ><br />

5 PS > 1 PS > ) 1 (PT > 1 PT ><br />

1 TS > )<br />

)<br />

5<br />

6. a.<br />

b. 0 > EC > PQ > 1 SQ ><br />

c. Yes, the diagonals of a rectangular prism are of<br />

equal length<br />

7. 5 3a > 2<br />

1 3b > 6b > 2<br />

524a > 2 3c > 15c > 2 6a > 1 12b > 2 6c > 2 a ><br />

8. a. 5 6i > 1 9b ><br />

2<br />

5 12i > 8j > 2 24c ><br />

1<br />

2 17j > 2k > 1<br />

1 5k > 6i > 2 9j > 1 3k ><br />

b.<br />

c.<br />

(a + b)<br />

b + a<br />

(b + c)<br />

a<br />

(a + b)+ c =<br />

= a + b + c<br />

5 3i > 2<br />

527i > 4j > 1<br />

5 2(3i > 1 11j > k > 2 10i ><br />

2 4j > 2<br />

23(26i > 1 k > 4k > 1 15j > 2 5k ><br />

1 8j > 1 6i ><br />

2 2k > 2 9j > 1<br />

1 14i > 3k > )<br />

2 21j > 1 7k > )<br />

526i > 1 13j > 2 7k ><br />

9. Solve the first equation for x > .<br />

x > 5 1 2 a> 2 3 2 y><br />

b<br />

c<br />

a<br />

a<br />

a + b<br />

a + (b+ c)<br />

b<br />

b<br />

6-10 <strong>Chapter</strong> 6: Introduction to Vectors


Substitute into the second equation.<br />

3<br />

2 TO> 5 TZ > 1 1 2 TX><br />

6b > 52a 1 2 a> 2 3 2 y> b 1 5y > 6-11<br />

y > 5 1<br />

13 a> 1 12<br />

Lastly, find x > 13 b><br />

in terms of a > and b > .<br />

x > 5 1 2 a> 2 3 2 a 1 13 a> 1 12<br />

13 b> b<br />

5 5<br />

10. a > 13 a> 2 18<br />

5 x > 2 y ><br />

11. a. AG ><br />

BH > 5 a > 1<br />

CE > 52a > b > 1<br />

DF > 52a > 1 b > c ><br />

b. @AG > 5 a > 2<br />

2<br />

@ 2 5 0 a > b > b > 1 c ><br />

1<br />

1 c > c ><br />

0 2<br />

5 0 2a > 1 @b > @ 2<br />

0 2 1 @b > 1 0 c > 2<br />

0<br />

@ 2 1 0 c > 0<br />

5 @BH > 2<br />

@<br />

Therefore, @AG > @ 5 @BH > @<br />

12. T<br />

X<br />

Z<br />

5 2 3 y> 1 1 3 z> 2 (b > 1 z > )<br />

5 2 3 y> 2 2 3 z> 2 b ><br />

5 2 3 (y> 2 z > ) 2 b ><br />

5 2 3 b> 2 b ><br />

52 1 3 b><br />

13 b><br />

O<br />

Applying the triangle law for adding vectors<br />

shows that<br />

TY > 5 TZ > 1 ZY ><br />

The given information states that<br />

TX > 5 2 ZY ><br />

1<br />

2 TX> 5 ZY ><br />

By the properties of trapezoids, this gives<br />

1<br />

and since TY > 5 TO > 1 OY > 2 TO > 5 OY > ,<br />

, the<br />

original equation gives<br />

TO > 1 1 2 TO> 5 TZ > 1 1 2 TX><br />

Y<br />

2<br />

TO > 5 2 3 TZ> 1 1 3 TX><br />

Mid-<strong>Chapter</strong> Review, pp. 308–309<br />

1. a. AB ><br />

BA > 5 DC ><br />

AD > 5 CD ><br />

CB > 5 BC ><br />

5 DA ><br />

There is not enough information to determine if<br />

there is a vector equal to<br />

b. @PD > @ 5 @DA > AP > .<br />

@<br />

(parallelogram)<br />

2. a. RV >5 @BC> @<br />

b. RV ><br />

c. PS ><br />

d. RU ><br />

e. PS ><br />

f. PQ ><br />

3. a. Find @a > 1 b > @ using the cosine law, and the<br />

supplement to the angle between the vectors.<br />

@a > 1 b > @ 2 5 0 a > 0 2 1 @b > @ 2 2 20 a > 0 @b > @ cos 60°<br />

1<br />

5 3 2 1 4 2 2 2(3)(4)<br />

2<br />

@a > 1 b > 5 3<br />

@ 5 "3<br />

b. Find u using the ratio<br />

tan u5 @b> @<br />

0 a > 0<br />

5 4 3<br />

u5tan 21 4 3<br />

8 53°<br />

4. t 5 4 or t 524<br />

5. In quadrilateral PQRS, look at ^PQR. Joining the<br />

midpoints B and C creates a vector that is parallel<br />

to PR > and half the length of PR > BC ><br />

. Look at ^SPR.<br />

Joining the midpoints A and D creates a vector<br />

that is parallel to and half the length of<br />

is parallel to AD > PR > PR > AD ><br />

and equal in length to AD > . BC ><br />

.<br />

Therefore, ABCD is a parallelogram.<br />

6. a. Find using the cosine law. Note<br />

0 2v > and the angle between and is<br />

0 u > 0 5<br />

2 v > 0 v > 0 u > 2 v > 0<br />

0<br />

0 2 5 0 u > 0 2 1 0 2v > 0 2 2 20 u > u ><br />

002v > 2v > 120°.<br />

0 cos 60°<br />

Calculus and Vectors <strong>Solutions</strong> Manual


5 8 2 1 10 2 2 2(8)(10)a 1 2 b<br />

0 u > 2 v > 0 5 2"21<br />

b. Find the direction of u > 2 v > using the sine law.<br />

sin u<br />

0 2v > sin 60°<br />

5<br />

0 0 u > 2 v > 0<br />

sin u5 5 sin 60°<br />

"21<br />

u5sin 21 5<br />

"28<br />

8 71°<br />

1<br />

c.<br />

0 u > 1 v > (u > 1 v > ) 5 1<br />

0<br />

d. Find using the cosine law.<br />

0 5u > 0 5u ><br />

1 2v > 1 2v > 2"21 (u> 1 v > )<br />

0 2 5 0 5u > 0<br />

0 2 1 0 2v > 0 2 2 20 5u > 002v > 0 cos 120°<br />

0 5u > 1 2v > 0<br />

7. Find using the cosine law.<br />

0 2p > 2 q > 0 2p > 5 20"7<br />

2 q ><br />

0 2 5 0 2p > 0<br />

0 2 1 0 2q > 0 2 2 20 2p > 002q > 0 cos 60°<br />

8. 0 m ><br />

9. BC > 1 n > 0 5<br />

DC > 52y > 0 m > 02 0 n > 0<br />

5 x ><br />

5 40 2 1 20 2 2 2(40)(20)a2 1 2 b<br />

5 2 2 1 1 2 2 2(2)(1)a 1 2 b 5 3<br />

BD ><br />

AC > 52x ><br />

5 x > 2<br />

2 y > y ><br />

10. Construct a parallelogram with sides OA ><br />

and .<br />

Since the diagonals bisect each other, is the<br />

diagonal equal to<br />

Or<br />

and So, And<br />

AC > AB > 5 1<br />

5 OC > 2 OA > OB > 5<br />

. Now OB > OA > 1 1<br />

Multiplying by 2 gives<br />

11.<br />

3x > AC > 1<br />

2 y > CD ><br />

1<br />

AB > 3x > 2y > 5 AD > 2OB > 5 OA > 2 AC > 2 AC > OA > 1 OC > 2OB > OC ><br />

. OB > 5 OA > 1 AB ><br />

.<br />

.<br />

5 OA > 1 1 2 (OC ><br />

1 OC > 2 OA > )<br />

.<br />

x > 1 BD > 1 y > 5 AD ><br />

5<br />

1 BD > 5 AD > AD ><br />

AB > BD > 5 3x ><br />

x > 1 BC > 5 2x > 1 y ><br />

1 BC > 5 AC > 1 y ><br />

BC > 5 3x ><br />

5 2x > 2 y ><br />

2 y ><br />

12. The air velocity of the airplane and the<br />

wind velocity (W > (V > air)<br />

) have opposite directions.<br />

V > ground 5 V > air 2 W ><br />

5 460 km><br />

h due south<br />

13. a.<br />

b. PT > PT><br />

c. SR ><br />

14. a. a<br />

b.<br />

c.<br />

d.<br />

15.<br />

1<br />

2<br />

a<br />

1<br />

3<br />

a<br />

1<br />

2<br />

b<br />

b<br />

1<br />

3<br />

–2 b<br />

a + b<br />

–b<br />

PS > 5 PQ ><br />

RS > 5 3b > 1<br />

5 QS > 2 a > QS ><br />

2<br />

523a > QR ><br />

6.5 Vectors in R 2 and R 3 , pp. 316–318<br />

1. No, as the y-coordinate is not a real number.<br />

2. a. We first arrange the x-, y-, and z-axes (each a<br />

copy of the real line) in a way so that each pair of<br />

axes are perpendicular to each other (i.e., the x- and<br />

y-axes are arranged in their usual way to form the<br />

xy-plane, and the z-axis passes through the origin of<br />

the xy-plane and is perpendicular to this plane).<br />

This is easiest viewed as a “right-handed system,”<br />

where, from the viewer’s perspective, the positive<br />

z-axis points upward, the positive x-axis points out<br />

of the page, and the positive y-axis points rightward<br />

in the plane of the page. Then, given point P(a, b, c),<br />

we locate this point’s unique position by moving a<br />

units along the x-axis, then from there b units<br />

parallel to the y-axis, and finally c units parallel to<br />

the z-axis. It’s associated unique position vector is<br />

determined by drawing a vector with tail at the<br />

origin O(0, 0, 0) and head at P.<br />

b. Since this position vector is unique, its<br />

coordinates are unique. Therefore a 524, b 523,<br />

and c 528.<br />

3. a. Since A and B are really the same point, we<br />

can equate their coordinates. Therefore a 5 5,<br />

b 523, and c 5 8.<br />

3<br />

2<br />

a<br />

6-12 <strong>Chapter</strong> 6: Introduction to Vectors


. From part a., A(5, 23, 8), so OA > 5 (5, 23, 8).<br />

Here is a depiction of this vector.<br />

z<br />

z<br />

OA (5, –3, 8)<br />

y<br />

O(0, 0, 0)<br />

(0, 4, 0)<br />

y<br />

(0, 0, –2)<br />

(4, 0, 0)<br />

(0, 4, –2)<br />

x<br />

(4, 4, 0)<br />

(4, 0, –2) C(4, 4, –2)<br />

x<br />

4. This is not an acceptable vector in I 3 as the<br />

z-coordinate is not an integer. However, since all of<br />

the coordinates are real numbers, this is acceptable<br />

as a vector in R 3 .<br />

5.<br />

z<br />

(4, –4, 0)<br />

x<br />

A(4, –4, –2)<br />

(0, –4, –2)<br />

(0, –4, 0)<br />

(4, 0, –2)<br />

z<br />

O(0, 0, 0)<br />

(0, 0, –2)<br />

(4, 0, 0)<br />

(–4, 0, 2)<br />

y<br />

B(–4, 4, 2)<br />

6. a. A(0, 21, 0) is located on the y-axis.<br />

B(0, 22, 0), C(0, 2, 0), and D(0, 10, 0) are three<br />

other points on this axis.<br />

b. OA > 5 (0, 21, 0), the vector with tail at the<br />

origin O(0, 0, 0) and head at A.<br />

7. a. Answers may vary. For example:<br />

OA ><br />

OC > 5 (0, 0, 1), OB > 5 (0, 0, 21),<br />

5 (0, 0, 25)<br />

b. Yes, these vectors are collinear (parallel), as they<br />

all lie on the same line, in this case the z-axis.<br />

c. A general vector lying on the z-axis would be of<br />

the form OA > 5 (0, 0, a) for any real number a.<br />

Therefore, this vector would be represented by<br />

placing the tail at O, and the head at the point<br />

(0, 0, a) on the z-axis.<br />

8.<br />

z<br />

A(1, 0, 0)<br />

E(2, 0, 3)<br />

B(0, –2, 0)<br />

F(0, 2, 3)<br />

D(2, 3, 0)<br />

y<br />

(0, 0, 2)<br />

(–4, 0, 0)<br />

x<br />

C(0, 0, –3)<br />

O(0, 0, 0)<br />

(0, 4, 2)<br />

(0, 4, 0)<br />

(–4, 4, 0)<br />

y<br />

9. a.<br />

z<br />

x<br />

y<br />

x<br />

A(3, 2, –4)<br />

C(0, 1, –4)<br />

B(1, 1, –4)<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-13


. Every point on the plane containing points A, B,<br />

and C has z-coordinate equal to 24. Therefore, the<br />

equation of the plane containing these points is<br />

z 524 (a plane parallel to the xy-plane through<br />

the point z 524).<br />

10. a.<br />

z A(1, 2, 3)<br />

(0, 0, 3)<br />

(0, 2, 3)<br />

(1, 0, 3)<br />

O(0, 0, 0)<br />

(0, 2, 0)<br />

y<br />

(1, 0, 0)<br />

(1, 2, 0)<br />

x<br />

d.<br />

e.<br />

z<br />

D(1, 1, 1)<br />

(0, 0, 1)<br />

(1, 0, 1)<br />

(0, 1, 1)<br />

y<br />

O(0, 0, 0)<br />

(0, 1, 0)<br />

(1, 1, 0)<br />

x (1, 0, 0)<br />

z<br />

(0, –1, 1)<br />

(0, 0, 1)<br />

E(1, –1, 1)<br />

b.<br />

z<br />

(–2, 0, 1)<br />

(–2, 0, 0)<br />

(0, 0, 1)<br />

B(–2, 1, 1)<br />

(–2, 1, 0)<br />

(0, 1, 1)<br />

(0, –1, 0)<br />

(1, –1, 0)<br />

(1, 0, 0)<br />

x<br />

(1, 0, 1)<br />

O(0, 0, 0)<br />

y<br />

(0, 1, 0)<br />

y<br />

f.<br />

z<br />

x<br />

O(0, 0, 0)<br />

(1, 0, 0)<br />

(0, –1, 0)<br />

c.<br />

z<br />

(1, –1, 0)<br />

(0, –1, –1)<br />

O(0, 0, 0)<br />

y<br />

(0, –2, 1)<br />

C(1, –2, 1)<br />

(0, –2, 0)<br />

(0, 0, 1)<br />

x<br />

F(1, –1, –1)<br />

(1, 0, –1)<br />

(0, 0, –1)<br />

(1, –2, 0)<br />

(1, 0, 0)<br />

O(0, 0, 0)<br />

(1, 0, 1)<br />

y<br />

11. a. OA > 5 (1, 2, 3)<br />

z<br />

(0, 0, 3)<br />

(1, 0, 3)<br />

A(1, 2, 3)<br />

(0, 2, 3)<br />

O(0, 0, 0)<br />

(1, 0, 0)<br />

OA<br />

(0, 2, 0)<br />

y<br />

(1, 2, 0)<br />

x<br />

6-14 <strong>Chapter</strong> 6: Introduction to Vectors


. OB > 5 (22, 1, 1)<br />

z<br />

(–2, 0, 1)<br />

B(–2, 1, 1)<br />

f. OF > 5 (1, 21, 21)<br />

z<br />

x<br />

(–2, 0, 0)<br />

(0, 0, 1)<br />

O(0, 0, 0)<br />

OB<br />

(0, 1, 0)<br />

(–2, 1, 0)<br />

(0, 1, 1)<br />

y<br />

(1, 0, 0)<br />

(0, –1, 0)<br />

(1, –1, 0)<br />

(0, –1, –1)<br />

F(1, –1, –1)<br />

x<br />

OF<br />

O(0, 0, 0)<br />

(0, 0, –1)<br />

(1, 0, –1)<br />

y<br />

c. OC > 5 (1, 22, 1)<br />

z<br />

(0, –2, 1)<br />

(0, –2, 0)<br />

(1, 0, 1)<br />

C(1, –2, 1)<br />

(1, –2, 0)<br />

x<br />

d. OD > 5 (1, 1, 1)<br />

e. OE > 5 (1, 21, 1)<br />

z<br />

E(1, –1, 1)<br />

(0, –1, 0)<br />

(1, –1, 0)<br />

x<br />

x<br />

(1, 0, 0)<br />

(1, 0, 1)<br />

(0, 0, 1)<br />

O(0, 0, 0)<br />

(1, 0, 0)<br />

(0, –1, 1)<br />

(1, 0, 0)<br />

(0, 0, 1)<br />

O(0, 0, 0)<br />

z<br />

OC<br />

D(1, 1, 1)<br />

(0, 1, 1)<br />

OD<br />

y<br />

(0, 1, 0)<br />

(1, 1, 0)<br />

(0, 0, 1)<br />

(1, 0, 1)<br />

O(0, 0, 0)<br />

OE<br />

y<br />

y<br />

12. a. Since P and Q represent the same point,<br />

we can equate their y- and z-coordinates to get the<br />

system of equations<br />

a 2 c 5 6<br />

a 5 11<br />

Substituting this second equation into the first gives<br />

11 2 c 5 6<br />

c 5 5<br />

So a 5 11 and c 5 5.<br />

b. Since P and Q represent the same point in R 3 ,<br />

they will have the same associated position vector,<br />

i.e. OP > 5 OQ > . So, since these vectors are equal,<br />

they will certainly have equal magnitudes,<br />

i.e. @OP > @ 5 @OQ > @ .<br />

13. P(x, y, 0) represents a general point on the<br />

xy-plane, since the z-coordinate is 0. Similarly,<br />

Q(x, 0, z) represents a general point in the xz-plane,<br />

and R(0, y, z) represents a general point in the<br />

yz-plane.<br />

14. a. Every point on the plane containing points M,<br />

N, and P has y-coordinate equal to 0. Therefore, the<br />

equation of the plane containing these points is<br />

y 5 0 (this is just the xz-plane).<br />

b. The plane y 5 0 contains the origin O(0, 0, 0),<br />

and so since it also contains the points M, N, and P<br />

as well, it will contain the position vectors associated<br />

with these points joining O (tail) to the given point<br />

(head). That is, the plane contains the vectors<br />

OM > , ON > , and OP > y 5 0<br />

.<br />

15. a. A(22, 0, 0), B(22, 4, 0), C(0, 4, 0),<br />

D(0, 0, 27), E(0, 4, 27), F(22, 0, 27)<br />

b. OA > 5 (22, 0, 0), OB > 5 (22, 4, 0),<br />

OC > 5 (0, 4, 0), OD > 5 (0, 0, 27),<br />

OE > 5 (0, 4, 27), OF > 5 (22, 0, 27)<br />

c. Rectangle DEPF is 7 units below the xy-plane.<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-15


d. Every point on the plane containing points B, C,<br />

E, and P has y-coordinate equal to 4. Therefore, the<br />

equation of the plane containing these points is<br />

y 5 4 (a plane parallel to the xz-plane through the<br />

point y 5 4).<br />

e. Every point contained in rectangle BCEP has<br />

y-coordinate equal to 4, and so is of the form<br />

(x, 4,z) where x and z are real numbers such that<br />

22 # x # 0 and 27 # z # 0.<br />

16. a.<br />

y<br />

d.<br />

x<br />

O(0, 0, 0)<br />

z<br />

M(–1, 3, –2)<br />

y<br />

O(0, 0)<br />

x<br />

e.<br />

z<br />

F(0, 0, 5)<br />

P(4, –2)<br />

O(0, 0, 0)<br />

y<br />

b.<br />

y<br />

x<br />

D(–3, 4)<br />

f.<br />

z<br />

O(0, 0)<br />

x<br />

J(–2, –2, 0)<br />

O(0, 0, 0)<br />

y<br />

c.<br />

z<br />

x<br />

O(0, 0, 0)<br />

C(2, 4, 5)<br />

y<br />

17. The following box illustrates the three dimensional<br />

solid consisting of the set of all points (x, y, z) such<br />

that 0 # x # 1, 0 # y # 1, and 0 # z # 1.<br />

z<br />

x<br />

(1, 0, 1)<br />

O(0, 0, 0)<br />

(0, 0, 1) (0, 1, 1)<br />

(1, 1, 1)<br />

(1, 0, 0)<br />

x<br />

(1, 1, 0)<br />

y<br />

(0, 1, 0)<br />

6-16 <strong>Chapter</strong> 6: Introduction to Vectors


18. First, by the triangle law of<br />

vector addition, where<br />

OB > OA ><br />

OP > 5 (5,<br />

and OA > 210, 0),<br />

5 (0, 0, 210),<br />

are drawn in<br />

standard position (starting from the origin<br />

and OB > O(0,<br />

is drawn starting from the head of<br />

Notice that OA > OA > 0, 0)),<br />

lies in the xy-plane, and OB > .<br />

is<br />

perpendicular to the xy-plane (so is perpendicular to<br />

OA > ). So, OP > , OA > , and OB ><br />

form a right triangle and,<br />

by the Pythagorean theorem,<br />

Similarly, OA > 5 a > 1 b > @OP > @ 2 5 @OA > @ 2 1 @OB > 2<br />

@<br />

by the triangle law of<br />

vector addition, where<br />

and<br />

b > a > 5 (5, 0, 0)<br />

5 (0, 210, 0), and these three vectors form a<br />

right triangle as well. So,<br />

@OA > @ 2 5 0 a > 0 2 1 @b > 2<br />

@<br />

5 25 1 100<br />

5 125<br />

Obviously @OB > @ 2 5 100, and so substituting gives<br />

@OP > @ 2 5 @OA > @ 2 1 @OB > 2<br />

@<br />

5 125 1 100<br />

@OP > 5 225<br />

@ 5 "225<br />

5 15<br />

19. To find a vector equivalent to<br />

OP > AB ><br />

5 (22, 3, 6), where B(4, 22, 8), we need to<br />

move 2 units to the right of the x-coordinate for B<br />

(to 4 1 2 5 6), 3 units to the left of the y-coordinate<br />

for B (to 22 2 3 525), and 6 units below the<br />

z-coordinate for B (to 8 2 6 5 2). So we get the<br />

point A(6, 25, 2). Indeed, notice that to get from A<br />

to B (which describes vector AB ><br />

), we move 2 units<br />

left in the x-coordinate, 3 units right in the<br />

y-coordinate, and 6 units up in the z-coordinate.<br />

This is equivalent to vector OP > 5 (22, 3, 6).<br />

6.6 Operations with Algebraic<br />

R 2<br />

Vectors in , pp. 324–326<br />

1.<br />

A(–1, 3)<br />

y<br />

B(2, 5)<br />

x<br />

a. AB > 5 (2, 5) 2 (21, 3)<br />

5 (3, 2)<br />

BA > 52AB ><br />

52(3, 2)<br />

5 (23, 22)<br />

Here is a sketch of these two vectors in the<br />

xy-coordinate plane.<br />

y<br />

AB<br />

x<br />

b. 0 OA > 0 5 "(21) 2 1 3 2<br />

5 "29<br />

c. 0 AB > 8 5.39<br />

0 5 "3 2 1 2 2<br />

5 "13<br />

8 3.61<br />

Also, since<br />

0BA > BA ><br />

0 5 0 2AB > 52AB > ,<br />

0<br />

5 0 21 0 ? 0 AB > 0<br />

2.<br />

A(–1, 3)<br />

5 "10<br />

0 OB > 8 3.16<br />

0 5 "2 2 1 5 2<br />

5 0 AB > 0<br />

BA<br />

B(2, 5)<br />

5 "13<br />

8 3.61<br />

y<br />

30<br />

25<br />

20<br />

15<br />

10<br />

A(6, 10)<br />

O(0, 0)<br />

5<br />

OA x<br />

–9 –6 –3 0<br />

–5<br />

3 6 9<br />

–10<br />

OP > 5 OA > 1 OB > 6-17<br />

Calculus and Vectors <strong>Solutions</strong> Manual


a.<br />

y<br />

(12, 20)<br />

20<br />

15<br />

10<br />

5<br />

O(0, 0)<br />

(3, 5)<br />

x<br />

–12–9<br />

–6 –3 0 3 6 9 12<br />

(–3, –5) –5<br />

–10<br />

–15<br />

–20<br />

(–12, –20)<br />

b. The vectors with the same magnitude are<br />

1<br />

2 OA> and 2 1<br />

2OA > 2 OA> ,<br />

and 22OA ><br />

3. @OA > @ 5 "3 2 1 (24) 2<br />

5 "25<br />

5<br />

4. a. The i > 5<br />

-component will be equal to the first<br />

coordinate in component form, and so<br />

Similarly, the j > a 523.<br />

-component will be equal to the<br />

second coordinate in component form, and so b 5 5.<br />

b. 0(23, b)0 5 0(23, 5)0<br />

5 "(23) 2 1 5 2<br />

5 "34<br />

5. a. 0 a > 8 5.83<br />

0 5 "(260) 2 1 11 2<br />

5 "3721<br />

@b > 5 61<br />

@ 5 "(240) 2 1 (29) 2<br />

5 "1681<br />

b. a > 1 b > 5 41<br />

5 (260, 11) 1 (240, 29)<br />

@a > 5 (2100, 2)<br />

1 b > @ 5 "(2100) 2 1 2 2<br />

5 "10 004<br />

a > 2 b > 8 100.02<br />

5 (260, 11) 2 (240, 29)<br />

@a > 5 (220, 20)<br />

2 b > @ 5 "(220) 2 1 20 2<br />

5 "800<br />

8 28.28<br />

6. a. 2(22, 3) 1 (2, 1) 5 (2(22) 1 2, 2(3) 1 1)<br />

5 (22, 7)<br />

b. 23(4, 29) 2 9(2, 3)<br />

5 (23(4) 2 9(2), 23(29) 2 9(3))<br />

5 (230, 0)<br />

c.<br />

2 1 2 (6, 22) 1 2 (6, 15)<br />

3<br />

5 a2 1 2 (6) 1 2 3 (6), 21 2 (22) 1 2 3 (15)b<br />

5<br />

7. x > (1, 11)<br />

5<br />

a. 3x > 2i ><br />

2 y > 2 j > , y > 52i ><br />

5 3(2i > 2 j > 1 5j ><br />

)<br />

5 (6 1 1)i > 2 (2i > 1 5j > )<br />

1 (23 2 5)j ><br />

b. 2 (x > 5<br />

524x > 1 2y > 7i > 2 8j ><br />

)<br />

2<br />

524(2i > 11y > 1 3(2x > 2 3y > )<br />

5 3i > 2<br />

c. 2(x > 2 51j > j > ) 2 11(2i > 1 5j > )<br />

1<br />

5213x > 3y > ) 2<br />

1<br />

5213(2i > 3y > 3(y > 1 5x > )<br />

5229i > 2 j ><br />

8. a. x > 1<br />

1 y > 28j > ) 1 3(2i > 1 5j > )<br />

5 (2i ><br />

0x > 1 y > 5 i > 2<br />

0 5 @i > 1 4j > j > ) 1 (2i > 1 5j > )<br />

1 4j > @<br />

5 "1 2 1 4 2<br />

5 "17<br />

b. x > 2 y > 8 4.12<br />

5 (2i ><br />

0x > 2 y > 5 3i > 2 j ><br />

0 5 @3i > 2 6j > ) 2 (2i > 1 5j > )<br />

2 6j > @<br />

5 "3 2 1 (26) 2<br />

5 "45<br />

c. 2x > 8<br />

2 3y > 6.71<br />

5 2(2i > 2 j > ) 2 3(2i > 1 5j > )<br />

0 2x > 2 3y > 5 7i ><br />

0 5 @7i > 2 17j ><br />

2 17j > @<br />

5 "7 2 1 (217) 2<br />

5 "338<br />

d. 0 3y > 2 2x > 8 18.38<br />

0 5 0 2 (2x > 2<br />

5 0 21<br />

5 0 2x > 002x > 3y > )<br />

2<br />

2 3y > 3y > 0<br />

0<br />

0<br />

so, from part c.,<br />

0 3y > 2 2x > 0 5 0 2x > 2 3y > 0<br />

5 "338<br />

8 18.38<br />

6-18 <strong>Chapter</strong> 6: Introduction to Vectors


9.<br />

y<br />

8<br />

6 D(4, 5)<br />

B(–4, 4)<br />

4<br />

2<br />

A(–8, 2) C(2, 1) x<br />

–8 –6 –4 –2 0 2 4 6 8<br />

F(–7, 0)<br />

–2<br />

H(6, –2)<br />

–4<br />

G(1, –2)<br />

E(–1, –4)<br />

–6<br />

–8<br />

so obviously we will have @OA > @ 5 @ BC > @ .<br />

(It turns out that their common magnitude is<br />

"6 2 1 3 2 5 "45.)<br />

11. a.<br />

y<br />

C(–4, 11)<br />

B(6, 6)<br />

A(2, 3)<br />

x<br />

a. AB > 5 (24, 4) 2 (28, 2)<br />

CD > 5 (4, 2)<br />

5 (4, 5) 2 (2, 1)<br />

EF > 5 (2, 4)<br />

5 (27, 0) 2 (21, 24)<br />

GH > 5 (26, 4)<br />

5 (6, 22) 2 (1, 22)<br />

5 (5, 0)<br />

b.<br />

@AB > @ 5 "4 2 1 2 2<br />

5 "20<br />

@CD > 8 4.47<br />

@ 5 "2 2 1 4 2<br />

5 "20<br />

@EF > 8 4.47<br />

@ 5 "(26) 2 1 4 2<br />

5 "52<br />

@GH > 8 7.21<br />

@ 5 "5 2 1 0 2<br />

5 "25<br />

5 5<br />

10. a. By the parallelogram law of vector addition,<br />

OC > 5 OA > 1 OB ><br />

5 (6, 3) 1 (11, 26)<br />

5 (17, 23)<br />

For the other vectors,<br />

BA > 5 OA > 2 OB ><br />

5 (6, 3) 2 (11, 26)<br />

BC > 5 (25,<br />

5 OC > 9)<br />

2 OB ><br />

5 (17, 23) 2 (11, 26)<br />

5<br />

b. OA > (6, 3)<br />

5 (6,<br />

5 BC > 3)<br />

,<br />

b.<br />

c.<br />

AB > 5 (6, 6) 2 (2, 3)<br />

5 (4, 3)<br />

@AB > @ 5 "4 2 1 3 2<br />

5 "25<br />

AC > 5 5<br />

5 (24, 11) 2 (2, 3)<br />

5 (26, 8)<br />

@AC > @ 5 "(26) 2 1 8 2<br />

5 "100<br />

CB > 5 10<br />

5 (6, 6) 2 (24, 11)<br />

5 (10, 25)<br />

@CB > @ 5 "10 2 1 (25) 2<br />

5 "125<br />

@CB > 8 11.18<br />

@ 2 5 125 @AC > @ 2 5 100, @AB > @ 2 5 25<br />

Since @CB > @ 2 5 @AC > @ 2 1 @AB > 2<br />

@ , the triangle is a right<br />

triangle.<br />

12. a.<br />

y<br />

A(–1, 2)<br />

C(2, 8)<br />

B(7, –2)<br />

x<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-19


.<br />

y<br />

X(–6, 12)<br />

A(–1, 2)<br />

C(2, 8)<br />

Z(10, 4)<br />

x<br />

B(7, –2)<br />

Substituting this into the last equation above, we<br />

can now solve for y.<br />

22(212) 2 5y 5 4<br />

y 5 4<br />

So x 5212 and y 5 4.<br />

14. a.<br />

y<br />

C(x, y)<br />

B(–6, 9) D(8, 11)<br />

Y(4, –8)<br />

A(2, 3)<br />

c. As a first possibility for the fourth vertex, there is<br />

X(x 1 , x 2 ). From the sketch in part b., we see that we b. Because ABCD is a rectangle, we will have<br />

would then have<br />

BC > 5 AD ><br />

CX > 5 BA ><br />

( x, y) 2 (26, 9) 5 (8, 11) 2 (2, 3)<br />

( x 1 2 2, x 2 2 8) 5 (21 2 7, 2 2 (22))<br />

( x 1 6, y 2 9) 5 (6, 8)<br />

5 (28, 4)<br />

x 1 6 5 6<br />

x 1 2 2 528<br />

y 2 9 5 8<br />

x 2 2 8 5 4<br />

So, x 5 0 and y i.e.,<br />

So<br />

By similar reasoning for the other 15. a. Since and<br />

points labelled in the sketch in part b.,<br />

PA > 0 PA > 5 17,<br />

0 5 0 PB > C(0, 17).<br />

X(26, 12).<br />

0,<br />

5 (5, 0) 2 (a, 0)<br />

AY > 5 CB ><br />

PB > 5 (5 2 a, 0),<br />

( y 1 2 (21), y 2 2 2) 5 (7 2 2, 22 2 8)<br />

5 (0, 2) 2 (a, 0)<br />

5 (5, 210)<br />

5 (2a, 2),<br />

y 1 1 1 5 5<br />

this means that<br />

y 2 2 2 5210<br />

(5 2 a)<br />

So Finally,<br />

2 5 (2a) 2 1 2<br />

Y(4, 28).<br />

2<br />

25 2 10a 1 a 2 5 a 2 1 4<br />

BZ > 5 AC ><br />

10a 5 21<br />

( z 1 27, z 2 2 (22)) 5 (2 2 (21), 8 22)<br />

5 (3, 6)<br />

a 5 21<br />

10<br />

z 1 2 7 5 3<br />

z So Pa 21<br />

10 , 0b. 2 1 2 5 6<br />

So Z(10, 4). In conclusion, the three possible b. This point Q on the y-axis will be of the form<br />

locations for a fourth vertex in a parallelogram Q(0, b) for some real number b. Reasoning<br />

with vertices A, B, and C are X(26, 12), Y(4, 28), similarly to part a., we have<br />

and Z(10, 4).<br />

QA > 5 (5, 0) 2 (0, b)<br />

13. a. 3(x, 1) 2 5(2, 3y) 5 (11, 33)<br />

QB > 5 (5, 2b)<br />

(3x 2 5(2), 3 2 5(3y)) 5 (11, 33)<br />

5 (0, 2) 2 (0, b)<br />

(3x 2 10, 3 2 15y) 5 (11, 33)<br />

5 (0, 2 2<br />

So since @QA > b)<br />

3 x 2 10 5 11<br />

@ 5 @QB > @<br />

3 2 15y 5 33<br />

,<br />

So x 5 7 and y 522.<br />

( 2b) 2 1 5 2 5 (2 2 b) 2<br />

b. 22(x, x 1 y) 2 3(6, y) 5 (6, 4)<br />

b 2 1 25 5 4 2 4b 1 b 2<br />

( 22x 2 18, 22x 2 5y) 5 (6, 4)<br />

4b 5221<br />

22x 2 18 5 6<br />

b 52 21<br />

22x 2 5y 5 4<br />

4<br />

To solve for x, use<br />

So Qa0, 2 21<br />

22x 2 18 5 6<br />

4 b.<br />

x 5212<br />

6-20 <strong>Chapter</strong> 6: Introduction to Vectors<br />

x


16. is in the direction opposite to and<br />

QP > 5 OP > 2 OQ > PQ > ,<br />

5 (11, 19) 2 (2, 221)<br />

5 (9, 40)<br />

@QP > @ 5 "9 2 1 40 2<br />

5 "1681<br />

5 41<br />

A unit vector in the direction of QP ><br />

is<br />

u > 5 1<br />

41 QP><br />

5 a 9 41 , 40<br />

Indeed, is obviously in the same direction as<br />

(since u > u > 41 b<br />

is a positive scalar multiple of QP > QP ><br />

), and<br />

notice that<br />

0 u > 0 5 Å<br />

a 9<br />

41 b 2<br />

1 a 40<br />

41 b 2<br />

81 1 1600<br />

5 Å 1681<br />

5 1<br />

17. a. O, P, and R can be thought of as the vertices<br />

of a triangle.<br />

PR > 5 OR > 2 OP ><br />

5 (28, 21) 2 (27, 24)<br />

@PR > 5 (21, 225)<br />

@ 2 5 (21) 2 1 (225) 2<br />

@OR > 5 626<br />

@ 2 5 (28) 2 1 (21) 2<br />

@OP > 5 65<br />

@ 2 5 (27) 2 1 24 2<br />

5 625<br />

By the cosine law, the angle, , between and<br />

OP > u OR ><br />

satisfies<br />

cos u5 @OR> @ 2 1 @OP > @ 2 2 @PR > 2<br />

@<br />

2 @OR > @ ? @OP > @<br />

65 1 625 2 626<br />

5<br />

2!65 ? !625<br />

65 1 625 2 626<br />

u5cos 21 a<br />

2!65 ? !625 b<br />

8 80.9°<br />

So the angle between OR ><br />

and is about .<br />

b. We found the vector in part a.,<br />

so RP > and<br />

@RP > 52PR > PR > OP ><br />

80.86°<br />

5 (21, 225)<br />

@ 2 5 @PR > 5 (1, 25)<br />

2<br />

@<br />

5 626<br />

Also, by the parallelogram law of vector addition,<br />

OQ > 5 OR > 1 OP ><br />

5 (28, 21) 1 (27, 24)<br />

@OQ > 5 (215, 23)<br />

@ 2 5 (215) 2 1 23 2<br />

5 754<br />

Placing RP > 5 (1, 25) and OQ > 5 (215, 23) with<br />

their tails at the origin, a triangle is formed by<br />

joining the heads of these two vectors. The third<br />

side of this triangle is the vector<br />

v > 5 RP > 2 OQ ><br />

5 (1, 25) 2 (215, 23)<br />

0 v > 5 (16, 2)<br />

0 2 5 16 2 1 2 2<br />

5 260<br />

Now by reasoning similar to part a., the cosine law<br />

implies that the angle, u, between RP ><br />

and OQ ><br />

satisfies<br />

cos u5 @RP> @ 2 1 @OQ > @ 2 2 @v > 2<br />

@<br />

2 @RP > @ ? @OQ > @<br />

626 1 754 2 260<br />

5<br />

2!626 ? !754<br />

u5cos 21 626 1 754 2 260<br />

a<br />

2!626 ? !754 b<br />

8 35.4°<br />

So the angle between RP ><br />

and OQ ><br />

is about 35.40° .<br />

However, since we are discussing the diagonals of<br />

parallelogram OPQR here, it would also have been<br />

appropriate to report the supplement of this angle,<br />

or about 180°235.40° 5 144.60°, as the angle<br />

between these vectors.<br />

6.7 Operations with Vectors in R 3 ,<br />

pp. 332–333<br />

OA > 521i > 1 2j > 1 4k ><br />

1. a.<br />

b. @OA > @<br />

2. OB > 5 "(21) 2 1 2 2 1 4 2 5 "21 8 4.58<br />

@OB > 5 (3, 4, 24)<br />

@ 5 "3 2 1 4 2 1 (24) 2 5 "41 8 6.40<br />

3. a > 1 1 3 b> 2 c > 5 (1, 3, 23) 1 (21, 2, 4)<br />

2 (0, 8, 1)<br />

5 (1 1 (21) 2 0, 3 1 2 2 8,<br />

(23) 1 4 2 1)<br />

5 (0, 23, 0)<br />

` a > 1 1 3 b> 2 c >` 5 "0 2 1 (23) 2 1 0 2<br />

QP > 6-21<br />

5 3<br />

Calculus and Vectors <strong>Solutions</strong> Manual


4. a. OP > 5 OA > 1 OB ><br />

5 ((23) 1 2, 4 1 2, 12 1 (21))<br />

5 (21, 6, 11)<br />

b.<br />

c.<br />

c.<br />

@OA > @ 5 "(23) 2 1 4 2 1 12 2 5 13<br />

@OB > @ 5 "2 2 1 2 2 1 (21) 2 5 3<br />

@OP > @ 5 "(21) 2 1 6 2 1 11 2 8 12.57<br />

AB > 5 OB > 2 OA ><br />

5 (2, 2, 21) 2 (23, 4, 12)<br />

5 (2 2 (23), 2 2 4, (21) 2 12)<br />

5 (5, 22, 213)<br />

@AB > @<br />

AB > 5 "5 2 1 (22) 2 1 (213) 2 5 "198 8 14.07<br />

represents the vector from the tip of to the tip<br />

of OB > OA ><br />

It is the difference between the two vectors.<br />

5. a. x > .<br />

2 2y > 2 z ><br />

5 (1, 4, 21) 2 2(1, 3, 22) 2 (22, 1, 0)<br />

5 (1, 4, 21) 2 (2, 6, 24) 2 (22, 1, 0)<br />

5 (1 2 2 2 (22), 4 2 6 2 1, 21 2 (24) 2 0)<br />

5 (1, 23, 3)<br />

b. 22x > 2 3y > 1 z ><br />

522(1, 4, 21) 2 3(1, 3, 22) 1 (22, 1, 0)<br />

5 (22, 28, 2) 2 (3, 9, 26) 1 (22, 1, 0)<br />

5 (22 2 3 2 2, 28 2 9 1 1, 2 1 6 1 0)<br />

5 (27, 216, 8)<br />

1<br />

2 x> 2 y > 1 3z ><br />

5 1 (1, 4, 21) 2 (1, 3, 22) 1 3(22, 1, 0)<br />

2<br />

5 a 1 2 , 2, 21 b 2 (1, 3, 22) 1 (26, 3, 0)<br />

2<br />

5 a 1 2 2 1 1 (26), 2 2 3 1 3, 21 2 (22) 1 0b<br />

2<br />

5 a2 13<br />

d. 3x > 2 , 2, 3<br />

1 5y > 2 b<br />

1 3z ><br />

5 3(1, 4, 21) 1 5(1, 3, 22) 1 3(22, 1, 0)<br />

5 (3, 12, 23) 1 (5, 15, 210) 1 (26, 3, 0)<br />

5 (3 1 5 2 6, 12 1 15 1 3, 23 2 10 1 0)<br />

5 (2, 30, 213)<br />

6. a. p > 1 q > 5 (2i > 2 j > 1<br />

5 (2<br />

b. p > 2 q > 5 i > 2<br />

5 (2i > 2 2j > 1)i > k > ) 1 (2i > 2<br />

1<br />

2 j > 1<br />

1<br />

5 (2<br />

5 3i > 1 1)i > k > 2k > (21 2 1)j > j > 1 k > )<br />

1 (1 1 1)k ><br />

) 2 (2i > 2<br />

1 0j > 1 (21<br />

1 0k > 1 1)j > j > 1 k > )<br />

1 (1 2 1)k ><br />

c. 2p > 2 5q > 5 2(2i ><br />

5 (4i > 2 j ><br />

2 2j > 1 k > )<br />

5 (4<br />

d. 22p > 5<br />

1<br />

5 (24i > 5q > 9i > 1 5)i > 1 2k > 2 5(2i > 2<br />

) 2 (25i > j > 1<br />

1 3j > 1 (22<br />

2<br />

1 2j > 522(2i > 3k > 1 5)j > 2 5j > k > )<br />

1 5k > )<br />

1 (2 2 5)k ><br />

5 (24<br />

529i > 2 5)i > 2 2k > 2 j > 1<br />

) 1 (25i > k > ) 1<br />

7. a. m > 2 3j > 1 (2<br />

2 n > 1 3k > 2 5)j > 2 5j > 5(2i ><br />

1 5k > 2 j > 1 k > )<br />

)<br />

1 (22 1 5)k ><br />

5 (2i ><br />

5 (2<br />

5<br />

0 m > 4i > 2<br />

2<br />

2<br />

b. m > n > j > (22))i > 2 k > )2<br />

2 3k > 1 (21)j > (22i > 1 j > 1 2k ><br />

1 (21 2 2)k > )<br />

0 5<br />

1 n > "4 2<br />

5 (2i > 1 (21) 2<br />

2 k > 1 (23) 2<br />

) 1 (22i ><br />

5 (2<br />

5 0i > 1<br />

1 j > (22))i ><br />

1 k > 1 j > 1 j > 5 "26<br />

1 2k > 8 5.10<br />

)<br />

1 (21 1 2)k ><br />

0 m > 1<br />

c. 2m > n > 0 5<br />

1 3n > "0 2 1<br />

5 2(2i > 1 2 1<br />

5 (4i > 2 k > 1 2 5 "2<br />

2 2k > ) 1 3(22i > 8 1.41<br />

) 1<br />

5 (4 1<br />

522i > (26))i > (26i > 1 j ><br />

1 3j > 1<br />

1 4k > 3j > 1 3j > 1 2k ><br />

1 6k > )<br />

)<br />

1 (22 1 6)k ><br />

0 2m > 1<br />

d. 25m > 3n > 0 5 "(22) 2<br />

525(2i > 2 k > 1 3 2 1<br />

) 5210i > 4 2 5<br />

1 5k > "29 8 5.39<br />

0 25m ><br />

8. x > 0 5<br />

1 x > 1 y > "(210) 2 1 (5) 2 5 "125 8 11.18<br />

2x > 2 y > 52i ><br />

5 3i > 1 2j ><br />

5 2i > 1 6j > 1 5k ><br />

1 8j > 2 7k ><br />

2 2k ><br />

Divide by 2 on both sides to get:<br />

x > 5 i > 1 4j > 2 k ><br />

Plug this equation into the first given equation:<br />

i > 1 4j > 2 k > 1 y > y > 52i ><br />

y > 52i > 1 2j ><br />

1 2j > 1 5k ><br />

1<br />

y > 5 (21<br />

522i > 2 1)i > 5k > 2 (i > 1<br />

9. a. The vectors OA > 2<br />

, OB > 2j > 1 (2<br />

1 6k > 2 4)j > 4j > 2 k > )<br />

1 (5 1 1)k ><br />

, and OC ><br />

represent the<br />

xy-plane, xz-plane, and yz-plane, respectively.<br />

They are also the vector from the origin to points<br />

(a, b, 0), (a, 0,c), and (0, b, c), respectively.<br />

b. OA ><br />

OB > 5 ai ><br />

OC > 5 ai > 1 bj ><br />

c. @OA > 5 0i > 1 0j > 1 0k ><br />

1 bj > 1 ck ><br />

1 ck ><br />

@ 5 "a 2 1 b 2<br />

@OB > @ 5 "a 2 1 c 2<br />

@OB > @ 5 "b 2 1 c 2<br />

6-22 <strong>Chapter</strong> 6: Introduction to Vectors


d. AB > AB > 5 (a, 0, c) 2 (a, b, 0) 5 (0, 2b, c)<br />

is a direction vector from A to B.<br />

10. a. @OA > @ 5 "(22) 2 1 (26) 2 1 3 2 5 "49 5 7<br />

b.<br />

c.<br />

d. @AB > @ 5 "5 2 1 2 2 1 9 2 5 "110 8 10.49<br />

e. BA > 5 OA > 2 OB ><br />

5 (22, 26, 3) 2 (3, 24, 12)<br />

5 (25, 22, 29)<br />

f.<br />

11.<br />

@OB > @<br />

AB > 5 "(3) 2<br />

5 OB > 1<br />

2 OA > (24) 2 1 12 2 5 "169 5 13<br />

5 (3, 24, 12) 2 (22, 26, 3)<br />

5 (3 2 (22), 24 2 (26), 12 2 3)<br />

5 (5, 2, 9)<br />

@BA > @ 5 "(25) 2 1 (22) 2 1 (29) 2<br />

5 "110 8 10.49<br />

z<br />

B(3, –1, 17)<br />

BC<br />

C(7, –3, 15)<br />

DC<br />

D(4, 1, 3)<br />

x<br />

AB<br />

A(0, 3, 5)<br />

AD<br />

In order to show that ABCD is a parallelogram, we<br />

must show that AB > 5 DC ><br />

or BC > 5 AD > . This will<br />

show they have the same direction, thus the opposite<br />

sides are parallel. By showing the vectors are<br />

equal they will have the same magnitude, implying<br />

the opposite sides having congruency.<br />

AB > 5 (3, 21, 17) 2 (0, 3, 5)<br />

5 (3 2 0, 21 2 3, 17 2 5)<br />

DC > 5 (3, 24, 12)<br />

5 (7, 23, 15) 2 (4, 1, 3)<br />

5 (7 2 4, 23 2 1, 15 2 3)<br />

5 (3,<br />

Thus AB > 24, 12)<br />

5 DC > . Do the calculations for the other<br />

pair as a check.<br />

BC > 5 (7, 23, 15) 2 (3, 21, 17)<br />

5 (7 2 3, 23 2 (21), 15 2 17)<br />

AD > 5 (4, 22, 22)<br />

5 (4, 1, 3) 2 (0, 3, 5)<br />

5 (4 2 0, 1 2 3, 3 2 5)<br />

5<br />

So BC > (4, 22,<br />

5 AD > 22)<br />

.<br />

We have shown AB > 5 DC ><br />

and BC > 5 AD > , so<br />

ABCD is a parallelogram.<br />

y<br />

12. 2x > 1 y > 2 2z ><br />

5 2(21, b, c) 1 (a, 22, c) 2 2(2a, 6, c)<br />

5 (22, 2b, 2c) 1 (a, 22, c) 2 (22a, 12, 2c)<br />

5 (22 1 a 1 2a, 2b 2 2 2 12, 2c 1 c 2 2c)<br />

5 (22 1 3a, 2b 2 14, c)<br />

5 (0, 0, 0)<br />

22 1 3a 5 0;<br />

3a 5 2; a 5 2 3<br />

2b 5 14; b 5 7<br />

c 5 0<br />

13. a.<br />

x<br />

2b 2 14 5 0; c 5 0<br />

z<br />

OB<br />

O<br />

OA<br />

OC<br />

b. V 1 5 (0, 0, 0), the origin<br />

V 2 5 end point of OA ><br />

V 3 5 end point of OB > 5 (22, 2, 5)<br />

V 4 5 end<br />

V 5 5 OA > point of<br />

1 OB > OC > 5 (0, 4, 1)<br />

5 (0, 5, 21)<br />

5 (22, 2, 5) 1 (0, 4, 1)<br />

5 (22 1 0, 2 1 4, 5 1 1)<br />

V 6 5 OA > 1 OC > 5 (22, 6, 6)<br />

5 (22, 2, 5) 1 (0, 5, 21)<br />

5 (22 1 0, 2 1 5, 5 2 1)<br />

V 7 5 OB > 1 OC > 5 (22, 7, 4)<br />

5 (0, 4, 1) 1 (0, 5, 21)<br />

5 (0 1 0, 4 1 5, 1 2 1)<br />

V 8 5 OA > 1 OB > 5 (0,<br />

1 OC > 9, 0)<br />

5 (22, 2, 5) 1 (0, 9, 0) (by V 7 )<br />

5 (22 1 0, 2 1 9, 5 1 0)<br />

5 (22, 11, 5)<br />

14. Any point on the x-axis has y-coordinate 0 and<br />

z-coordinate 0. The z-coordinate of each of A and B<br />

is 3, so the z-component of the distance from the<br />

desired point is the same for each of A and B. The<br />

y-component of the distance from the desired point<br />

will be 1 for each of A and B, 12 5 (21) 2 . So, the<br />

x-coordinate of the desired point has to be halfway<br />

between the x-coordinates of A and B. The desired<br />

point is (1, 0, 0).<br />

A<br />

B<br />

C<br />

y<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-23


15.<br />

a<br />

A<br />

a – b<br />

b<br />

To solve this problem, we must first consider the<br />

triangle formed by a > , b > , and a > 1 b > . We will use<br />

their magnitudes to solve for angle A, which will be<br />

1<br />

used to solve for in the triangle formed by<br />

1<br />

1<br />

b > , and 2 a > 2 b > 2 a > 1 b > 2 a > 2 b ><br />

,<br />

.<br />

Using the cosine law, we see that:<br />

cos (A) 5 @b> @ 2 1 @a > 1 b > @ 2 2 @a > 2<br />

@<br />

2 @b > @@a > 1 b > @<br />

25 1 49 2 9<br />

5<br />

5 13<br />

70<br />

14<br />

Now, consider the triangle formed by ,<br />

1<br />

and 2 a > 2 b > 1<br />

b > , 2 a > 1 b ><br />

. Using the cosine law again:<br />

cos (A) 5 @b> @ 2 1 ( 1 2 @a> 1 b > @ ) 2 2 ( 1 2 @a> 2 b > @ ) 2<br />

@b > @@a > 1 b > @<br />

149<br />

13<br />

14 5 4 2 (1 2 @a> 2 b > @) 2<br />

35<br />

@a > 2 b > @ 2 524a 65<br />

@a > 2 2 149<br />

4 b<br />

@a > 2 b > @ 2 5 19<br />

2 b > @ 5 "19 or 4.36<br />

6.8 Linear Combinations and Spanning<br />

Sets, pp. 340–341<br />

1. They are collinear, thus a linear combination is<br />

not applicable.<br />

2. It is not possible to use 0 ><br />

in a spanning set.<br />

Therefore, the remaining vectors only span R 2 .<br />

3. The set of vectors spanned by (0, 1) is m(0, 1).<br />

If we let m 521, then m(0, 1) 5 (0, 21).<br />

4. i > spans the set m(1, 0, 0). This is any vector<br />

along the x-axis. Examples: (2, 0, 0), (221, 0,<br />

5. As in question 2, it is not possible to use 0 > 0)<br />

in a<br />

spanning set.<br />

b<br />

a + b<br />

C<br />

B<br />

a<br />

6. 5(21, 2), (21, 1)6, 5(2, 24), (21, 1)6,<br />

5(21, 1), (23, 6)6 are all the possible spanning sets<br />

for R 2 with 2 vectors.<br />

7. a.<br />

5 4i > 2(2a ><br />

5 6i > 2 8j > 2 3b ><br />

2<br />

2<br />

4(2a > 20j > 6j > 1 c > )<br />

1<br />

524i > 1 b > 1 22k > 18k > 5 4a ><br />

1 2i > 2 6b ><br />

2 6j > 1 2c ><br />

1 4k<br />

2<br />

528i > 1 8j > c > )<br />

1<br />

(a > 2 c > 1 24j > 4j > 524a ><br />

2<br />

) 5 i > 2 20k > 12k > 1 4b ><br />

2 4i > 2 4c ><br />

1 12j > 2 8k<br />

2(2a > 5<br />

5 6i > 2 3b > j > 2 2j ><br />

2<br />

2<br />

5 14i > 20j > 1 c > 2k > 2 i > 1 3j > 2 2k ><br />

)24(2a ><br />

1<br />

2 43j > 22k > 1<br />

1 40k > 8i > 1 b ><br />

2 24j > 2 c > ) 1<br />

1 20k > (a ><br />

1 j > 2 c > )<br />

2 2k ><br />

1<br />

b.<br />

2 (2a> 2 4b > 2 8c > ) 5 a > 2 2b > 2 4c ><br />

5 i > 2<br />

523i > 2j > 2<br />

1 8j > 2j > 1 6k ><br />

2 2k > 2 4i > 1 12j > 2 8k ><br />

1<br />

3 (3a> 2 6b > 1 9c > ) 5 a > 2 2b > 1 3c ><br />

5 i > 2<br />

5 4i > 2j > 2<br />

2 15j > 2j > 1 6k ><br />

1 12k > 1 3i > 2 9j > 1 6k ><br />

1<br />

2 (2a> 2 4b > 2 8c > )2 1<br />

523i ><br />

527i > 1 8j > 2<br />

1 23j > 2k > 3 (3a> 2 6b > 1 9c > )<br />

2<br />

2 14k > 4i > 1 15j > 2 12k ><br />

8. 5(1, 0, 0), (0, 1, 0)6:<br />

(21, 2, 0) 521(1, 0, 0) 1 2(0, 1, 0)<br />

(3, 4, 0) 5 3(1, 0, 0) 1 4(0, 1, 0)<br />

5(1, 1, 0), (0, 1, 0)6<br />

(21, 2, 0) 521(1, 1, 0) 1 3(0, 1, 0)<br />

(3, 4, 0) 5 3(1, 1, 0) 1 (0, 1, 0)<br />

9. a. It is the set of vectors in the xy-plane.<br />

b. (22, 4, 0) 522(1, 0, 0) 1 4(0, 1, 0)<br />

c. By part a. the vector is not in the xy-plane.<br />

There is no combination that would produce a<br />

number other than 0 for the z-component.<br />

d. It would still only span the xy-plane. There<br />

would be no need for that vector.<br />

10. Looking at the x-component:<br />

2a 1 3c 5 5<br />

The y-component:<br />

6 1 21 5 b 1 c<br />

The z-component:<br />

2c 1 3c 5 15<br />

5c 5 15<br />

c 5 3<br />

6-24 <strong>Chapter</strong> 6: Introduction to Vectors


Substituting this into the first and second equation:<br />

2a 1 9 5 5<br />

a 522<br />

27 5 b 1 3<br />

b 5 24<br />

11. (210, 234) 5 a(21, 3) 1 b(1, 5)<br />

Looking at the x-component:<br />

210 52a 1 b a 5 10 1 b<br />

Looking at the y-component:<br />

234 5 3a 1 5b<br />

Substituting in a:<br />

234 5 30 1 3b 1 5b<br />

b 528<br />

Substituting b into x-component equation:<br />

210 52a 1 (28)<br />

a 522<br />

(210, 234) 5 22(21, 3) 2 8(1, 5)<br />

12. a. a(2, 21) 1 b(21, 1) 5 (x, y)<br />

x 5 2a 2 b<br />

b 5 2a 2 x<br />

y 52a 1 b<br />

Substitute in b:<br />

y 52a 1 2a 2 x<br />

a 5 x 1 y<br />

Substitute this back into the first equation:<br />

b 5 2x 1 2y 2 x<br />

b 5 x 1 2y<br />

b. Using the formulas in part a:<br />

For (2, 23):<br />

a 5 x 1 y 5 2 2 3 521<br />

b 5 x 1 2y 5 2 2 6 524<br />

(2, 23) 521(2, 21) 2 4(21, 1)<br />

For (124, 25):<br />

a 5 124 2 5 5 119<br />

b 5 124 2 10 5 114<br />

(124, 25) 5 119(2, 21) 1 114(21, 1)<br />

For (4, 211)<br />

a 5 4 2 11 527<br />

b 5 4 2 22 5218<br />

(4, 211) 527(2, 21) 2 18(21, 1)<br />

13. Try: a(21, 2, 3) 1 b(4, 1, 22)<br />

5 (214, 21, 16)<br />

x components:<br />

2a 1 4b 5214<br />

a 5 14 1 4b<br />

y components:<br />

2a 1 b 521<br />

Substitute in a:<br />

28 1 8b 1 b 521<br />

b 523<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

Substitute this result into the x-components:<br />

a 5 14 2 3 5 11<br />

Check by substituting into z-components:<br />

3a 2 2b 5 16<br />

33 1 5 5 16<br />

Therefore:<br />

a(21, 2, 3) 1 b(4, 1, 22) 2 (214, 21, 16) for<br />

any a and b. They do not lie on the same plane.<br />

b. a(21, 3, 4) 1 b(0, 21, 1) 5 (23, 14, 7)<br />

x components:<br />

2a 523<br />

a 5 3<br />

y components:<br />

3a 2 b 5 14<br />

Substitute in a:<br />

9 2 b 5 14<br />

b 525<br />

Check with z components:<br />

4a 1 b 5 7<br />

12 2 5 5 7<br />

Since there exists an a and b to form a linear<br />

combination of 2 of the vectors to form the third,<br />

they lie on the same plane.<br />

3(21, 3, 4) 2 5(0,<br />

14. Let vector a > 21, 1) 5 (23, 14,<br />

and b > 7)<br />

5 (21, 3, 4) 5 (22, 3, 21)<br />

(vectors from the origin to points A and B,<br />

respectively). To determine x, we let c > (vector from<br />

origin to C) be a linear combination of a > and b > .<br />

a(21, 3, 4) 1 b(22, 3, 21) 5 (25, 6, x)<br />

x components:<br />

2a 2 2b 525<br />

a 5 5 2 2b<br />

y components:<br />

3a 1 3b 5 6<br />

Substitute in a:<br />

15 2 6b 1 3b 5 6<br />

b 5 3<br />

Substitute in b in x component equation:<br />

a 5 5 2 6 521<br />

z components:<br />

4a 2 b 5 x<br />

Substitute in a and b:<br />

x 524 2 3 527<br />

15. m 5 2, n 5 3. Non-parallel vectors cannot be<br />

equal, unless their magnitudes equal 0.<br />

16. Answers may vary. For example:<br />

Try linear combinations of the 2 vectors such that<br />

6-25


the z component equals 5. Then calculate what p<br />

and q would equal.<br />

21(4, 1, 7) 1 2(21, 1, 6) 5 (26, 1, 5)<br />

So p 526 and q 5 1<br />

5(4, 1, 5) 2 5(21, 1, 6) 5 (25, 0, 5)<br />

So p 5 25 and q 5 0<br />

(4, 1, 7) 2 1 (21, 1, 6) 5 a13<br />

3 3 , 2 3 , 5b<br />

So p 5 13 and q 5 2 3 3<br />

17. As in question 15, non-parallel vectors.<br />

Their magnitudes must be 0 again to make the<br />

equality true.<br />

m 2 1 2m 2 3 5 (m 2 1)(m 1 3)<br />

m 5 1, 23<br />

m 2 1 m 2 6 5 (m 2 2)(m 1 3)<br />

m 5 2, 23<br />

So, when m 523, their sum will be 0.<br />

Review Exercise, pp. 344–347<br />

1. a. false; Let then:<br />

@a > 1 b > @ 5 0 a > b > 52a ><br />

1 (2a > 2 0<br />

) 0<br />

5 0 0 0<br />

5<br />

b. true; @a > 0 , 0 a > 0<br />

1 b > @ and 0 a > 1 c > 0 both represent the<br />

lengths of the diagonal of a parallelogram, the first<br />

with sides and and the second with sides and<br />

c > a > b ><br />

; since both parallelograms have as a side and<br />

diagonals of equal length .<br />

c. true; Subtracting from both sides shows that<br />

b > 5 c > a > @b > @ 5 0 c > a > a ><br />

0<br />

d. true; Draw the parallelogram formed by and<br />

SW > . FW ><br />

and RS ><br />

RF ><br />

are the opposite sides of a parallelogram<br />

and must be equal.<br />

e. true; The distributive law for scalars<br />

f. false; Let and let Then,<br />

0 a > 0 5 and<br />

but @a > 0 2a > b > 52a ><br />

0 5 @b > @<br />

1 b > @ 5 0 a > 0 c > c ><br />

0<br />

1 (2 a > 5 @d > 5 d > 2 0.<br />

@<br />

)0 5 0<br />

@c > 1 d > @ 5 0 c > 1 c > 0 5 0 2c > 0<br />

so @a > 1 b > @ 2 @c > 1 d > @<br />

2. a. Substitute the given values of and into<br />

the expression<br />

2x > 2 3y ><br />

5 2(2a > 1 5z > 2x> 2 3y > 1 5z > x > , y > , z ><br />

2<br />

1 5(2a > 3b > 2<br />

2 3b > 4c > ) 2<br />

1 5c > 3 (22a > 1 3b > 1 3c > )<br />

)<br />

5 4a > 2<br />

1<br />

5 4a > 25c > 6b > 2 8c > 1 6a > 2 9b > 2 9c > 1 10a > 2 15b ><br />

1<br />

1 25c > 6a > 1 10a > 2 6b > 2 9b ><br />

2 15b > 2 8c > 2 9c ><br />

5 20a > 2 30b > 1 8c ><br />

b. Simplify the expression before substituting the<br />

given values of and<br />

3(22x > 2 4y > x > ,<br />

2 2(24x > 1 z > y > , z ><br />

5 26x > 2<br />

2<br />

1 10y > 12y > 5y > ) 2<br />

1<br />

5 26x > 2 2z > 1 3z > z > (2x > 2 y > 1 z > )<br />

)<br />

2 2x > 1 y > 2 z > 1 8x ><br />

2<br />

5 0x > z > 2 2x ><br />

2<br />

5 2y > 2 y > 2z > 1 8x > 2 12y > 1 y > 1 10y > 1 3z ><br />

1 0z ><br />

5 2a > 2<br />

3. a. XY > 3b > 2 3c ><br />

5 OY > 2 OX ><br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (24 2 (22), 4 2 1, 8 2 2)<br />

5 (22, 3, 6)<br />

@XY > @ 5 "(x 2 2 x 1 ) 2 1 (y 2 2 y 1 ) 2 1 (z 2 2 z 1 ) 2<br />

5 "(22) 2 1 (3) 2 1 (6) 2<br />

5 "4 1 9 1 36<br />

5 "49<br />

5 7<br />

b. The components of a unit vector in the same<br />

1<br />

direction as XY ><br />

are 7(22, 3, 6) 5 (2 2 7, 3 7, 6 7) .<br />

4. a. The position vector is equivalent to<br />

OP > 5 YX > OP ><br />

YX > .<br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (21 2 5, 2 2 5, 6 2 12)<br />

5 (26, 23, 26)<br />

b. @YX > @ 5 "(26) 2 1 (23) 2 1 (26) 2<br />

5 "81<br />

5 9<br />

The components of a unit vector in the same<br />

1<br />

direction as are<br />

5. 2MN > YX ><br />

5 NM ><br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (2 28, 3 2 1, 5 2 2)<br />

5 (26, 2, 3)<br />

9(26, 23, 26) 5 (2 2 3, 2 1 3, 2 2 3)<br />

6-26 <strong>Chapter</strong> 6: Introduction to Vectors


@NM > @ 5 "(26) 2 1 (2) 2 1 (3) 2 6-27<br />

5 "49<br />

5 7<br />

The components of the unit vector with the opposite<br />

1<br />

direction to MN ><br />

are 7(26, 2, 3) 5 (2 6 7, 2 7, 3 7)<br />

6. a. The two diagonals can be found by calculating<br />

OA > 1 OB > and OA > 2 OB > .<br />

O<br />

OA > 1 OB > 5 (3, 2, 26) 1 (26, 6, 22)<br />

5 (3 126, 2 1 6, 26 122)<br />

OA > 2 OB > 5 (23, 8, 28)<br />

5 (3, 2, 26) 1 (26, 6, 22)<br />

5 (3 2 (26), 2 2 6, 26 2 (22))<br />

5 (9, 24, 24)<br />

b. To determine the angle between the sides of the<br />

parallelogram, calculate<br />

and<br />

@OA > 2 OB > @OA > @ , @OB > @ ,<br />

@ and apply<br />

the cosine law.<br />

@OA > @ 5 "(3) 2 1 (2) 2 1 (26) 2<br />

@OB > @ 5 "(26) 2 1 (6) 2 1 (22) 2<br />

@OA > 2 OB > @ 5 "(9) 2 1 (24) 2 1 (24) 2<br />

cos u5 @OA> @ 2 1 @OB > @ 2 2 @OA > 2 OB > @<br />

2<br />

2 @OA > @@OB > @<br />

cos u5 (7)2 1 (2"19) 2 2 ("113) 2<br />

2(7)(2"19)<br />

cos u 8 0.098<br />

u 8 84.4°<br />

7. a.<br />

5 "49<br />

5 7<br />

5 "76<br />

5 2"19<br />

A<br />

5 "113<br />

@AB > @ 5 "(2 2 (21)) 2 1 (0 2 1) 2 1 (3 2 1) 2<br />

5 "(3) 2 1 (21) 2 1 (2) 2<br />

5 "9 1 1 1 4<br />

5 "14<br />

OA – OB<br />

B<br />

OA + OB<br />

@BC > @ 5 "(3 2 2) 2 1 (3 2 0) 2 1 (24 2 3) 2<br />

5 "(1) 2 1 (3) 2 1 (27) 2<br />

5 "1 1 9 1 49<br />

5 "59<br />

0 CA > 0 5 "(21 2 3) 2 1 (1 2 3) 2 1 (1 2 (24)) 2<br />

5 "(24) 2 1 (22) 2 1 (5) 2<br />

5 "16 1 4 1 25<br />

5 "45<br />

Triangle ABC is a right triangle if and only if<br />

@AB > @ 2 1 @CA > @ 2 5 @BC > @ 2 .<br />

@AB > @ 2 1 @CA > @ 2 5 ("14) 2 1 ("45) 2<br />

5 14 1 45<br />

@BC > 5 59<br />

@ 2 5 ("59) 2<br />

5 59<br />

So triangle ABC is a right triangle.<br />

b. Area of a triangle For triangle ABC the<br />

longest side BC > 5 1 2bh.<br />

is the hypotenuse, so AB ><br />

and CA ><br />

are the base and height of the triangle.<br />

Area 5 1 2 ( 0 AB> 0 )(0 CA > 0 )<br />

5 1 2 "14"45<br />

5 1 2 "630<br />

5 3 or 12.5<br />

2 "70<br />

c. Perimeter of a triangle equals the sum of the sides.<br />

Perimeter 5 @AB > @ 1 @BC > @ 1 @CA > @<br />

5 "14 1 "59 1 "45<br />

8 18.13<br />

d. The fourth vertex D is the head of the diagonal<br />

vector from A. To find take .<br />

AB > AD ><br />

AB > 1 AC ><br />

5 (2 2 (21), 0 2 1, 3 2 1) 5 (3, 21, 2)<br />

AC > 5 (3 2 (21), 3 2 1, 24 2 1) 5 (4, 2, 25)<br />

AD > 5 AB > 1 AC ><br />

5 (3 1 4, 21 1 2, 2 1 (25))<br />

5 (7, 1, 23)<br />

So the fourth vertex is D(21 1 7, 1 1 1, 1 1 (23))<br />

or D(6, 2, 22).<br />

Calculus and Vectors <strong>Solutions</strong> Manual


8. a.<br />

b<br />

a – b<br />

a – b + c<br />

a<br />

c<br />

b. Since the vectors and are perpendicular,<br />

0 a > 1 b > 0 2 5 0 a > a ><br />

0 2 1 0 b > b ><br />

0 2 . So,<br />

0 a > 1 b > 0 2 5 (4) 2 1 (3) 2<br />

5 16 1 9<br />

0 a > 1 b > 5 25<br />

0 5 "25 5 5<br />

9. Express r > as a linear combination of p > and q > :<br />

Solve for a and b:<br />

r > 5 ap > 1 bq ><br />

( 21, 2) 5 a(211, 7) 1 b(23, 1)<br />

( 21, 2) 5 (211a, 7a) 1 (23b, b)<br />

( 21, 2) 5 (211a 2 3b, 7a 1 b)<br />

Solve the system of equations:<br />

21 5211a 2 3b<br />

2 5 7a 1 b<br />

Use the method of elimination:<br />

3(2) 5 3(7a 1 b)<br />

6 5 21a 1 3b<br />

1 21 5211a 2 3b<br />

5 5 10a<br />

a – b<br />

1<br />

2 5 a<br />

By substitution, b 52 3 2<br />

1<br />

Therefore<br />

Express q > 2(211, 7) 12 3 2(23, 1) 5 (21,<br />

as a linear combination of p > 2)<br />

and r > .<br />

Solve for a and b:<br />

q > 5 ap > 1 br ><br />

( 23, 1) 5 a(211, 7) 1 b(21, 2)<br />

( 23, 1) 5 (211a, 7a) 1 (2b, 2b)<br />

( 23, 1) 5 (211a 2 b, 7a 1 2b)<br />

Solve the system of equations:<br />

23 5211a 2 b<br />

1 5 7a 1 2b<br />

Use the method of elimination:<br />

2(23) 5 2(211a 2 b)<br />

26 5222a 2 2b<br />

1 1 5 7a 1 2b<br />

25 5215a<br />

1<br />

3 5 a<br />

By substitution, 2 2 3 5 b<br />

1<br />

Therefore<br />

Express p > 3(211, 7) 1 2 2 3(21, 2) 5 (23,<br />

as a linear combination of q > 1)<br />

and r > .<br />

Solve for a and b:<br />

p > 5 aq > 1 br ><br />

( 211, 7) 5 a(23, 1) 1 b(21, 2)<br />

( 211, 7) 5 (23a, a) 1 (2b, 2b)<br />

( 211, 7) 5 (23a 2 b, a 1 2b)<br />

Solve the system of equations:<br />

211 523a 2 b<br />

7 5 a 1 2b<br />

Use the method of elimination:<br />

2(211) 5 2(23a 2 b)<br />

222 526a 2 2b<br />

1 7 5 a 1 2b<br />

215 525a<br />

3 5 a<br />

By substitution, 2 5 b<br />

Therefore 3(23, 1) 1 2(21, 2) 5 (211, 7)<br />

10. a. Let P(x, y, z) be a point equidistant from A<br />

and B. Then @PA > @ 5 @PB > @ .<br />

(x 2 2) 2 1 (y 2 (21)) 2 1 (z 2 3) 2<br />

5 (x 2 1) 2 1 (y 2 2) 2 1 (z 2 (23)) 2<br />

x 2 2 4x 1 4 1 y 2 1 2y 1 1 1 z 2 2 6z 1 9<br />

5 x 2 2 2x 1 1 1 y 2 2 4y 1 4 1 z 2 1 6z 1 9<br />

22x 1 6y 2 12z 5 0<br />

x 2 3y 1 6z 5 0<br />

b. (0, 0, 0) and (1, 1 3, 0) clearly satisfy the equation<br />

and are equidistant from A and B.<br />

11. a.<br />

(224, 3, 25) 5 2(a, b, 4) 1 1 (6, 8, c) 2 3(7, c, 24)<br />

2<br />

(224, 3, 25) 5 (2a, 2b, 8) 1 a3, 4, c 2 b<br />

(224, 3, 25) 5 a2a 2 18, 2b 1 4 2 3c, c 2 1 20b<br />

Solve the equations:<br />

i. 224 5 2a 2 18<br />

26 5 2a<br />

23 5 a<br />

ii. 25 5 c 2 1 20<br />

5 5 c 2<br />

10 5 c<br />

iii. 3 5 2b 1 4 2 3c<br />

3 5 2b 1 4 2 3(18)<br />

3 5 2b 2 50<br />

53 5 2b<br />

26.5 5 b<br />

2 (21, 3c, 212)<br />

6-28 <strong>Chapter</strong> 6: Introduction to Vectors


. (3, 222, 54)<br />

5 2aa, a, 1 2 ab 1 (3b, 0, 25c) 1 2ac, 3 c, 0b<br />

2<br />

(3, 222, 54)<br />

5 (2a, 2a, a) 1 (3b, 0, 25c) 1 (2c, 3c, 0)<br />

(3, 222, 54) 5 (2a 1 3b 1 2c, 2a 1 3c, a 2 5c)<br />

Solve the system of equations:<br />

222 5 2a 1 3c<br />

54 5 a 2 5c<br />

Use the method of elimination:<br />

22(54) 522(a 2 5c)<br />

2108 522a 1 10c<br />

1222 5 2a 1 3c<br />

2130 5 13c<br />

210 5 c<br />

By substitution, 8 5 a<br />

Solve the equation:<br />

3 5 2a 1 3b 1 2c<br />

3 5 2(8) 1 3b 1 2(210)<br />

3 5 16 1 3b 2 20<br />

3 5 3b 2 4<br />

7 5 3b<br />

7<br />

3 5 b<br />

12. a. Find @AB > @ , @BC > @ ,<br />

Test<br />

5 "11<br />

5 "21<br />

5 "(3) 2 1 (21) 2<br />

5 "10<br />

@AB > @ ,<br />

@BC > @ ,<br />

@CA > @<br />

@CA > @<br />

@AB > @ 5 "(2 2 1) 2 1 (2 2 (21)) 2 1 (2 2 1) 2<br />

5 "(1) 2 1 (3) 2 1 (1) 2<br />

@BC > @ 5 "(4 2 2) 2 1 (22 2 2) 2 1 (1 2 2) 2<br />

5 "(2) 2 1 (24) 2 1 (21) 2<br />

@CA > @ 5 "(4 2 1) 2 1 (22 2 (21)) 2 1 (1 2 1) 2<br />

in the Pythagorean theorem:<br />

@AB > @ 2 1 @CA > @ 2 5 ("11) 2 1 ("10) 2<br />

5 11 1 10<br />

@BC > 5 21<br />

@ 2 5 ("21) 2<br />

5 21<br />

So triangle ABC is a right triangle.<br />

b. Yes, P(1, 2, 3), Q(2, 4, 6), and R(21, 22, 23)<br />

are collinear because:<br />

2P 5 (2, 4, 6)<br />

1Q 5 (2, 4, 6)<br />

22R 5 (2, 4, 6)<br />

13. a. Find 0 AB > 0, 0 BC > 0, 0 CA > 0<br />

@AB > @ 5 "(1 2 3) 2 1 (2 2 0) 2 1 (5 2 4) 2<br />

5 "9<br />

@BC > 5 3<br />

@ 5 "(2 2 1) 2 1 (1 2 2) 2 1 (3 2 5) 2<br />

@CA > @ 5 "(2 2 3) 2 1 (1 2 0) 2 1 (3 2 4) 2<br />

Test<br />

in the Pythagorean theorem:<br />

@BC > @ 2 1 @CA > @ 2 5 A"6B 2 1 A"3B 2<br />

5 6 1 3<br />

@AB > 5 9<br />

@ 2 5 (3) 2<br />

5 9<br />

So triangle ABC is a right triangle.<br />

b. Since triangle ABC is a right triangle,<br />

6<br />

cos/ABC 5 Å 3<br />

14. a. DA > , BC ><br />

and EB > , ED ><br />

b. DC > , AB ><br />

and CE > , EA ><br />

c. @AD > @ 2 1 @DC > @ 2 5 @AC > @ 2 ,<br />

But @AC > @ 2 5 @DB > 2<br />

@<br />

Therefore, @AD > @ 2 1 @DC > @ 2 5 @DB > 2<br />

@<br />

15. a. C(3, 0, 5); P(3, 4, 5); E(0, 4, 5); F(0, 4, 0)<br />

b. DB > 5 (3 2 0, 4 2 0, 0 2 5)<br />

5 (3, 4, 25)<br />

CF > 5 (0 2 3, 4 2 0, 0 2 5)<br />

5 (23, 4, 25)<br />

c. D<br />

P<br />

O<br />

5 "(22) 2 1 (2) 2 1 (1) 2<br />

5 "(1) 2 1 (21) 2 1 (22) 2<br />

5 "6<br />

5 "(21) 2 1 (1) 2 1 (21) 2<br />

5 "3<br />

@AB > @ ,<br />

@BC > @ ,<br />

X<br />

u<br />

@CA > @<br />

B<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-29


0 OD > 0 5 5<br />

0 DP > 0 5 5 by the Pythagorean theorem<br />

Thus ODPB is a square and cos u50,<br />

so the angle<br />

between the vectors is 90°.<br />

d. E<br />

P<br />

O<br />

u<br />

5 180° 2 2acos 21 a 3<br />

!50 bb<br />

8 50.2°<br />

16. a.<br />

d + e<br />

150° 30° d<br />

e<br />

Use the cosine law to evaluate 0 d > 1 e > 0<br />

0 d > 1 e > 0 2 5 0 d > 0 2 1 0 e > 0 2 2 20 d > 00e > 0 cos u<br />

5 (3) 2 1 (5) 3 2 2(3)(5) cos 150°<br />

5 9 1 25 2 30 2"3<br />

2<br />

0 d > 1 e > 8 59.98<br />

0 8 "59.98<br />

8 7.74<br />

b.<br />

d d – e<br />

e<br />

Use the cosine law to evaluate 0 d > 2 e > 0<br />

0 d > 2 e > 0 2 5 0 d > 0 2 1 0 e > 0 2 2 20 d > 00e > 0 cos u<br />

5 (3) 2 1 (5) 3 2 2(3)(5) cos 30°<br />

5 9 1 25 2 30 "3<br />

2<br />

0 d > 2 e > 8 8.02<br />

0 8 "8.02<br />

c. 0 e > 2 d > 8 2.83<br />

0 5 0 2 (d > 2 e > ) 0 5 0 d > 2 e > 0 8 2.83<br />

17. a.<br />

u<br />

A<br />

@OA > @ 5 3, @OP > @ 5 "5<br />

u5180° 2 2(m/POA)<br />

A:<br />

400 km/h<br />

Let represent the air speed of the airplane and let<br />

W > A ><br />

represent the velocity of the wind. In one hour,<br />

the plane will travel 0 A > 1 W > 0 kilometers. Because<br />

A > and W ><br />

make a right angle, use the Pythagorean<br />

theorem:<br />

0 A > 1 W > 0 2 5 0 A > 0 2 1 0 W > 2<br />

0<br />

0 A > 1 W > 0 5 "170000<br />

8 412.3 km<br />

So in 3 hours, the plane will travel<br />

3(412.3)km 8 1236.9 km<br />

b.<br />

5 170 000<br />

tan u5 0 W> 0<br />

0 A > 0<br />

5 100<br />

400<br />

5 (400) 2 1 (100) 2<br />

u5tan 21 a 1 4 b<br />

8 14.0°<br />

The direction of the airplane is S14.0°<br />

W.<br />

18. a. Any pair of nonzero, noncollinear vectors<br />

will span R 2 . To show that (2, 3) and (3, 5) are<br />

noncollinear, show that there does not exist any<br />

number k such that k(2, 3) 5 (3, 5) . Solve the<br />

system of equations:<br />

2k 5 3<br />

3k 5 5<br />

Solving both equations gives two different values<br />

3 5<br />

for k, 2 and 3, so (2, 3) and (3, 5) are noncollinear<br />

and thus span R 2<br />

b. (323, 795) 5 m(2, 3) 1 n(3, 5)<br />

(323, 795) 5 (2m, 3m) 1 (3n, 5n)<br />

(323, 795) 5 (2m 1 3n, 3m 1 5n)<br />

Solve the system of equations:<br />

323 5 2m 1 3n<br />

795 5 3m 1 5n<br />

Use the method of elimination:<br />

23(323) 523(2m 1 3n)<br />

2(795) 5 2(3m 1 5n)<br />

2969 526m 2 9n<br />

1 1590 5 6m 1 10n<br />

621 5 n<br />

By substitution, m 52770.<br />

W: 100 km/h<br />

6-30 <strong>Chapter</strong> 6: Introduction to Vectors


19. a. Find a and b such that<br />

(5, 9, 14) 5 a(22, 3, 1) 1 b(3, 1, 4)<br />

(5, 9, 14) 5 (22a, 3a, a) 1 (3b, b, 4b)<br />

(5, 9, 14) 5 (22a 1 3b, 3a 1 b, a 1 4b)<br />

i. 5 522a 1 3b<br />

ii. 9 5 3a 1 b<br />

iii. 14 5 a 1 4b<br />

Use the method of elimination with i. and iii.<br />

2(14) 5 2(a 1 4b)<br />

28 5 2a 1 8b<br />

1 5 522a 1 3b<br />

33 5 11b<br />

3 5 b<br />

By substitution, .<br />

a > a 5 2<br />

lies in the plane determined by b ><br />

and c > because it<br />

can be written as a linear combination of and<br />

b. If vector a > is in the span of b ><br />

and c > b ><br />

then can<br />

be written as a linear combination of b > a > c > .<br />

,<br />

and c > . Find<br />

m and n such that<br />

(213, 36, 23) 5 m(22, 3, 1) 1 n(3, 1, 4)<br />

5 (22m, 3m, m) 1 (3n, n, 4n)<br />

5 (22m 1 3n, 3m 1 n, m 1 4n)<br />

Solve the system of equations:<br />

213 522m 1 3n<br />

36 5 3m 1 n<br />

23 5 m 1 4n<br />

Use the method of elimination:<br />

2(23) 5 2(m 1 4n)<br />

46 5 2m 1 8n<br />

1 213 522m 1 3n<br />

33 5 11n<br />

3 5 n<br />

By substitution, .<br />

So, vector a > m 5 11<br />

is in the span of b ><br />

and c > .<br />

20. a.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 0, 4)<br />

(4, 4, 4)<br />

b.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 4, 4)<br />

(4, 0, 4)<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

x (4, 0, 0) (4, 4, 0)<br />

PO > 5 (4, 4, 4) so,<br />

OP > 52PO > 52(4, 4, 4) 5 (24, 24, 24)<br />

c.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 0, 4)<br />

(4, 4, 4)<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

x (4, 0, 0) (4, 4, 0)<br />

The vector PQ ><br />

from P(4, 4, 4) to Q(0, 4, 0) can be<br />

written as PQ > 5 (24, 0, 24) .<br />

d.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 0, 4)<br />

(4, 4, 4)<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

x (4, 0, 0) (4, 4, 0)<br />

x<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

(4, 0, 0) (4, 4, 0)<br />

The vector with the coordinates (4, 4, 0).<br />

21. 0 2(a ><br />

5 0 2a > 1 b ><br />

5 0 4a > 1 2b > 2 c > )<br />

2 3b > 2 2c > 2 (a ><br />

1 c > 2 a > 1 2b ><br />

2 2b > ) 1<br />

1 3a > 3(a > 2<br />

2 3b > b > 1<br />

1 3c > c > ) 0<br />

0<br />

0<br />

5 0 4(1, 1, 21) 2 3(2, 21, 3) 1 (2, 0, 13) 0<br />

5 0 (4, 4, 24) 1 (26, 3, 29) 1 (2, 0, 13) 0<br />

5 0 (0, 7, 0) 0<br />

5 7<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-31


22.<br />

a. @AB > @ 5 10 because it is the diameter of the circle.<br />

5 2"5 or 4.47<br />

5 "80 or 8.94<br />

b. If A, B, and C are vertices of a right triangle, then<br />

5 100<br />

So, triangle ABC is a right triangle.<br />

23. a. FL ><br />

b. MK > 5<br />

5 JK > FG > 1<br />

2 JM > GH ><br />

5 a > 1 HL ><br />

2 b > 5 a > 1 b > 1 c ><br />

c.<br />

d.<br />

e.<br />

24.<br />

5 "20<br />

A(–3, 4)<br />

@BC > @ 5 "(5 2 3) 2 1 (0 2 (24)) 2<br />

@CA > @ 5 "(5 2 (23)) 2 1 (0 2 4) 2<br />

@BC > @ 2 1 @CA > @ 2 5 @AB > @<br />

2<br />

@BC > @ 2 1 @CA > @ 2 5 A2"5B 2 1 A"80B 2<br />

b<br />

y<br />

5 "(2) 2 1 (4) 2<br />

5 "(8) 2 1 (24) 2<br />

B(3, –4)<br />

5 20 1 80<br />

@AB > 5 100<br />

@ 2 5 10 2<br />

HJ ><br />

IH > 5 HG ><br />

IK > 1 KJ > 1 GF ><br />

2 IH > 5 FG > 1 FJ ><br />

5 HK > 1 GF > 52b > 2 a > 1 c ><br />

5 IJ > 5<br />

5 b > 0<br />

2 c ><br />

a<br />

C(5, 0)<br />

25. a. "0a > 0 2 1 0 b > 2<br />

0 by the Pythagorean theorem<br />

b. "0a > 0 2 1 0 b > 2<br />

0 by the Pythagorean theorem<br />

c. "40 a > 0 2 1 90 b > 0<br />

2 by the Pythagorean theorem<br />

26. Case 1 If b ><br />

and c > are collinear, then is<br />

also collinear with both and But is perpendicular<br />

to b > and c > , so a > b > c > . a > 2b > 1 4c ><br />

is perpendicular to 2b > 1 4c > .<br />

x<br />

Case 2 If and are not collinear, then by spanning<br />

sets, b > b ><br />

and c > c ><br />

span a plane in and is in<br />

that plane. If a > R 3 ,<br />

is perpendicular to b > 2b > 1<br />

and c > 4c ><br />

, then it is<br />

perpendicular to the plane and all vectors in the<br />

plane. So, a > is perpendicular to 2b > 1 4c > .<br />

<strong>Chapter</strong> 6 Test, p. 348<br />

1. Let P be the tail of and let Q be the head of<br />

The vector sums 3a > a ><br />

1 (b > 1 c > and 3(a > 1 b > c ><br />

) 1 c > .<br />

)4<br />

4<br />

can be depicted as in the diagram below, using the<br />

triangle law of addition. We see that<br />

PQ > 5 a > 1 (b > 1 c > ) 5 (a > 1 b > ) 1 c > . This is the<br />

associative property for vector addition.<br />

P<br />

a<br />

(a + b)<br />

PQ= (a + b) + c = a + (b + c)<br />

2. a. AB ><br />

b. @AB > 5 (6 2 (22), 7 2 3, 3 2 (25)) 5 (8, 4, 8)<br />

@<br />

c. BA > 5 "8 2 1<br />

5 (21)AB > 4 2 1 8 2 5 12<br />

5 (28, 24, 28);<br />

@BA > @ 5 @AB > @ 5 12; unit vector in direction of<br />

@BA > @ 5 1<br />

@BA > BA ><br />

@<br />

5 1 (28, 24, 28)<br />

12<br />

5 a2 2 3 2 1 3 , 22 3 b<br />

b<br />

(b + c)<br />

3. Let x > 5 PQ > , y > 5 QR > , and as in the<br />

diagram below. Note that 0 RS > 2y > 5<br />

0 5 0 2y > QS > ,<br />

0 5 6 and that<br />

triangle PQR and triangle PRS share angle u.<br />

c<br />

Q<br />

6-32 <strong>Chapter</strong> 6: Introduction to Vectors


P<br />

x + y<br />

x – y<br />

x<br />

By the cosine law:<br />

cos u5 0 y> 0<br />

2 1 0 x > 1 y > 0<br />

2 2 0 x > 2<br />

0<br />

and<br />

20 y > 00x > 1 y > 0<br />

cos u5 0 2y> 0<br />

2 1 0 x > 1 y > 0<br />

2 2 0 x > 2 y > 2<br />

0<br />

20 2y > 00x > 1 y > .<br />

0<br />

Hence,<br />

0 y > 0 2 1 0 x > 1 y > 0 2 2 0 x > 2<br />

0<br />

20 y > 00x > 1 y > 0<br />

5 0 2y> 0<br />

2 1 0 x > 1 y > 0<br />

2 2 0 x > 2 y > 2<br />

0<br />

20 2y > 00x > 1 y > 0<br />

2(0 y > 0 2<br />

5<br />

1 20 x > 2<br />

0 x > 0 2y > 1 0 x > 1<br />

0 2<br />

2 y > 1 0 x > y > 0 2<br />

1<br />

0 2 5 20 y > y > 2 0 x > 0 2<br />

0 2 2<br />

0 2 2 0 x > 0 x > )<br />

2<br />

1 y > y > 2<br />

0<br />

2<br />

0 0<br />

1 "20 x > 2<br />

0 x > 2 y > 0 5 "20 y > 0 2 2 0 x > 1 y > 2<br />

0 0<br />

0 x > 2 y > 0 5 "2(3) 2 2 ("17) 2 1 "2(3) 2<br />

0 x > 2 y > 0 5 "19<br />

u<br />

R<br />

y<br />

Q<br />

S<br />

– y<br />

4. a. We have 3x > 2 2y > 5 a > and 5x > 2 3y > 5 b > .<br />

Multiplying the first equation by and the second<br />

equation by 2 yields:<br />

and<br />

10x > Adding these equations, we have:<br />

x > 2<br />

5 2b > 6y > 5<br />

2 3a > 2b > 29x > 1 6y > 23<br />

523a ><br />

.<br />

Substituting this into the first equation<br />

yields:<br />

Simplifying, we<br />

have: y > 3(2b > .<br />

2 3a > )<br />

5 3b > 2 5a > 2 2y > 5 a > .<br />

.<br />

b. First, conduct scalar multiplication on the third<br />

vector, yielding:<br />

(2, 21, c) 1 (a, b, 1) 2 (6, 3a, 12) 5 (23, 1, 2c).<br />

Now, each of the three components corresponds to<br />

an equation. First, 2 1 a 2 6 523, which implies<br />

a 5 1. Second, 21 1 b 2 3a 5 1. Substituting<br />

a 5 1 and simplifying yields b 5 5. Third,<br />

c 1 1 so<br />

5. a. a > 2 12 5<br />

and b > 2c, c 5211.<br />

span R 2 , because any vector (x, y) in<br />

R 2 can be written as a linear combination of a > and b > .<br />

These two vectors are not multiples of each other.<br />

b. First, conduct scalar multiplication on the vectors,<br />

yielding: (22p, 3p) 1 (3q, 2q) 5 (13, 29).<br />

Now, each component corresponds to an equation.<br />

First, 22p 1 3q 5 13. Second, 3p 2 q 529.<br />

Multiplying the second equation by 3 and adding<br />

the result to the first equation yields: 7p 5214,<br />

which implies p 522. Substituting this into the<br />

first equation and simplifying yields<br />

6. a. a > 5 mb > 1 nc > q 5 3.<br />

(1, 12, 229) 5 m(3, 1, 4) 1 n(1, 2, 23)<br />

(1, 12, 229) 5 (3m, m, 4m) 1 (n, 2n, 23n)<br />

Each of the three components corresponds to an<br />

equation. First, 1 5 3m 1 n. Second, 12 5 m 1 2n.<br />

Third, 229 5 4m 2 3n. Multiplying the first<br />

equation by 22 and adding the result to the second<br />

equation yields m 522. Substituting m 522 into<br />

the first equation yields n 5 7. Since m 522 and<br />

n also solves the third component’s equation,<br />

a > 5 7<br />

5 mb > 1 nc > for m 522 and n 5 7. Hence, can<br />

be written as a linear combination of and<br />

b. r > 5 mp > 1 nq > b > c > a ><br />

.<br />

(16, 11, 224) 5 m(22, 3, 4) 1 n(4, 1, 26)<br />

(16, 11, 224) 5 (22m, 3m, 4m) 1 (4n, n, 26n)<br />

Each of the three components corresponds to an<br />

equation. First, 16 522m 1 4n. Second,<br />

11 5 3m 1 n. Third, 224 5 4m 2 6n. Multiplying<br />

the first equation by 2 and adding the result to the<br />

third equation yields n 5 4. Substituting n 5 4 into<br />

the first equation yields m 5 0. We have that n 5 4<br />

and m 5 0 is the unique solution to the first and<br />

third equations, but n 5 4 and m 5 0 does not<br />

solve the second equation. Hence, this system of<br />

equations has no solution, and cannot be written<br />

as a linear combination of and In other words,<br />

r > p > r > q > .<br />

does not lie in the plane determined by and<br />

7. x > and y ><br />

p > q > .<br />

have magnitudes of 1 and 2, respectively,<br />

and have an angle of 120° between them, as depicted<br />

in the picture below.<br />

y<br />

120˚<br />

x<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-33


Since 60° is the complement of 120° 3x > 1 2y > can<br />

be depicted as below.<br />

3x + 2y<br />

u<br />

3x<br />

60˚<br />

2y<br />

By the cosine law:<br />

0 3x ><br />

0 3x > 1 2y ><br />

0 3x > 1 2y > 0 2 5 0 3x ><br />

1 2y > 0 2 5 90 x > 0 2 1 0 2y ><br />

0 2 1 40 y > 0 2 2 20 3x ><br />

0 2 2 60 x > 002y ><br />

00y > 0 cos 60<br />

0<br />

0 2 5 9 1 16 2 12<br />

0 3x > 1 2y > 0 5 "13 or 3.61<br />

The direction of 3x > 1 2y > is u, the angle from x > .<br />

This can be computed from the sine law:<br />

0 3x > 1 2y > 0<br />

5 0 2y> 0<br />

sin 60 sin u<br />

sin u5 0 2y> 0 sin 60<br />

0 3x > 1 2y > 0<br />

u5sin 21 (4) sin 60<br />

a b<br />

"13<br />

relative to x<br />

8. DE > u 8<br />

DE > 5 CE > 73.9°<br />

5 b > 2<br />

2 a > CD ><br />

Also,<br />

BA ><br />

BA > 5 CA ><br />

5 2b > 2 CB ><br />

2 2a ><br />

Thus,<br />

DE > 5 1 2 BA><br />

u5sin 21 a 0 2y> 0 sin 60<br />

0 3x > 1 2y > b<br />

0<br />

6-34 <strong>Chapter</strong> 6: Introduction to Vectors


CHAPTER 6<br />

Introduction to Vectors<br />

Review of Prerequisite Skills, p. 273<br />

"3<br />

1<br />

"2<br />

1. a. c. e.<br />

2<br />

2<br />

2<br />

"3<br />

b. 2"3 d. f. 1<br />

2<br />

2. Find BC using the Pythagorean theorem,<br />

AC 2 5 AB 2 1 BC 2 .<br />

BC 2 5 AC 2 2 AB 2<br />

5 10 2 2 6 2<br />

5 64<br />

BC 5 8<br />

Next, use the ratio tan A 5 opposite<br />

tan A 5 BC<br />

adjacent .<br />

AB<br />

5 8 6<br />

5 4 3<br />

3. a. To solve DABC, find measures of the sides and<br />

angles whose values are not given: AB, /B, and<br />

/C. Find AB using the Pythagorean theorem,<br />

BC 2 5 AB 2 1 AC 2 .<br />

AB 2 5 BC 2 2 AC 2<br />

5 (37.0) 2 2 (22.0) 2<br />

5 885<br />

AB 5 "885<br />

8 29.7<br />

Find /B using the ratio sin B 5<br />

opposite<br />

sin B 5 AC<br />

hypotenuse .<br />

BC<br />

5 22.0<br />

37.0<br />

/B 8 36.5°<br />

/C 5 90° 2 /B<br />

/C 5 90° 2 36.5°<br />

/C 8 53.5°<br />

b. Find measures of the angles whose values are not<br />

given. Find /A using the cosine law,<br />

cos A 5 b2 1 c 2 2 a 2<br />

.<br />

2bc<br />

5 52 1 8 2 2 10 2<br />

2(5)(8)<br />

5 211<br />

80<br />

/A 8 97.9°<br />

Find /B using the sine law.<br />

sin B<br />

5 sin A<br />

b a<br />

sin B sin (97.9°)<br />

5<br />

5 10<br />

sin B 8 0.5<br />

/B 8 29.7°<br />

Find /C using the sine law.<br />

sin C<br />

5 sin A<br />

c a<br />

sin C sin (97.9°)<br />

5<br />

8 10<br />

sin C 8 0.8<br />

/C 8 52.4°<br />

4. Since the sum of the internal angles of a triangle<br />

equals 180°, determine the measure of /Z using<br />

/X 5 60° and /Y 5 70°.<br />

/Z 5 180°2 (/X 1 /Y)<br />

5 180°2 (60° 1 70°)<br />

5 50°<br />

Find XY and YZ using the sine law.<br />

XY<br />

sin Y 5 XY<br />

sin Z<br />

XY<br />

sin 70° 5 6<br />

sin 50°<br />

XZ 8 7.36<br />

YZ<br />

sin X 5 XY<br />

sin Z<br />

YZ<br />

sin 60° 5 6<br />

sin 50°<br />

YZ 8 6.78<br />

5. Find each angle using the cosine law.<br />

cos R 5 RS2 1 RT 2 2 ST 2<br />

2(RS)(RT)<br />

5 42 1 7 2 2 5 2<br />

2(4)(7)<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-1


5 5 7<br />

/R 8 44°<br />

cos S 5 RS2 1 ST 2 2 RT 2<br />

2(RS)(ST)<br />

5 42 1 5 2 2 7 2<br />

2(4)(5)<br />

52 1 5<br />

/S 8 102°<br />

cos T 5 RT 2 1 ST 2 2 RS 2<br />

2(RT)(ST)<br />

5 72 1 5 2 2 4 2<br />

2(7)(5)<br />

5 29<br />

35<br />

/T 8 34°<br />

6.<br />

A<br />

3.5 km<br />

T<br />

70°<br />

8.<br />

C<br />

100 km/h<br />

48°<br />

A<br />

T<br />

80 km/h<br />

Find AC and AT using the speed of each vehicle and<br />

the elapsed time (in hours) until it was located,<br />

distance 5 speed 3 time.<br />

AC 5 100 km>h 3 1 4 h<br />

5 25 km<br />

AT 5 80 km>h 3 1 3 h<br />

5 26 2 3 km<br />

Find CT using the cosine law.<br />

CT 2 5 AC 2 1 AT 2 2 2(AC)(AT) cos A<br />

5 (25 km) 2 1 a26 2 2<br />

3 kmb<br />

6 km<br />

2 2(25 km) a26 2 kmb cos 48°<br />

3<br />

8 443.94 km 2<br />

CT 8 21.1 km<br />

9.<br />

A<br />

Find AB (the distance between the airplanes) using<br />

the cosine law.<br />

AB 2 5 AT 2 1 BT 2 2 2(AT)(BT)cos T<br />

5 (3.5 km) 2 1 (6 km) 2<br />

2 2(3.5 km)(6 km) cos 70°<br />

8 33.89 km 2<br />

AB 8 5.82 km<br />

7. P<br />

2 km 7 km<br />

Q 142°<br />

R<br />

Find QR using the cosine law.<br />

QR 2 5 PQ 2 1 PR 2 2 2(PQ)(PR) cos P<br />

5 (2 km) 2 1 (7 km) 2<br />

2 2(2 km)(7 km) cos 142°<br />

8 75.06 km 2<br />

QR 8 8.66 km<br />

B<br />

5 cm<br />

5 cm<br />

B<br />

C<br />

C<br />

The pentagon can be divided into 10 congruent<br />

right triangles with height AC and base BC.<br />

10 3 /A 5 360°<br />

/A 5 36°<br />

Find AC and BC using trigonometric ratios.<br />

AC 5 AB 3 cos A<br />

5 5 cos 36°<br />

8 4.0 cm<br />

BC 5 AB 3 sin A<br />

5 5 sin 36°<br />

8 2.9 cm<br />

The area of the pentagon is the sum of the areas of<br />

the 10 right triangles. Use the area of ^ABC to<br />

determine the area of the pentagon.<br />

6-2 <strong>Chapter</strong> 6: Introduction to Vectors


Area pentagon 5 10 3 1 2 (BC)(AC) 6-3<br />

5 10 3 1 (2.9 cm)(4.0 cm)<br />

2<br />

5 59.4 cm 2<br />

6.1 An Introduction to Vectors,<br />

pp. 279–281<br />

1. a. False. Two vectors with the same magnitude<br />

can have different directions, so they are not equal.<br />

b. True. Equal vectors have the same direction and<br />

the same magnitude.<br />

c. False. Equal or opposite vectors must be parallel<br />

and have the same magnitude. If two parallel vectors<br />

have different magnitude, they cannot be equal or<br />

opposite.<br />

d. False. Equal or opposite vectors must be parallel<br />

and have the same magnitude. Two vectors with the<br />

same magnitude can have directions that are not<br />

parallel, so they are not equal or opposite.<br />

2. Vectors must have a magnitude and direction. For<br />

some scalars, it is clear what is meant by just the<br />

number. Other scalars are related to the magnitude<br />

of a vector.<br />

• Height is a scalar. Height is the distance (see below)<br />

from one end to the other end. No direction is given.<br />

• Temperature is a scalar. Negative temperatures are<br />

below freezing, but this is not a direction.<br />

• Weight is a vector. It is the force (see below) of<br />

gravity acting on your mass.<br />

• Mass is a scalar. There is no direction given.<br />

• Area is a scalar. It is the amount space inside a<br />

two-dimensional object. It does not have<br />

direction.<br />

• Volume is a scalar. It is the amount of space inside<br />

a three-dimensional object. No direction is given.<br />

• Distance is a scalar. The distance between two<br />

points does not have direction.<br />

• Displacement is a vector. Its magnitude is related<br />

to the scalar distance, but it gives a direction.<br />

• Speed is a scalar. It is the rate of change of<br />

distance (a scalar) with respect to time, but does<br />

not give a direction.<br />

• Force is a vector. It is a push or pull in a certain<br />

direction.<br />

• Velocity is a vector. It is the rate of change of<br />

displacement (a vector) with respect to time. Its<br />

magnitude is related to the scalar speed.<br />

3. Answers may vary. For example: Friction resists<br />

the motion between two surfaces in contact by<br />

acting in the opposite direction of motion.<br />

• A rolling ball stops due to friction which resists<br />

the direction of motion.<br />

• A swinging pendulum stops due to friction<br />

resisting the swinging pendulum.<br />

4. Answers may vary. For example:<br />

a. AD ><br />

b. AD > 5 BC > ; AB ><br />

ED > 52CB > 5<br />

52EB > AB > DC > ; AE ><br />

c. AC > & DB > DA > 52CD > 5 EC ><br />

;<br />

; AE > 52BC > AE > ; DE > 5<br />

;<br />

; 52CE > EB ><br />

;<br />

& EB > ; EC ><br />

& DE > ; AB ><br />

& CB ><br />

5.<br />

B<br />

H<br />

D<br />

E<br />

A<br />

C<br />

G<br />

a. AB ><br />

b. AB > 5 CD ><br />

c. @AB > 52EF ><br />

@ but<br />

d. GH > 5 @ EF > @<br />

e. AB > 5 2AB > AB > 2 EF ><br />

522JI ><br />

7. a. 100 km><br />

h, south<br />

b. 50 km><br />

h, west<br />

c. 100 km><br />

h, northeast<br />

d. 25 km><br />

h, northwest<br />

e. 60 km><br />

h, east<br />

F<br />

6. a. b. c. d. e.<br />

I<br />

J<br />

Calculus and Vectors <strong>Solutions</strong> Manual


Area pentagon 5 10 3 1 2 (BC)(AC) 6-3<br />

5 10 3 1 (2.9 cm)(4.0 cm)<br />

2<br />

5 59.4 cm 2<br />

6.1 An Introduction to Vectors,<br />

pp. 279–281<br />

1. a. False. Two vectors with the same magnitude<br />

can have different directions, so they are not equal.<br />

b. True. Equal vectors have the same direction and<br />

the same magnitude.<br />

c. False. Equal or opposite vectors must be parallel<br />

and have the same magnitude. If two parallel vectors<br />

have different magnitude, they cannot be equal or<br />

opposite.<br />

d. False. Equal or opposite vectors must be parallel<br />

and have the same magnitude. Two vectors with the<br />

same magnitude can have directions that are not<br />

parallel, so they are not equal or opposite.<br />

2. Vectors must have a magnitude and direction. For<br />

some scalars, it is clear what is meant by just the<br />

number. Other scalars are related to the magnitude<br />

of a vector.<br />

• Height is a scalar. Height is the distance (see below)<br />

from one end to the other end. No direction is given.<br />

• Temperature is a scalar. Negative temperatures are<br />

below freezing, but this is not a direction.<br />

• Weight is a vector. It is the force (see below) of<br />

gravity acting on your mass.<br />

• Mass is a scalar. There is no direction given.<br />

• Area is a scalar. It is the amount space inside a<br />

two-dimensional object. It does not have<br />

direction.<br />

• Volume is a scalar. It is the amount of space inside<br />

a three-dimensional object. No direction is given.<br />

• Distance is a scalar. The distance between two<br />

points does not have direction.<br />

• Displacement is a vector. Its magnitude is related<br />

to the scalar distance, but it gives a direction.<br />

• Speed is a scalar. It is the rate of change of<br />

distance (a scalar) with respect to time, but does<br />

not give a direction.<br />

• Force is a vector. It is a push or pull in a certain<br />

direction.<br />

• Velocity is a vector. It is the rate of change of<br />

displacement (a vector) with respect to time. Its<br />

magnitude is related to the scalar speed.<br />

3. Answers may vary. For example: Friction resists<br />

the motion between two surfaces in contact by<br />

acting in the opposite direction of motion.<br />

• A rolling ball stops due to friction which resists<br />

the direction of motion.<br />

• A swinging pendulum stops due to friction<br />

resisting the swinging pendulum.<br />

4. Answers may vary. For example:<br />

a. AD ><br />

b. AD > 5 BC > ; AB ><br />

ED > 52CB > 5<br />

52EB > AB > DC > ; AE ><br />

c. AC > & DB > DA > 52CD > 5 EC ><br />

;<br />

; AE > 52BC > AE > ; DE > 5<br />

;<br />

; 52CE > EB ><br />

;<br />

& EB > ; EC ><br />

& DE > ; AB ><br />

& CB ><br />

5.<br />

B<br />

H<br />

D<br />

E<br />

A<br />

C<br />

G<br />

a. AB ><br />

b. AB > 5 CD ><br />

c. @AB > 52EF ><br />

@ but<br />

d. GH > 5 @ EF > @<br />

e. AB > 5 2AB > AB > 2 EF ><br />

522JI ><br />

7. a. 100 km><br />

h, south<br />

b. 50 km><br />

h, west<br />

c. 100 km><br />

h, northeast<br />

d. 25 km><br />

h, northwest<br />

e. 60 km><br />

h, east<br />

F<br />

6. a. b. c. d. e.<br />

I<br />

J<br />

Calculus and Vectors <strong>Solutions</strong> Manual


8. a. 400 km><br />

h, due south<br />

b. 70 km><br />

h, southwesterly<br />

c. 30 km><br />

h southeasterly<br />

d. 25 km><br />

h, due east<br />

9. a. i. False. They have equal magnitude, but<br />

opposite direction.<br />

ii. True. They have equal magnitude.<br />

iii. True. The base has sides of equal length, so the<br />

vectors have equal magnitude.<br />

iv. True. They have equal magnitude and direction.<br />

b. E<br />

H<br />

To calculate @BD > @ , @BE > @ and @BH > @ , find the lengths<br />

of their corresponding line segments BD, BE and<br />

BH using the Pythagorean theorem.<br />

BD 2 5 AB 2 1 AD 2<br />

5 3 2 1 3 2<br />

BD 5 "18<br />

BE 2 5 AB 2 1 AE 2<br />

5 3 2 1 8 2<br />

BE 5 "73<br />

BH 2 5 BD 2 1 DH 2<br />

5 ("18) 2 1 8 2<br />

BH 5 "82<br />

10. a. The tangent vector describes James’s velocity<br />

at that moment. At point A his speed is 15 km><br />

h<br />

and he is heading north. The tangent vector shows<br />

his velocity is 15 km><br />

h, north.<br />

b. The length of the vector represents the magnitude<br />

of James’s velocity at that point. James’s speed is<br />

the same as the magnitude of James’s velocity.<br />

c. The magnitude of James’s velocity (his speed) is<br />

constant, but the direction of his velocity changes at<br />

every point.<br />

d. Point C<br />

e. This point is halfway between D and A, which is<br />

of the way around the circle. Since he is running<br />

7<br />

8<br />

A<br />

B<br />

F<br />

D<br />

G<br />

C<br />

15 km><br />

h and the track is 1 km in circumference, he<br />

can run around the track 15 times in one hour. That<br />

7<br />

means each lap takes him 4 minutes. 8 of 4 minutes<br />

is 3.5 minutes.<br />

3<br />

f. When he has travelled 8 of a lap, James will be<br />

halfway between B and C and will be heading<br />

southwest.<br />

11. a. Find the magnitude of AB ><br />

using the distance<br />

formula.<br />

@AB > @ 5 "(x A 2 x B ) 2 1 (y B 2 y A ) 2<br />

5 "(24 1 1) 2 1 (3 2 2) 2<br />

5 or 3.16<br />

b. CD > "10<br />

5 AB > . AB ><br />

moves from A(24, 2) to<br />

B(21, 3) or (x B , y B ) 5 (x A 1 3, y A 1 1). Use this<br />

to find point D.<br />

( x D , y D ) 5 (x C 1 3, y C 1 1)<br />

5 (26 1 3, 0 1 1)<br />

c. EF > 5 (23,<br />

5 AB > 1)<br />

. Find point E using<br />

( x A , y A ) 5 (x B 2 3, y B 2 1).<br />

( x E , y E ) 5 (x F 2 3, y F 2 1)<br />

5 (3 2 3, 22 2 1)<br />

d. GH > 5 (0, 23)<br />

52AB > , and moves in the opposite<br />

direction as AB > .<br />

( x H , y H ) 5 (x G 2 3, y G 2 1).<br />

( x H , y H ) 5 (x G 2 3, y G 2 1)<br />

5 (3 2 3, 1 2 1)<br />

5 (0, 0)<br />

6.2 Vector Addition, pp. 290–292<br />

1. a.<br />

b.<br />

c.<br />

x – y<br />

y – x<br />

x<br />

–x<br />

x + y<br />

x<br />

y<br />

–y<br />

y<br />

6-4 <strong>Chapter</strong> 6: Introduction to Vectors


8. a. 400 km><br />

h, due south<br />

b. 70 km><br />

h, southwesterly<br />

c. 30 km><br />

h southeasterly<br />

d. 25 km><br />

h, due east<br />

9. a. i. False. They have equal magnitude, but<br />

opposite direction.<br />

ii. True. They have equal magnitude.<br />

iii. True. The base has sides of equal length, so the<br />

vectors have equal magnitude.<br />

iv. True. They have equal magnitude and direction.<br />

b. E<br />

H<br />

To calculate @BD > @ , @BE > @ and @BH > @ , find the lengths<br />

of their corresponding line segments BD, BE and<br />

BH using the Pythagorean theorem.<br />

BD 2 5 AB 2 1 AD 2<br />

5 3 2 1 3 2<br />

BD 5 "18<br />

BE 2 5 AB 2 1 AE 2<br />

5 3 2 1 8 2<br />

BE 5 "73<br />

BH 2 5 BD 2 1 DH 2<br />

5 ("18) 2 1 8 2<br />

BH 5 "82<br />

10. a. The tangent vector describes James’s velocity<br />

at that moment. At point A his speed is 15 km><br />

h<br />

and he is heading north. The tangent vector shows<br />

his velocity is 15 km><br />

h, north.<br />

b. The length of the vector represents the magnitude<br />

of James’s velocity at that point. James’s speed is<br />

the same as the magnitude of James’s velocity.<br />

c. The magnitude of James’s velocity (his speed) is<br />

constant, but the direction of his velocity changes at<br />

every point.<br />

d. Point C<br />

e. This point is halfway between D and A, which is<br />

of the way around the circle. Since he is running<br />

7<br />

8<br />

A<br />

B<br />

F<br />

D<br />

G<br />

C<br />

15 km><br />

h and the track is 1 km in circumference, he<br />

can run around the track 15 times in one hour. That<br />

7<br />

means each lap takes him 4 minutes. 8 of 4 minutes<br />

is 3.5 minutes.<br />

3<br />

f. When he has travelled 8 of a lap, James will be<br />

halfway between B and C and will be heading<br />

southwest.<br />

11. a. Find the magnitude of AB ><br />

using the distance<br />

formula.<br />

@AB > @ 5 "(x A 2 x B ) 2 1 (y B 2 y A ) 2<br />

5 "(24 1 1) 2 1 (3 2 2) 2<br />

5 or 3.16<br />

b. CD > "10<br />

5 AB > . AB ><br />

moves from A(24, 2) to<br />

B(21, 3) or (x B , y B ) 5 (x A 1 3, y A 1 1). Use this<br />

to find point D.<br />

( x D , y D ) 5 (x C 1 3, y C 1 1)<br />

5 (26 1 3, 0 1 1)<br />

c. EF > 5 (23,<br />

5 AB > 1)<br />

. Find point E using<br />

( x A , y A ) 5 (x B 2 3, y B 2 1).<br />

( x E , y E ) 5 (x F 2 3, y F 2 1)<br />

5 (3 2 3, 22 2 1)<br />

d. GH > 5 (0, 23)<br />

52AB > , and moves in the opposite<br />

direction as AB > .<br />

( x H , y H ) 5 (x G 2 3, y G 2 1).<br />

( x H , y H ) 5 (x G 2 3, y G 2 1)<br />

5 (3 2 3, 1 2 1)<br />

5 (0, 0)<br />

6.2 Vector Addition, pp. 290–292<br />

1. a.<br />

b.<br />

c.<br />

x – y<br />

y – x<br />

x<br />

–x<br />

x + y<br />

x<br />

y<br />

–y<br />

y<br />

6-4 <strong>Chapter</strong> 6: Introduction to Vectors


d.<br />

–x<br />

–y –x<br />

2. a. BA > A<br />

C<br />

b. 0 > A<br />

–y<br />

B<br />

c.<br />

d.<br />

a<br />

a<br />

–b<br />

–c<br />

a – b – c<br />

–b<br />

c<br />

c. CB ><br />

C<br />

A<br />

B<br />

4. a.<br />

a – b + c<br />

b<br />

d. CA ><br />

3. a.<br />

C<br />

C<br />

c<br />

A<br />

B<br />

B<br />

b.<br />

a<br />

a<br />

(b + c)<br />

a + (b + c)<br />

b<br />

(a + b)<br />

c<br />

c<br />

b.<br />

b<br />

a + b + c<br />

–c<br />

a<br />

b<br />

(a + b) + c<br />

c. The resultant vectors are the same. The order in<br />

which you add vectors does not matter.<br />

Aa > 1 b > B 1 c > 5 a > 1 Ab > 1 c > B<br />

5. a. PS ><br />

R –RQ Q<br />

RS<br />

PR PQ<br />

S<br />

PS<br />

b. 0 > P<br />

Q<br />

a + b – c<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

a<br />

S<br />

RQ<br />

–RS<br />

R –PQ<br />

6. x > 1 y > PS<br />

5 MR > P<br />

z > 1 t > 5 MS > 1 RS ><br />

5 ST ><br />

5 SQ > 1 TQ ><br />

so<br />

( x > 1 y > ) 1 (z > 1 t > ) 5 MS ><br />

5 MQ > 1 SQ ><br />

6-5


7. a.<br />

b. y > 2x ><br />

c. x > 1<br />

d. 2x > y ><br />

e. x > 1<br />

1<br />

f. 2x > y > y ><br />

1<br />

g. 2x > 2 y > z ><br />

h. 2x > 1 y ><br />

2z > 1 z ><br />

8. a.<br />

y<br />

b. See the figure in part a. for the drawn vectors.<br />

0 y > 2 and<br />

so 0 y > 2 x > 0 2 5 0 x > 2 y > 2<br />

0 2x > x > 0 2 5<br />

0 5 0 x > 0 y > 0 2 1 0 x > 0 2 2 20 y > 002x > 0 cos (u)<br />

0,<br />

0<br />

9. a. Maria’s velocity is 11 km><br />

h downstream.<br />

b.<br />

4 km/h<br />

c.<br />

y –x<br />

7 km/h<br />

4 km/h<br />

A<br />

7 km/h<br />

Maria’s speed is 3 km><br />

h.<br />

10. a.<br />

x<br />

u<br />

11 km/h<br />

3 km/h<br />

f 1 + f 2<br />

u<br />

f 1<br />

ship<br />

f 2<br />

b. The vectors form a triangle with side lengths<br />

S S S S S S<br />

@ f1 @ , @ f2 @ and @ f1 1 f2 @ . Find @ f 1 1 f2 @ using the<br />

cosine law.<br />

S S<br />

@ f 1 1 f2 @ 2 S<br />

5 @ f 1<br />

@ 2 S<br />

1 @ f 2<br />

@ 2 S S<br />

2 2 @ f @@f2<br />

1<br />

@ cos (u)<br />

S S S<br />

@ f 1 1 f2 0 5 $ @ f 1<br />

@ 2 S<br />

1 @ f 2<br />

@ 2 S S<br />

2 2 @ f @@f2<br />

1<br />

@ cos (u)<br />

B<br />

y<br />

y<br />

x – y<br />

D<br />

u<br />

x<br />

C<br />

11.<br />

a + w<br />

Find using the Pythagorean theorem.<br />

0 a > 0 a ><br />

1 w > 1 w > 0<br />

0 2 5 0 a > 0 2 1 0 w > 2<br />

0<br />

5 (150 km>h) 2 1 (80 km>h) 2<br />

0 a > 1 w > 5 28900<br />

0 2 5 170<br />

Find the direction of a > 1 w > using the ratio<br />

tan(u) 5 0 w> 0<br />

0 a > 0<br />

u5tan 21<br />

150 km>h<br />

a > 1<br />

12. and form a right triangle. Find<br />

0 x > x > w > 8 N 28.1° W<br />

y > 5 170<br />

using the Pythagorean theorem.<br />

0 x > 1 y > x > km>h,<br />

1 y > N 28.1° W<br />

, ,<br />

1 y > 0<br />

0 2 5 0 x > 0 2 1 0 y > 2<br />

0<br />

5 7 2 1 24 2<br />

0 x > 1 y > 5 625<br />

0 5 25<br />

Find the angle between x > and x > 1 y > using the ratio<br />

tan (u) 5 0 y> 0<br />

0 x > 0<br />

21<br />

24<br />

u5tan<br />

7<br />

8 73.7°<br />

13. Find @AB > 1 AC > @ using the cosine law and the<br />

supplement to the angle between and<br />

@AB > 1<br />

5 @AB > AC > AB ><br />

AC > .<br />

2<br />

@<br />

@ 2 1 @AC > @ 2 2 2 @AB > @ @AC > @ cos (30°)<br />

5 1 2 1 1 2 "3<br />

2 2(1)(1)<br />

2<br />

5 2<br />

@AB > 2 "3<br />

1 AC > @ 8 0.52<br />

14. D<br />

A<br />

w<br />

u<br />

a<br />

80 km>h<br />

E<br />

B<br />

The diagonals of a parallelogram bisect each other.<br />

So EA > 52EC ><br />

and<br />

Therefore, EA > ED ><br />

1 EB > 52EB ><br />

1 EC > .<br />

1 ED > 5 0 > .<br />

C<br />

6-6 <strong>Chapter</strong> 6: Introduction to Vectors


15. P<br />

T<br />

b<br />

R<br />

b S<br />

Multiple applications of the Triangle Law for<br />

adding vectors show that<br />

RM > 1 b > 5 a > 1 TP ><br />

(since both are equal to the<br />

undrawn vector and that<br />

RM > 1 a > 5 b > TM ><br />

1 SQ > ),<br />

(since both are equal to the<br />

undrawn vector RQ > )<br />

Adding these two equations gives<br />

2 RM > 1 a > 1 b > 5 a > 1> b > 1 TP > ><br />

1 SQ ><br />

16. a > 1 b >2 RM> 5 TP 1 SQ<br />

and a > 2 b ><br />

represent the diagonals of a<br />

parallelogram with sides a > and b > .<br />

b<br />

a<br />

M<br />

a – b<br />

Since @a > a<br />

1 b > @ 5 @a > 2 b > @ and the only parallelogram<br />

with equal diagonals is a rectangle, the parallelogram<br />

must also be a rectangle.<br />

17. P<br />

G<br />

a<br />

a + b<br />

Q<br />

M<br />

R<br />

Let point M be defined as shown. Two applications<br />

of the Triangle Law for adding vectors show that<br />

GQ ><br />

GR > 1 QM ><br />

1 RM > 1 MG ><br />

1 MG > 5 0 ><br />

5 0 ><br />

Adding these two equations gives<br />

GQ > 1 QM > 1 2 MG > 1 GR > 1 RM > 5 0 ><br />

From the given information,<br />

2 MG > and<br />

QM > 5 GP ><br />

1 RM > 5 0 ><br />

(since they are opposing vectors of<br />

equal length), so<br />

GQ > 1 GP > 1 GR > 5 0 > , as desired.<br />

Q<br />

6.3 Multiplication of a Vector by a<br />

Scalar, pp. 298–301<br />

1. A vector cannot equal a scalar.<br />

2. a.<br />

3 cm<br />

b.<br />

c.<br />

d.<br />

3. E25°N describes a direction that is 25° toward the<br />

north of due east ( 90° east of north), in other words<br />

90° 2 25° 5 65° toward the east of due north. N65°E<br />

and “a bearing of 65° ” both describe a direction that<br />

is 65° toward the east of due north. So, each is<br />

describing the same direction in a different way.<br />

4. Answers may vary. For example:<br />

a.<br />

2 cm<br />

v ><br />

6 cm<br />

9 cm<br />

2v ><br />

b. 1<br />

c.<br />

2 v><br />

d. e. 1<br />

0 v > v ><br />

0<br />

22v > 2 2 3 v><br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-7


15. P<br />

T<br />

b<br />

R<br />

b S<br />

Multiple applications of the Triangle Law for<br />

adding vectors show that<br />

RM > 1 b > 5 a > 1 TP ><br />

(since both are equal to the<br />

undrawn vector and that<br />

RM > 1 a > 5 b > TM ><br />

1 SQ > ),<br />

(since both are equal to the<br />

undrawn vector RQ > )<br />

Adding these two equations gives<br />

2 RM > 1 a > 1 b > 5 a > 1> b > 1 TP > ><br />

1 SQ ><br />

16. a > 1 b >2 RM> 5 TP 1 SQ<br />

and a > 2 b ><br />

represent the diagonals of a<br />

parallelogram with sides a > and b > .<br />

b<br />

a<br />

M<br />

a – b<br />

Since @a > a<br />

1 b > @ 5 @a > 2 b > @ and the only parallelogram<br />

with equal diagonals is a rectangle, the parallelogram<br />

must also be a rectangle.<br />

17. P<br />

G<br />

a<br />

a + b<br />

Q<br />

M<br />

R<br />

Let point M be defined as shown. Two applications<br />

of the Triangle Law for adding vectors show that<br />

GQ ><br />

GR > 1 QM ><br />

1 RM > 1 MG ><br />

1 MG > 5 0 ><br />

5 0 ><br />

Adding these two equations gives<br />

GQ > 1 QM > 1 2 MG > 1 GR > 1 RM > 5 0 ><br />

From the given information,<br />

2 MG > and<br />

QM > 5 GP ><br />

1 RM > 5 0 ><br />

(since they are opposing vectors of<br />

equal length), so<br />

GQ > 1 GP > 1 GR > 5 0 > , as desired.<br />

Q<br />

6.3 Multiplication of a Vector by a<br />

Scalar, pp. 298–301<br />

1. A vector cannot equal a scalar.<br />

2. a.<br />

3 cm<br />

b.<br />

c.<br />

d.<br />

3. E25°N describes a direction that is 25° toward the<br />

north of due east ( 90° east of north), in other words<br />

90° 2 25° 5 65° toward the east of due north. N65°E<br />

and “a bearing of 65° ” both describe a direction that<br />

is 65° toward the east of due north. So, each is<br />

describing the same direction in a different way.<br />

4. Answers may vary. For example:<br />

a.<br />

2 cm<br />

v ><br />

6 cm<br />

9 cm<br />

2v ><br />

b. 1<br />

c.<br />

2 v><br />

d. e. 1<br />

0 v > v ><br />

0<br />

22v > 2 2 3 v><br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-7


5. a.<br />

y<br />

b. c.<br />

3b<br />

–3b<br />

y<br />

y<br />

x + 3y<br />

x<br />

b.<br />

–y<br />

x<br />

–y<br />

x – 3y<br />

c.<br />

–2x + y<br />

–x<br />

y –x<br />

d.<br />

–x<br />

–x<br />

–2x – y<br />

–y<br />

6. Answers may vary. For example:<br />

a<br />

b<br />

–y<br />

d.<br />

e.<br />

7. a.<br />

m(2a > ) 1 n a 3 2 a> b 5 0 ><br />

m(4a > ) 1 n (3a > ) 5 0 ><br />

m 5 3 and n 524 satisfy the equation, as does any<br />

multiple of the pair (3, 24). There are infinitely<br />

many values possible.<br />

b.<br />

da > 1 ea 3 2 a> b 1 f(2a > ) 5 0 ><br />

2da > 1 3ea > 1 4f a > 5 0 ><br />

d 5 2, e 5 0, and f 521 satisfy the equation, as<br />

does any multiple of the triple (2, 0, 21). There are<br />

infinitely many values possible.<br />

8. or<br />

a > and b ><br />

are collinear, so where k is a nonzero<br />

scalar. Since 0 a > 0 5 @b > a > 5 kb > ,<br />

@ , k can only be 21 or 1.<br />

9.<br />

b<br />

a<br />

2a – 3b<br />

c > 5 2a > , b > 5 3<br />

mc > 1 nb > 2 a><br />

5 0 ><br />

c > 5 2a > , b > 5 3<br />

da > 1 eb > 1 fc > 5 0 > 2 a><br />

a<br />

a<br />

4a<br />

a<br />

2a + 3b<br />

a<br />

–b<br />

–b<br />

–b<br />

b<br />

b<br />

b<br />

a.<br />

2a<br />

–2b<br />

Yes<br />

6-8 <strong>Chapter</strong> 6: Introduction to Vectors


10. Two vectors are collinear if and only if they can<br />

be related by a scalar multiple. In this case<br />

a. collinear<br />

b. not collinear<br />

c. not collinear<br />

d. collinear<br />

1<br />

11. a. is a vector with length 1 unit in the same<br />

direction as x > .<br />

b. 2 1<br />

0 x > 0 x > is a vector with length 1 unit in the<br />

opposite direction of x > .<br />

12.<br />

a<br />

a<br />

b<br />

b<br />

b<br />

1<br />

2<br />

sin u 8 (1)<br />

2.91<br />

u 8 9.9° from x > towards y ><br />

16. b > 5 1<br />

0 a > a > 0<br />

@ b > 1<br />

@ 5 `<br />

0 a > a >` 0<br />

@b > @ 5 1<br />

b > @b > 0 a > 0 a > 0<br />

0<br />

@ 5 1<br />

is a positive multiple of so it points in the<br />

same direction as a > a > ,<br />

and has magnitude 1. It is a<br />

unit vector in the same direction as a > .<br />

17.<br />

A<br />

0 x > 0 x > a > 2 kb > 6-9<br />

m 5 2 3<br />

13. a. 2 2 3 a><br />

1<br />

b.<br />

3 a> 1<br />

c.<br />

3 0 a> 0<br />

2<br />

d.<br />

3 0 a> 0<br />

4<br />

e.<br />

3 a><br />

14. and make an angle of , so you may find<br />

0 2x > x ><br />

using the Pythagorean theorem.<br />

0 2x > 1 y > y ><br />

90°<br />

1 y > 0<br />

0 2 5 0 2x > 0 2 1 0 y > 2<br />

0<br />

0 2x > 1 y > 5 2 2 1 1 2<br />

0 5 "5 or 2.24<br />

Find the direction of 2x > 1 y > using the ratio<br />

tan (u) 5 0 y> 0<br />

0 2x > 0<br />

u5tan 1 21 2<br />

8 26.6° from towards 2<br />

15. Find 0 2x > 1 y > x > x > 1 y ><br />

0 using the cosine law, and the<br />

supplement to the angle between and<br />

0 2x > 1 y > 0 2 5 0 2x > 0 2 1 0 y > 0 2 2 20 2x > x ><br />

00y > y > .<br />

0 cos (150°)<br />

2"3<br />

5 2 2 1 1 2 2 2(2)(1)<br />

0 2x > 1 y > 2<br />

0 8 2.91<br />

Find the direction of 2x > 1 y > using the sine law.<br />

sin u<br />

0 y > sin (150°)<br />

5<br />

0 0 2x > 1 y > 0<br />

B<br />

D<br />

C<br />

AD ><br />

AD > 5 c ><br />

2 AD > 5 b > 1 CD ><br />

5<br />

But CD > c > 1 BD ><br />

1 b ><br />

So 2 AD > 1 BD > 1 CD > 1 BD ><br />

5 c > 5<br />

1 b > 0.<br />

, or AD > 5 1<br />

18. PM > and<br />

so MN > 5 a ><br />

5 PN > PN ><br />

PQ > 5<br />

5 and<br />

so QR > 2a > b > 2<br />

2 a > PM > 5 b > 2 c> 1 1 2 b> .<br />

5 PR > PR > 5<br />

5 2b > 2 PQ > 2b ><br />

2 2a ><br />

Notice that<br />

2MN > 5 2b > 2<br />

5 QR > 2a ><br />

We can conclude that is parallel to and<br />

@QR > @ 5 2 @MN > QR ><br />

MN ><br />

@ .<br />

19. A B<br />

C<br />

c<br />

AD<br />

E<br />

b<br />

D<br />

Calculus and Vectors <strong>Solutions</strong> Manual


Answers may vary. For example:<br />

a. u > and<br />

b. u > 5 AB ><br />

and<br />

c. u > 5 AD > v ><br />

and<br />

d. u > 5 AC > v > 5 CD ><br />

5 ED > v > 5 AE ><br />

and v > 5 DB ><br />

5 AD ><br />

20. a. Since the magnitude of is three times the<br />

magnitude of y > x ><br />

and because the given sum is<br />

must be in the opposite direction of ny > 0, mx ><br />

and<br />

0 n 0 5 30 m 0.<br />

b. Whether x > and y > are collinear or not, m 5 0 and<br />

n 5 0 will make the given equation true.<br />

21. a.<br />

b. BE > CD > 5<br />

5 2b > b > 2<br />

2<br />

5 2(b > 2a > a ><br />

2<br />

5 2CD > a > )<br />

The two are therefore parallel (collinear) and<br />

@BE > @ 5 2 @CD > @<br />

22. A<br />

B<br />

E<br />

D<br />

C<br />

Applying the triangle law for adding vectors shows<br />

that<br />

AC > 5 AD > 1 DC ><br />

The given information states that<br />

AB > 5 2 3 DC><br />

3<br />

2 AB> 5 DC ><br />

By the properties of trapezoids, this gives<br />

3<br />

2 AE > 5 EC > , and since<br />

AC > 5 AE > 1 EC ><br />

, the original equation gives<br />

AE > 1 3 2 AE> 5 AD > 1 3 2 AB><br />

5<br />

2 AE> 5 AD > 1 3 2 AB><br />

6.4 Properties of Vectors, pp. 306–307<br />

AE > 5 2 5 AD> 1 3 5 AB><br />

1. a. 0<br />

b. 1<br />

c. 0 ><br />

d. 1<br />

2. a > 1 b > 5 b > 1 a ><br />

b<br />

3.<br />

4. Answers may vary. For example:<br />

a<br />

a + b<br />

k(a + b)<br />

k a<br />

k b<br />

5. PQ > 5 RQ > 1<br />

5 (RQ > SR > 1<br />

5 (SR > 1 SR > TS > 1<br />

5 SQ > 1 RQ > ) 1 (TS > PT ><br />

5 PS > 1 PS > ) 1 (PT > 1 PT ><br />

1 TS > )<br />

)<br />

5<br />

6. a.<br />

b. 0 > EC > PQ > 1 SQ ><br />

c. Yes, the diagonals of a rectangular prism are of<br />

equal length<br />

7. 5 3a > 2<br />

1 3b > 6b > 2<br />

524a > 2 3c > 15c > 2 6a > 1 12b > 2 6c > 2 a ><br />

8. a. 5 6i > 1 9b ><br />

2<br />

5 12i > 8j > 2 24c ><br />

1<br />

2 17j > 2k > 1<br />

1 5k > 6i > 2 9j > 1 3k ><br />

b.<br />

c.<br />

(a + b)<br />

b + a<br />

(b + c)<br />

a<br />

(a + b)+ c =<br />

= a + b + c<br />

5 3i > 2<br />

527i > 4j > 1<br />

5 2(3i > 1 11j > k > 2 10i ><br />

2 4j > 2<br />

23(26i > 1 k > 4k > 1 15j > 2 5k ><br />

1 8j > 1 6i ><br />

2 2k > 2 9j > 1<br />

1 14i > 3k > )<br />

2 21j > 1 7k > )<br />

526i > 1 13j > 2 7k ><br />

9. Solve the first equation for x > .<br />

x > 5 1 2 a> 2 3 2 y><br />

b<br />

c<br />

a<br />

a<br />

a + b<br />

a + (b+ c)<br />

b<br />

b<br />

6-10 <strong>Chapter</strong> 6: Introduction to Vectors


Answers may vary. For example:<br />

a. u > and<br />

b. u > 5 AB ><br />

and<br />

c. u > 5 AD > v ><br />

and<br />

d. u > 5 AC > v > 5 CD ><br />

5 ED > v > 5 AE ><br />

and v > 5 DB ><br />

5 AD ><br />

20. a. Since the magnitude of is three times the<br />

magnitude of y > x ><br />

and because the given sum is<br />

must be in the opposite direction of ny > 0, mx ><br />

and<br />

0 n 0 5 30 m 0.<br />

b. Whether x > and y > are collinear or not, m 5 0 and<br />

n 5 0 will make the given equation true.<br />

21. a.<br />

b. BE > CD > 5<br />

5 2b > b > 2<br />

2<br />

5 2(b > 2a > a ><br />

2<br />

5 2CD > a > )<br />

The two are therefore parallel (collinear) and<br />

@BE > @ 5 2 @CD > @<br />

22. A<br />

B<br />

E<br />

D<br />

C<br />

Applying the triangle law for adding vectors shows<br />

that<br />

AC > 5 AD > 1 DC ><br />

The given information states that<br />

AB > 5 2 3 DC><br />

3<br />

2 AB> 5 DC ><br />

By the properties of trapezoids, this gives<br />

3<br />

2 AE > 5 EC > , and since<br />

AC > 5 AE > 1 EC ><br />

, the original equation gives<br />

AE > 1 3 2 AE> 5 AD > 1 3 2 AB><br />

5<br />

2 AE> 5 AD > 1 3 2 AB><br />

6.4 Properties of Vectors, pp. 306–307<br />

AE > 5 2 5 AD> 1 3 5 AB><br />

1. a. 0<br />

b. 1<br />

c. 0 ><br />

d. 1<br />

2. a > 1 b > 5 b > 1 a ><br />

b<br />

3.<br />

4. Answers may vary. For example:<br />

a<br />

a + b<br />

k(a + b)<br />

k a<br />

k b<br />

5. PQ > 5 RQ > 1<br />

5 (RQ > SR > 1<br />

5 (SR > 1 SR > TS > 1<br />

5 SQ > 1 RQ > ) 1 (TS > PT ><br />

5 PS > 1 PS > ) 1 (PT > 1 PT ><br />

1 TS > )<br />

)<br />

5<br />

6. a.<br />

b. 0 > EC > PQ > 1 SQ ><br />

c. Yes, the diagonals of a rectangular prism are of<br />

equal length<br />

7. 5 3a > 2<br />

1 3b > 6b > 2<br />

524a > 2 3c > 15c > 2 6a > 1 12b > 2 6c > 2 a ><br />

8. a. 5 6i > 1 9b ><br />

2<br />

5 12i > 8j > 2 24c ><br />

1<br />

2 17j > 2k > 1<br />

1 5k > 6i > 2 9j > 1 3k ><br />

b.<br />

c.<br />

(a + b)<br />

b + a<br />

(b + c)<br />

a<br />

(a + b)+ c =<br />

= a + b + c<br />

5 3i > 2<br />

527i > 4j > 1<br />

5 2(3i > 1 11j > k > 2 10i ><br />

2 4j > 2<br />

23(26i > 1 k > 4k > 1 15j > 2 5k ><br />

1 8j > 1 6i ><br />

2 2k > 2 9j > 1<br />

1 14i > 3k > )<br />

2 21j > 1 7k > )<br />

526i > 1 13j > 2 7k ><br />

9. Solve the first equation for x > .<br />

x > 5 1 2 a> 2 3 2 y><br />

b<br />

c<br />

a<br />

a<br />

a + b<br />

a + (b+ c)<br />

b<br />

b<br />

6-10 <strong>Chapter</strong> 6: Introduction to Vectors


Substitute into the second equation.<br />

3<br />

2 TO> 5 TZ > 1 1 2 TX><br />

6b > 52a 1 2 a> 2 3 2 y> b 1 5y > 6-11<br />

y > 5 1<br />

13 a> 1 12<br />

Lastly, find x > 13 b><br />

in terms of a > and b > .<br />

x > 5 1 2 a> 2 3 2 a 1 13 a> 1 12<br />

13 b> b<br />

5 5<br />

10. a > 13 a> 2 18<br />

5 x > 2 y ><br />

11. a. AG ><br />

BH > 5 a > 1<br />

CE > 52a > b > 1<br />

DF > 52a > 1 b > c ><br />

b. @AG > 5 a > 2<br />

2<br />

@ 2 5 0 a > b > b > 1 c ><br />

1<br />

1 c > c ><br />

0 2<br />

5 0 2a > 1 @b > @ 2<br />

0 2 1 @b > 1 0 c > 2<br />

0<br />

@ 2 1 0 c > 0<br />

5 @BH > 2<br />

@<br />

Therefore, @AG > @ 5 @BH > @<br />

12. T<br />

X<br />

Z<br />

5 2 3 y> 1 1 3 z> 2 (b > 1 z > )<br />

5 2 3 y> 2 2 3 z> 2 b ><br />

5 2 3 (y> 2 z > ) 2 b ><br />

5 2 3 b> 2 b ><br />

52 1 3 b><br />

13 b><br />

O<br />

Applying the triangle law for adding vectors<br />

shows that<br />

TY > 5 TZ > 1 ZY ><br />

The given information states that<br />

TX > 5 2 ZY ><br />

1<br />

2 TX> 5 ZY ><br />

By the properties of trapezoids, this gives<br />

1<br />

and since TY > 5 TO > 1 OY > 2 TO > 5 OY > ,<br />

, the<br />

original equation gives<br />

TO > 1 1 2 TO> 5 TZ > 1 1 2 TX><br />

Y<br />

2<br />

TO > 5 2 3 TZ> 1 1 3 TX><br />

Mid-<strong>Chapter</strong> Review, pp. 308–309<br />

1. a. AB ><br />

BA > 5 DC ><br />

AD > 5 CD ><br />

CB > 5 BC ><br />

5 DA ><br />

There is not enough information to determine if<br />

there is a vector equal to<br />

b. @PD > @ 5 @DA > AP > .<br />

@<br />

(parallelogram)<br />

2. a. RV >5 @BC> @<br />

b. RV ><br />

c. PS ><br />

d. RU ><br />

e. PS ><br />

f. PQ ><br />

3. a. Find @a > 1 b > @ using the cosine law, and the<br />

supplement to the angle between the vectors.<br />

@a > 1 b > @ 2 5 0 a > 0 2 1 @b > @ 2 2 20 a > 0 @b > @ cos 60°<br />

1<br />

5 3 2 1 4 2 2 2(3)(4)<br />

2<br />

@a > 1 b > 5 3<br />

@ 5 "3<br />

b. Find u using the ratio<br />

tan u5 @b> @<br />

0 a > 0<br />

5 4 3<br />

u5tan 21 4 3<br />

8 53°<br />

4. t 5 4 or t 524<br />

5. In quadrilateral PQRS, look at ^PQR. Joining the<br />

midpoints B and C creates a vector that is parallel<br />

to PR > and half the length of PR > BC ><br />

. Look at ^SPR.<br />

Joining the midpoints A and D creates a vector<br />

that is parallel to and half the length of<br />

is parallel to AD > PR > PR > AD ><br />

and equal in length to AD > . BC ><br />

.<br />

Therefore, ABCD is a parallelogram.<br />

6. a. Find using the cosine law. Note<br />

0 2v > and the angle between and is<br />

0 u > 0 5<br />

2 v > 0 v > 0 u > 2 v > 0<br />

0<br />

0 2 5 0 u > 0 2 1 0 2v > 0 2 2 20 u > u ><br />

002v > 2v > 120°.<br />

0 cos 60°<br />

Calculus and Vectors <strong>Solutions</strong> Manual


Substitute into the second equation.<br />

3<br />

2 TO> 5 TZ > 1 1 2 TX><br />

6b > 52a 1 2 a> 2 3 2 y> b 1 5y > 6-11<br />

y > 5 1<br />

13 a> 1 12<br />

Lastly, find x > 13 b><br />

in terms of a > and b > .<br />

x > 5 1 2 a> 2 3 2 a 1 13 a> 1 12<br />

13 b> b<br />

5 5<br />

10. a > 13 a> 2 18<br />

5 x > 2 y ><br />

11. a. AG ><br />

BH > 5 a > 1<br />

CE > 52a > b > 1<br />

DF > 52a > 1 b > c ><br />

b. @AG > 5 a > 2<br />

2<br />

@ 2 5 0 a > b > b > 1 c ><br />

1<br />

1 c > c ><br />

0 2<br />

5 0 2a > 1 @b > @ 2<br />

0 2 1 @b > 1 0 c > 2<br />

0<br />

@ 2 1 0 c > 0<br />

5 @BH > 2<br />

@<br />

Therefore, @AG > @ 5 @BH > @<br />

12. T<br />

X<br />

Z<br />

5 2 3 y> 1 1 3 z> 2 (b > 1 z > )<br />

5 2 3 y> 2 2 3 z> 2 b ><br />

5 2 3 (y> 2 z > ) 2 b ><br />

5 2 3 b> 2 b ><br />

52 1 3 b><br />

13 b><br />

O<br />

Applying the triangle law for adding vectors<br />

shows that<br />

TY > 5 TZ > 1 ZY ><br />

The given information states that<br />

TX > 5 2 ZY ><br />

1<br />

2 TX> 5 ZY ><br />

By the properties of trapezoids, this gives<br />

1<br />

and since TY > 5 TO > 1 OY > 2 TO > 5 OY > ,<br />

, the<br />

original equation gives<br />

TO > 1 1 2 TO> 5 TZ > 1 1 2 TX><br />

Y<br />

2<br />

TO > 5 2 3 TZ> 1 1 3 TX><br />

Mid-<strong>Chapter</strong> Review, pp. 308–309<br />

1. a. AB ><br />

BA > 5 DC ><br />

AD > 5 CD ><br />

CB > 5 BC ><br />

5 DA ><br />

There is not enough information to determine if<br />

there is a vector equal to<br />

b. @PD > @ 5 @DA > AP > .<br />

@<br />

(parallelogram)<br />

2. a. RV >5 @BC> @<br />

b. RV ><br />

c. PS ><br />

d. RU ><br />

e. PS ><br />

f. PQ ><br />

3. a. Find @a > 1 b > @ using the cosine law, and the<br />

supplement to the angle between the vectors.<br />

@a > 1 b > @ 2 5 0 a > 0 2 1 @b > @ 2 2 20 a > 0 @b > @ cos 60°<br />

1<br />

5 3 2 1 4 2 2 2(3)(4)<br />

2<br />

@a > 1 b > 5 3<br />

@ 5 "3<br />

b. Find u using the ratio<br />

tan u5 @b> @<br />

0 a > 0<br />

5 4 3<br />

u5tan 21 4 3<br />

8 53°<br />

4. t 5 4 or t 524<br />

5. In quadrilateral PQRS, look at ^PQR. Joining the<br />

midpoints B and C creates a vector that is parallel<br />

to PR > and half the length of PR > BC ><br />

. Look at ^SPR.<br />

Joining the midpoints A and D creates a vector<br />

that is parallel to and half the length of<br />

is parallel to AD > PR > PR > AD ><br />

and equal in length to AD > . BC ><br />

.<br />

Therefore, ABCD is a parallelogram.<br />

6. a. Find using the cosine law. Note<br />

0 2v > and the angle between and is<br />

0 u > 0 5<br />

2 v > 0 v > 0 u > 2 v > 0<br />

0<br />

0 2 5 0 u > 0 2 1 0 2v > 0 2 2 20 u > u ><br />

002v > 2v > 120°.<br />

0 cos 60°<br />

Calculus and Vectors <strong>Solutions</strong> Manual


5 8 2 1 10 2 2 2(8)(10)a 1 2 b<br />

0 u > 2 v > 0 5 2"21<br />

b. Find the direction of u > 2 v > using the sine law.<br />

sin u<br />

0 2v > sin 60°<br />

5<br />

0 0 u > 2 v > 0<br />

sin u5 5 sin 60°<br />

"21<br />

u5sin 21 5<br />

"28<br />

8 71°<br />

1<br />

c.<br />

0 u > 1 v > (u > 1 v > ) 5 1<br />

0<br />

d. Find using the cosine law.<br />

0 5u > 0 5u ><br />

1 2v > 1 2v > 2"21 (u> 1 v > )<br />

0 2 5 0 5u > 0<br />

0 2 1 0 2v > 0 2 2 20 5u > 002v > 0 cos 120°<br />

0 5u > 1 2v > 0<br />

7. Find using the cosine law.<br />

0 2p > 2 q > 0 2p > 5 20"7<br />

2 q ><br />

0 2 5 0 2p > 0<br />

0 2 1 0 2q > 0 2 2 20 2p > 002q > 0 cos 60°<br />

8. 0 m ><br />

9. BC > 1 n > 0 5<br />

DC > 52y > 0 m > 02 0 n > 0<br />

5 x ><br />

5 40 2 1 20 2 2 2(40)(20)a2 1 2 b<br />

5 2 2 1 1 2 2 2(2)(1)a 1 2 b 5 3<br />

BD ><br />

AC > 52x ><br />

5 x > 2<br />

2 y > y ><br />

10. Construct a parallelogram with sides OA ><br />

and .<br />

Since the diagonals bisect each other, is the<br />

diagonal equal to<br />

Or<br />

and So, And<br />

AC > AB > 5 1<br />

5 OC > 2 OA > OB > 5<br />

. Now OB > OA > 1 1<br />

Multiplying by 2 gives<br />

11.<br />

3x > AC > 1<br />

2 y > CD ><br />

1<br />

AB > 3x > 2y > 5 AD > 2OB > 5 OA > 2 AC > 2 AC > OA > 1 OC > 2OB > OC ><br />

. OB > 5 OA > 1 AB ><br />

.<br />

.<br />

5 OA > 1 1 2 (OC ><br />

1 OC > 2 OA > )<br />

.<br />

x > 1 BD > 1 y > 5 AD ><br />

5<br />

1 BD > 5 AD > AD ><br />

AB > BD > 5 3x ><br />

x > 1 BC > 5 2x > 1 y ><br />

1 BC > 5 AC > 1 y ><br />

BC > 5 3x ><br />

5 2x > 2 y ><br />

2 y ><br />

12. The air velocity of the airplane and the<br />

wind velocity (W > (V > air)<br />

) have opposite directions.<br />

V > ground 5 V > air 2 W ><br />

5 460 km><br />

h due south<br />

13. a.<br />

b. PT > PT><br />

c. SR ><br />

14. a. a<br />

b.<br />

c.<br />

d.<br />

15.<br />

1<br />

2<br />

a<br />

1<br />

3<br />

a<br />

1<br />

2<br />

b<br />

b<br />

1<br />

3<br />

–2 b<br />

a + b<br />

–b<br />

PS > 5 PQ ><br />

RS > 5 3b > 1<br />

5 QS > 2 a > QS ><br />

2<br />

523a > QR ><br />

6.5 Vectors in R 2 and R 3 , pp. 316–318<br />

1. No, as the y-coordinate is not a real number.<br />

2. a. We first arrange the x-, y-, and z-axes (each a<br />

copy of the real line) in a way so that each pair of<br />

axes are perpendicular to each other (i.e., the x- and<br />

y-axes are arranged in their usual way to form the<br />

xy-plane, and the z-axis passes through the origin of<br />

the xy-plane and is perpendicular to this plane).<br />

This is easiest viewed as a “right-handed system,”<br />

where, from the viewer’s perspective, the positive<br />

z-axis points upward, the positive x-axis points out<br />

of the page, and the positive y-axis points rightward<br />

in the plane of the page. Then, given point P(a, b, c),<br />

we locate this point’s unique position by moving a<br />

units along the x-axis, then from there b units<br />

parallel to the y-axis, and finally c units parallel to<br />

the z-axis. It’s associated unique position vector is<br />

determined by drawing a vector with tail at the<br />

origin O(0, 0, 0) and head at P.<br />

b. Since this position vector is unique, its<br />

coordinates are unique. Therefore a 524, b 523,<br />

and c 528.<br />

3. a. Since A and B are really the same point, we<br />

can equate their coordinates. Therefore a 5 5,<br />

b 523, and c 5 8.<br />

3<br />

2<br />

a<br />

6-12 <strong>Chapter</strong> 6: Introduction to Vectors


5 8 2 1 10 2 2 2(8)(10)a 1 2 b<br />

0 u > 2 v > 0 5 2"21<br />

b. Find the direction of u > 2 v > using the sine law.<br />

sin u<br />

0 2v > sin 60°<br />

5<br />

0 0 u > 2 v > 0<br />

sin u5 5 sin 60°<br />

"21<br />

u5sin 21 5<br />

"28<br />

8 71°<br />

1<br />

c.<br />

0 u > 1 v > (u > 1 v > ) 5 1<br />

0<br />

d. Find using the cosine law.<br />

0 5u > 0 5u ><br />

1 2v > 1 2v > 2"21 (u> 1 v > )<br />

0 2 5 0 5u > 0<br />

0 2 1 0 2v > 0 2 2 20 5u > 002v > 0 cos 120°<br />

0 5u > 1 2v > 0<br />

7. Find using the cosine law.<br />

0 2p > 2 q > 0 2p > 5 20"7<br />

2 q ><br />

0 2 5 0 2p > 0<br />

0 2 1 0 2q > 0 2 2 20 2p > 002q > 0 cos 60°<br />

8. 0 m ><br />

9. BC > 1 n > 0 5<br />

DC > 52y > 0 m > 02 0 n > 0<br />

5 x ><br />

5 40 2 1 20 2 2 2(40)(20)a2 1 2 b<br />

5 2 2 1 1 2 2 2(2)(1)a 1 2 b 5 3<br />

BD ><br />

AC > 52x ><br />

5 x > 2<br />

2 y > y ><br />

10. Construct a parallelogram with sides OA ><br />

and .<br />

Since the diagonals bisect each other, is the<br />

diagonal equal to<br />

Or<br />

and So, And<br />

AC > AB > 5 1<br />

5 OC > 2 OA > OB > 5<br />

. Now OB > OA > 1 1<br />

Multiplying by 2 gives<br />

11.<br />

3x > AC > 1<br />

2 y > CD ><br />

1<br />

AB > 3x > 2y > 5 AD > 2OB > 5 OA > 2 AC > 2 AC > OA > 1 OC > 2OB > OC ><br />

. OB > 5 OA > 1 AB ><br />

.<br />

.<br />

5 OA > 1 1 2 (OC ><br />

1 OC > 2 OA > )<br />

.<br />

x > 1 BD > 1 y > 5 AD ><br />

5<br />

1 BD > 5 AD > AD ><br />

AB > BD > 5 3x ><br />

x > 1 BC > 5 2x > 1 y ><br />

1 BC > 5 AC > 1 y ><br />

BC > 5 3x ><br />

5 2x > 2 y ><br />

2 y ><br />

12. The air velocity of the airplane and the<br />

wind velocity (W > (V > air)<br />

) have opposite directions.<br />

V > ground 5 V > air 2 W ><br />

5 460 km><br />

h due south<br />

13. a.<br />

b. PT > PT><br />

c. SR ><br />

14. a. a<br />

b.<br />

c.<br />

d.<br />

15.<br />

1<br />

2<br />

a<br />

1<br />

3<br />

a<br />

1<br />

2<br />

b<br />

b<br />

1<br />

3<br />

–2 b<br />

a + b<br />

–b<br />

PS > 5 PQ ><br />

RS > 5 3b > 1<br />

5 QS > 2 a > QS ><br />

2<br />

523a > QR ><br />

6.5 Vectors in R 2 and R 3 , pp. 316–318<br />

1. No, as the y-coordinate is not a real number.<br />

2. a. We first arrange the x-, y-, and z-axes (each a<br />

copy of the real line) in a way so that each pair of<br />

axes are perpendicular to each other (i.e., the x- and<br />

y-axes are arranged in their usual way to form the<br />

xy-plane, and the z-axis passes through the origin of<br />

the xy-plane and is perpendicular to this plane).<br />

This is easiest viewed as a “right-handed system,”<br />

where, from the viewer’s perspective, the positive<br />

z-axis points upward, the positive x-axis points out<br />

of the page, and the positive y-axis points rightward<br />

in the plane of the page. Then, given point P(a, b, c),<br />

we locate this point’s unique position by moving a<br />

units along the x-axis, then from there b units<br />

parallel to the y-axis, and finally c units parallel to<br />

the z-axis. It’s associated unique position vector is<br />

determined by drawing a vector with tail at the<br />

origin O(0, 0, 0) and head at P.<br />

b. Since this position vector is unique, its<br />

coordinates are unique. Therefore a 524, b 523,<br />

and c 528.<br />

3. a. Since A and B are really the same point, we<br />

can equate their coordinates. Therefore a 5 5,<br />

b 523, and c 5 8.<br />

3<br />

2<br />

a<br />

6-12 <strong>Chapter</strong> 6: Introduction to Vectors


. From part a., A(5, 23, 8), so OA > 5 (5, 23, 8).<br />

Here is a depiction of this vector.<br />

z<br />

z<br />

OA (5, –3, 8)<br />

y<br />

O(0, 0, 0)<br />

(0, 4, 0)<br />

y<br />

(0, 0, –2)<br />

(4, 0, 0)<br />

(0, 4, –2)<br />

x<br />

(4, 4, 0)<br />

(4, 0, –2) C(4, 4, –2)<br />

x<br />

4. This is not an acceptable vector in I 3 as the<br />

z-coordinate is not an integer. However, since all of<br />

the coordinates are real numbers, this is acceptable<br />

as a vector in R 3 .<br />

5.<br />

z<br />

(4, –4, 0)<br />

x<br />

A(4, –4, –2)<br />

(0, –4, –2)<br />

(0, –4, 0)<br />

(4, 0, –2)<br />

z<br />

O(0, 0, 0)<br />

(0, 0, –2)<br />

(4, 0, 0)<br />

(–4, 0, 2)<br />

y<br />

B(–4, 4, 2)<br />

6. a. A(0, 21, 0) is located on the y-axis.<br />

B(0, 22, 0), C(0, 2, 0), and D(0, 10, 0) are three<br />

other points on this axis.<br />

b. OA > 5 (0, 21, 0), the vector with tail at the<br />

origin O(0, 0, 0) and head at A.<br />

7. a. Answers may vary. For example:<br />

OA ><br />

OC > 5 (0, 0, 1), OB > 5 (0, 0, 21),<br />

5 (0, 0, 25)<br />

b. Yes, these vectors are collinear (parallel), as they<br />

all lie on the same line, in this case the z-axis.<br />

c. A general vector lying on the z-axis would be of<br />

the form OA > 5 (0, 0, a) for any real number a.<br />

Therefore, this vector would be represented by<br />

placing the tail at O, and the head at the point<br />

(0, 0, a) on the z-axis.<br />

8.<br />

z<br />

A(1, 0, 0)<br />

E(2, 0, 3)<br />

B(0, –2, 0)<br />

F(0, 2, 3)<br />

D(2, 3, 0)<br />

y<br />

(0, 0, 2)<br />

(–4, 0, 0)<br />

x<br />

C(0, 0, –3)<br />

O(0, 0, 0)<br />

(0, 4, 2)<br />

(0, 4, 0)<br />

(–4, 4, 0)<br />

y<br />

9. a.<br />

z<br />

x<br />

y<br />

x<br />

A(3, 2, –4)<br />

C(0, 1, –4)<br />

B(1, 1, –4)<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-13


. Every point on the plane containing points A, B,<br />

and C has z-coordinate equal to 24. Therefore, the<br />

equation of the plane containing these points is<br />

z 524 (a plane parallel to the xy-plane through<br />

the point z 524).<br />

10. a.<br />

z A(1, 2, 3)<br />

(0, 0, 3)<br />

(0, 2, 3)<br />

(1, 0, 3)<br />

O(0, 0, 0)<br />

(0, 2, 0)<br />

y<br />

(1, 0, 0)<br />

(1, 2, 0)<br />

x<br />

d.<br />

e.<br />

z<br />

D(1, 1, 1)<br />

(0, 0, 1)<br />

(1, 0, 1)<br />

(0, 1, 1)<br />

y<br />

O(0, 0, 0)<br />

(0, 1, 0)<br />

(1, 1, 0)<br />

x (1, 0, 0)<br />

z<br />

(0, –1, 1)<br />

(0, 0, 1)<br />

E(1, –1, 1)<br />

b.<br />

z<br />

(–2, 0, 1)<br />

(–2, 0, 0)<br />

(0, 0, 1)<br />

B(–2, 1, 1)<br />

(–2, 1, 0)<br />

(0, 1, 1)<br />

(0, –1, 0)<br />

(1, –1, 0)<br />

(1, 0, 0)<br />

x<br />

(1, 0, 1)<br />

O(0, 0, 0)<br />

y<br />

(0, 1, 0)<br />

y<br />

f.<br />

z<br />

x<br />

O(0, 0, 0)<br />

(1, 0, 0)<br />

(0, –1, 0)<br />

c.<br />

z<br />

(1, –1, 0)<br />

(0, –1, –1)<br />

O(0, 0, 0)<br />

y<br />

(0, –2, 1)<br />

C(1, –2, 1)<br />

(0, –2, 0)<br />

(0, 0, 1)<br />

x<br />

F(1, –1, –1)<br />

(1, 0, –1)<br />

(0, 0, –1)<br />

(1, –2, 0)<br />

(1, 0, 0)<br />

O(0, 0, 0)<br />

(1, 0, 1)<br />

y<br />

11. a. OA > 5 (1, 2, 3)<br />

z<br />

(0, 0, 3)<br />

(1, 0, 3)<br />

A(1, 2, 3)<br />

(0, 2, 3)<br />

O(0, 0, 0)<br />

(1, 0, 0)<br />

OA<br />

(0, 2, 0)<br />

y<br />

(1, 2, 0)<br />

x<br />

6-14 <strong>Chapter</strong> 6: Introduction to Vectors


. OB > 5 (22, 1, 1)<br />

z<br />

(–2, 0, 1)<br />

B(–2, 1, 1)<br />

f. OF > 5 (1, 21, 21)<br />

z<br />

x<br />

(–2, 0, 0)<br />

(0, 0, 1)<br />

O(0, 0, 0)<br />

OB<br />

(0, 1, 0)<br />

(–2, 1, 0)<br />

(0, 1, 1)<br />

y<br />

(1, 0, 0)<br />

(0, –1, 0)<br />

(1, –1, 0)<br />

(0, –1, –1)<br />

F(1, –1, –1)<br />

x<br />

OF<br />

O(0, 0, 0)<br />

(0, 0, –1)<br />

(1, 0, –1)<br />

y<br />

c. OC > 5 (1, 22, 1)<br />

z<br />

(0, –2, 1)<br />

(0, –2, 0)<br />

(1, 0, 1)<br />

C(1, –2, 1)<br />

(1, –2, 0)<br />

x<br />

d. OD > 5 (1, 1, 1)<br />

e. OE > 5 (1, 21, 1)<br />

z<br />

E(1, –1, 1)<br />

(0, –1, 0)<br />

(1, –1, 0)<br />

x<br />

x<br />

(1, 0, 0)<br />

(1, 0, 1)<br />

(0, 0, 1)<br />

O(0, 0, 0)<br />

(1, 0, 0)<br />

(0, –1, 1)<br />

(1, 0, 0)<br />

(0, 0, 1)<br />

O(0, 0, 0)<br />

z<br />

OC<br />

D(1, 1, 1)<br />

(0, 1, 1)<br />

OD<br />

y<br />

(0, 1, 0)<br />

(1, 1, 0)<br />

(0, 0, 1)<br />

(1, 0, 1)<br />

O(0, 0, 0)<br />

OE<br />

y<br />

y<br />

12. a. Since P and Q represent the same point,<br />

we can equate their y- and z-coordinates to get the<br />

system of equations<br />

a 2 c 5 6<br />

a 5 11<br />

Substituting this second equation into the first gives<br />

11 2 c 5 6<br />

c 5 5<br />

So a 5 11 and c 5 5.<br />

b. Since P and Q represent the same point in R 3 ,<br />

they will have the same associated position vector,<br />

i.e. OP > 5 OQ > . So, since these vectors are equal,<br />

they will certainly have equal magnitudes,<br />

i.e. @OP > @ 5 @OQ > @ .<br />

13. P(x, y, 0) represents a general point on the<br />

xy-plane, since the z-coordinate is 0. Similarly,<br />

Q(x, 0, z) represents a general point in the xz-plane,<br />

and R(0, y, z) represents a general point in the<br />

yz-plane.<br />

14. a. Every point on the plane containing points M,<br />

N, and P has y-coordinate equal to 0. Therefore, the<br />

equation of the plane containing these points is<br />

y 5 0 (this is just the xz-plane).<br />

b. The plane y 5 0 contains the origin O(0, 0, 0),<br />

and so since it also contains the points M, N, and P<br />

as well, it will contain the position vectors associated<br />

with these points joining O (tail) to the given point<br />

(head). That is, the plane contains the vectors<br />

OM > , ON > , and OP > y 5 0<br />

.<br />

15. a. A(22, 0, 0), B(22, 4, 0), C(0, 4, 0),<br />

D(0, 0, 27), E(0, 4, 27), F(22, 0, 27)<br />

b. OA > 5 (22, 0, 0), OB > 5 (22, 4, 0),<br />

OC > 5 (0, 4, 0), OD > 5 (0, 0, 27),<br />

OE > 5 (0, 4, 27), OF > 5 (22, 0, 27)<br />

c. Rectangle DEPF is 7 units below the xy-plane.<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-15


d. Every point on the plane containing points B, C,<br />

E, and P has y-coordinate equal to 4. Therefore, the<br />

equation of the plane containing these points is<br />

y 5 4 (a plane parallel to the xz-plane through the<br />

point y 5 4).<br />

e. Every point contained in rectangle BCEP has<br />

y-coordinate equal to 4, and so is of the form<br />

(x, 4,z) where x and z are real numbers such that<br />

22 # x # 0 and 27 # z # 0.<br />

16. a.<br />

y<br />

d.<br />

x<br />

O(0, 0, 0)<br />

z<br />

M(–1, 3, –2)<br />

y<br />

O(0, 0)<br />

x<br />

e.<br />

z<br />

F(0, 0, 5)<br />

P(4, –2)<br />

O(0, 0, 0)<br />

y<br />

b.<br />

y<br />

x<br />

D(–3, 4)<br />

f.<br />

z<br />

O(0, 0)<br />

x<br />

J(–2, –2, 0)<br />

O(0, 0, 0)<br />

y<br />

c.<br />

z<br />

x<br />

O(0, 0, 0)<br />

C(2, 4, 5)<br />

y<br />

17. The following box illustrates the three dimensional<br />

solid consisting of the set of all points (x, y, z) such<br />

that 0 # x # 1, 0 # y # 1, and 0 # z # 1.<br />

z<br />

x<br />

(1, 0, 1)<br />

O(0, 0, 0)<br />

(0, 0, 1) (0, 1, 1)<br />

(1, 1, 1)<br />

(1, 0, 0)<br />

x<br />

(1, 1, 0)<br />

y<br />

(0, 1, 0)<br />

6-16 <strong>Chapter</strong> 6: Introduction to Vectors


18. First, by the triangle law of<br />

vector addition, where<br />

OB > OA ><br />

OP > 5 (5,<br />

and OA > 210, 0),<br />

5 (0, 0, 210),<br />

are drawn in<br />

standard position (starting from the origin<br />

and OB > O(0,<br />

is drawn starting from the head of<br />

Notice that OA > OA > 0, 0)),<br />

lies in the xy-plane, and OB > .<br />

is<br />

perpendicular to the xy-plane (so is perpendicular to<br />

OA > ). So, OP > , OA > , and OB ><br />

form a right triangle and,<br />

by the Pythagorean theorem,<br />

Similarly, OA > 5 a > 1 b > @OP > @ 2 5 @OA > @ 2 1 @OB > 2<br />

@<br />

by the triangle law of<br />

vector addition, where<br />

and<br />

b > a > 5 (5, 0, 0)<br />

5 (0, 210, 0), and these three vectors form a<br />

right triangle as well. So,<br />

@OA > @ 2 5 0 a > 0 2 1 @b > 2<br />

@<br />

5 25 1 100<br />

5 125<br />

Obviously @OB > @ 2 5 100, and so substituting gives<br />

@OP > @ 2 5 @OA > @ 2 1 @OB > 2<br />

@<br />

5 125 1 100<br />

@OP > 5 225<br />

@ 5 "225<br />

5 15<br />

19. To find a vector equivalent to<br />

OP > AB ><br />

5 (22, 3, 6), where B(4, 22, 8), we need to<br />

move 2 units to the right of the x-coordinate for B<br />

(to 4 1 2 5 6), 3 units to the left of the y-coordinate<br />

for B (to 22 2 3 525), and 6 units below the<br />

z-coordinate for B (to 8 2 6 5 2). So we get the<br />

point A(6, 25, 2). Indeed, notice that to get from A<br />

to B (which describes vector AB ><br />

), we move 2 units<br />

left in the x-coordinate, 3 units right in the<br />

y-coordinate, and 6 units up in the z-coordinate.<br />

This is equivalent to vector OP > 5 (22, 3, 6).<br />

6.6 Operations with Algebraic<br />

R 2<br />

Vectors in , pp. 324–326<br />

1.<br />

A(–1, 3)<br />

y<br />

B(2, 5)<br />

x<br />

a. AB > 5 (2, 5) 2 (21, 3)<br />

5 (3, 2)<br />

BA > 52AB ><br />

52(3, 2)<br />

5 (23, 22)<br />

Here is a sketch of these two vectors in the<br />

xy-coordinate plane.<br />

y<br />

AB<br />

x<br />

b. 0 OA > 0 5 "(21) 2 1 3 2<br />

5 "29<br />

c. 0 AB > 8 5.39<br />

0 5 "3 2 1 2 2<br />

5 "13<br />

8 3.61<br />

Also, since<br />

0BA > BA ><br />

0 5 0 2AB > 52AB > ,<br />

0<br />

5 0 21 0 ? 0 AB > 0<br />

2.<br />

A(–1, 3)<br />

5 "10<br />

0 OB > 8 3.16<br />

0 5 "2 2 1 5 2<br />

5 0 AB > 0<br />

BA<br />

B(2, 5)<br />

5 "13<br />

8 3.61<br />

y<br />

30<br />

25<br />

20<br />

15<br />

10<br />

A(6, 10)<br />

O(0, 0)<br />

5<br />

OA x<br />

–9 –6 –3 0<br />

–5<br />

3 6 9<br />

–10<br />

OP > 5 OA > 1 OB > 6-17<br />

Calculus and Vectors <strong>Solutions</strong> Manual


18. First, by the triangle law of<br />

vector addition, where<br />

OB > OA ><br />

OP > 5 (5,<br />

and OA > 210, 0),<br />

5 (0, 0, 210),<br />

are drawn in<br />

standard position (starting from the origin<br />

and OB > O(0,<br />

is drawn starting from the head of<br />

Notice that OA > OA > 0, 0)),<br />

lies in the xy-plane, and OB > .<br />

is<br />

perpendicular to the xy-plane (so is perpendicular to<br />

OA > ). So, OP > , OA > , and OB ><br />

form a right triangle and,<br />

by the Pythagorean theorem,<br />

Similarly, OA > 5 a > 1 b > @OP > @ 2 5 @OA > @ 2 1 @OB > 2<br />

@<br />

by the triangle law of<br />

vector addition, where<br />

and<br />

b > a > 5 (5, 0, 0)<br />

5 (0, 210, 0), and these three vectors form a<br />

right triangle as well. So,<br />

@OA > @ 2 5 0 a > 0 2 1 @b > 2<br />

@<br />

5 25 1 100<br />

5 125<br />

Obviously @OB > @ 2 5 100, and so substituting gives<br />

@OP > @ 2 5 @OA > @ 2 1 @OB > 2<br />

@<br />

5 125 1 100<br />

@OP > 5 225<br />

@ 5 "225<br />

5 15<br />

19. To find a vector equivalent to<br />

OP > AB ><br />

5 (22, 3, 6), where B(4, 22, 8), we need to<br />

move 2 units to the right of the x-coordinate for B<br />

(to 4 1 2 5 6), 3 units to the left of the y-coordinate<br />

for B (to 22 2 3 525), and 6 units below the<br />

z-coordinate for B (to 8 2 6 5 2). So we get the<br />

point A(6, 25, 2). Indeed, notice that to get from A<br />

to B (which describes vector AB ><br />

), we move 2 units<br />

left in the x-coordinate, 3 units right in the<br />

y-coordinate, and 6 units up in the z-coordinate.<br />

This is equivalent to vector OP > 5 (22, 3, 6).<br />

6.6 Operations with Algebraic<br />

R 2<br />

Vectors in , pp. 324–326<br />

1.<br />

A(–1, 3)<br />

y<br />

B(2, 5)<br />

x<br />

a. AB > 5 (2, 5) 2 (21, 3)<br />

5 (3, 2)<br />

BA > 52AB ><br />

52(3, 2)<br />

5 (23, 22)<br />

Here is a sketch of these two vectors in the<br />

xy-coordinate plane.<br />

y<br />

AB<br />

x<br />

b. 0 OA > 0 5 "(21) 2 1 3 2<br />

5 "29<br />

c. 0 AB > 8 5.39<br />

0 5 "3 2 1 2 2<br />

5 "13<br />

8 3.61<br />

Also, since<br />

0BA > BA ><br />

0 5 0 2AB > 52AB > ,<br />

0<br />

5 0 21 0 ? 0 AB > 0<br />

2.<br />

A(–1, 3)<br />

5 "10<br />

0 OB > 8 3.16<br />

0 5 "2 2 1 5 2<br />

5 0 AB > 0<br />

BA<br />

B(2, 5)<br />

5 "13<br />

8 3.61<br />

y<br />

30<br />

25<br />

20<br />

15<br />

10<br />

A(6, 10)<br />

O(0, 0)<br />

5<br />

OA x<br />

–9 –6 –3 0<br />

–5<br />

3 6 9<br />

–10<br />

OP > 5 OA > 1 OB > 6-17<br />

Calculus and Vectors <strong>Solutions</strong> Manual


a.<br />

y<br />

(12, 20)<br />

20<br />

15<br />

10<br />

5<br />

O(0, 0)<br />

(3, 5)<br />

x<br />

–12–9<br />

–6 –3 0 3 6 9 12<br />

(–3, –5) –5<br />

–10<br />

–15<br />

–20<br />

(–12, –20)<br />

b. The vectors with the same magnitude are<br />

1<br />

2 OA> and 2 1<br />

2OA > 2 OA> ,<br />

and 22OA ><br />

3. @OA > @ 5 "3 2 1 (24) 2<br />

5 "25<br />

5<br />

4. a. The i > 5<br />

-component will be equal to the first<br />

coordinate in component form, and so<br />

Similarly, the j > a 523.<br />

-component will be equal to the<br />

second coordinate in component form, and so b 5 5.<br />

b. 0(23, b)0 5 0(23, 5)0<br />

5 "(23) 2 1 5 2<br />

5 "34<br />

5. a. 0 a > 8 5.83<br />

0 5 "(260) 2 1 11 2<br />

5 "3721<br />

@b > 5 61<br />

@ 5 "(240) 2 1 (29) 2<br />

5 "1681<br />

b. a > 1 b > 5 41<br />

5 (260, 11) 1 (240, 29)<br />

@a > 5 (2100, 2)<br />

1 b > @ 5 "(2100) 2 1 2 2<br />

5 "10 004<br />

a > 2 b > 8 100.02<br />

5 (260, 11) 2 (240, 29)<br />

@a > 5 (220, 20)<br />

2 b > @ 5 "(220) 2 1 20 2<br />

5 "800<br />

8 28.28<br />

6. a. 2(22, 3) 1 (2, 1) 5 (2(22) 1 2, 2(3) 1 1)<br />

5 (22, 7)<br />

b. 23(4, 29) 2 9(2, 3)<br />

5 (23(4) 2 9(2), 23(29) 2 9(3))<br />

5 (230, 0)<br />

c.<br />

2 1 2 (6, 22) 1 2 (6, 15)<br />

3<br />

5 a2 1 2 (6) 1 2 3 (6), 21 2 (22) 1 2 3 (15)b<br />

5<br />

7. x > (1, 11)<br />

5<br />

a. 3x > 2i ><br />

2 y > 2 j > , y > 52i ><br />

5 3(2i > 2 j > 1 5j ><br />

)<br />

5 (6 1 1)i > 2 (2i > 1 5j > )<br />

1 (23 2 5)j ><br />

b. 2 (x > 5<br />

524x > 1 2y > 7i > 2 8j ><br />

)<br />

2<br />

524(2i > 11y > 1 3(2x > 2 3y > )<br />

5 3i > 2<br />

c. 2(x > 2 51j > j > ) 2 11(2i > 1 5j > )<br />

1<br />

5213x > 3y > ) 2<br />

1<br />

5213(2i > 3y > 3(y > 1 5x > )<br />

5229i > 2 j ><br />

8. a. x > 1<br />

1 y > 28j > ) 1 3(2i > 1 5j > )<br />

5 (2i ><br />

0x > 1 y > 5 i > 2<br />

0 5 @i > 1 4j > j > ) 1 (2i > 1 5j > )<br />

1 4j > @<br />

5 "1 2 1 4 2<br />

5 "17<br />

b. x > 2 y > 8 4.12<br />

5 (2i ><br />

0x > 2 y > 5 3i > 2 j ><br />

0 5 @3i > 2 6j > ) 2 (2i > 1 5j > )<br />

2 6j > @<br />

5 "3 2 1 (26) 2<br />

5 "45<br />

c. 2x > 8<br />

2 3y > 6.71<br />

5 2(2i > 2 j > ) 2 3(2i > 1 5j > )<br />

0 2x > 2 3y > 5 7i ><br />

0 5 @7i > 2 17j ><br />

2 17j > @<br />

5 "7 2 1 (217) 2<br />

5 "338<br />

d. 0 3y > 2 2x > 8 18.38<br />

0 5 0 2 (2x > 2<br />

5 0 21<br />

5 0 2x > 002x > 3y > )<br />

2<br />

2 3y > 3y > 0<br />

0<br />

0<br />

so, from part c.,<br />

0 3y > 2 2x > 0 5 0 2x > 2 3y > 0<br />

5 "338<br />

8 18.38<br />

6-18 <strong>Chapter</strong> 6: Introduction to Vectors


9.<br />

y<br />

8<br />

6 D(4, 5)<br />

B(–4, 4)<br />

4<br />

2<br />

A(–8, 2) C(2, 1) x<br />

–8 –6 –4 –2 0 2 4 6 8<br />

F(–7, 0)<br />

–2<br />

H(6, –2)<br />

–4<br />

G(1, –2)<br />

E(–1, –4)<br />

–6<br />

–8<br />

so obviously we will have @OA > @ 5 @ BC > @ .<br />

(It turns out that their common magnitude is<br />

"6 2 1 3 2 5 "45.)<br />

11. a.<br />

y<br />

C(–4, 11)<br />

B(6, 6)<br />

A(2, 3)<br />

x<br />

a. AB > 5 (24, 4) 2 (28, 2)<br />

CD > 5 (4, 2)<br />

5 (4, 5) 2 (2, 1)<br />

EF > 5 (2, 4)<br />

5 (27, 0) 2 (21, 24)<br />

GH > 5 (26, 4)<br />

5 (6, 22) 2 (1, 22)<br />

5 (5, 0)<br />

b.<br />

@AB > @ 5 "4 2 1 2 2<br />

5 "20<br />

@CD > 8 4.47<br />

@ 5 "2 2 1 4 2<br />

5 "20<br />

@EF > 8 4.47<br />

@ 5 "(26) 2 1 4 2<br />

5 "52<br />

@GH > 8 7.21<br />

@ 5 "5 2 1 0 2<br />

5 "25<br />

5 5<br />

10. a. By the parallelogram law of vector addition,<br />

OC > 5 OA > 1 OB ><br />

5 (6, 3) 1 (11, 26)<br />

5 (17, 23)<br />

For the other vectors,<br />

BA > 5 OA > 2 OB ><br />

5 (6, 3) 2 (11, 26)<br />

BC > 5 (25,<br />

5 OC > 9)<br />

2 OB ><br />

5 (17, 23) 2 (11, 26)<br />

5<br />

b. OA > (6, 3)<br />

5 (6,<br />

5 BC > 3)<br />

,<br />

b.<br />

c.<br />

AB > 5 (6, 6) 2 (2, 3)<br />

5 (4, 3)<br />

@AB > @ 5 "4 2 1 3 2<br />

5 "25<br />

AC > 5 5<br />

5 (24, 11) 2 (2, 3)<br />

5 (26, 8)<br />

@AC > @ 5 "(26) 2 1 8 2<br />

5 "100<br />

CB > 5 10<br />

5 (6, 6) 2 (24, 11)<br />

5 (10, 25)<br />

@CB > @ 5 "10 2 1 (25) 2<br />

5 "125<br />

@CB > 8 11.18<br />

@ 2 5 125 @AC > @ 2 5 100, @AB > @ 2 5 25<br />

Since @CB > @ 2 5 @AC > @ 2 1 @AB > 2<br />

@ , the triangle is a right<br />

triangle.<br />

12. a.<br />

y<br />

A(–1, 2)<br />

C(2, 8)<br />

B(7, –2)<br />

x<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-19


.<br />

y<br />

X(–6, 12)<br />

A(–1, 2)<br />

C(2, 8)<br />

Z(10, 4)<br />

x<br />

B(7, –2)<br />

Substituting this into the last equation above, we<br />

can now solve for y.<br />

22(212) 2 5y 5 4<br />

y 5 4<br />

So x 5212 and y 5 4.<br />

14. a.<br />

y<br />

C(x, y)<br />

B(–6, 9) D(8, 11)<br />

Y(4, –8)<br />

A(2, 3)<br />

c. As a first possibility for the fourth vertex, there is<br />

X(x 1 , x 2 ). From the sketch in part b., we see that we b. Because ABCD is a rectangle, we will have<br />

would then have<br />

BC > 5 AD ><br />

CX > 5 BA ><br />

( x, y) 2 (26, 9) 5 (8, 11) 2 (2, 3)<br />

( x 1 2 2, x 2 2 8) 5 (21 2 7, 2 2 (22))<br />

( x 1 6, y 2 9) 5 (6, 8)<br />

5 (28, 4)<br />

x 1 6 5 6<br />

x 1 2 2 528<br />

y 2 9 5 8<br />

x 2 2 8 5 4<br />

So, x 5 0 and y i.e.,<br />

So<br />

By similar reasoning for the other 15. a. Since and<br />

points labelled in the sketch in part b.,<br />

PA > 0 PA > 5 17,<br />

0 5 0 PB > C(0, 17).<br />

X(26, 12).<br />

0,<br />

5 (5, 0) 2 (a, 0)<br />

AY > 5 CB ><br />

PB > 5 (5 2 a, 0),<br />

( y 1 2 (21), y 2 2 2) 5 (7 2 2, 22 2 8)<br />

5 (0, 2) 2 (a, 0)<br />

5 (5, 210)<br />

5 (2a, 2),<br />

y 1 1 1 5 5<br />

this means that<br />

y 2 2 2 5210<br />

(5 2 a)<br />

So Finally,<br />

2 5 (2a) 2 1 2<br />

Y(4, 28).<br />

2<br />

25 2 10a 1 a 2 5 a 2 1 4<br />

BZ > 5 AC ><br />

10a 5 21<br />

( z 1 27, z 2 2 (22)) 5 (2 2 (21), 8 22)<br />

5 (3, 6)<br />

a 5 21<br />

10<br />

z 1 2 7 5 3<br />

z So Pa 21<br />

10 , 0b. 2 1 2 5 6<br />

So Z(10, 4). In conclusion, the three possible b. This point Q on the y-axis will be of the form<br />

locations for a fourth vertex in a parallelogram Q(0, b) for some real number b. Reasoning<br />

with vertices A, B, and C are X(26, 12), Y(4, 28), similarly to part a., we have<br />

and Z(10, 4).<br />

QA > 5 (5, 0) 2 (0, b)<br />

13. a. 3(x, 1) 2 5(2, 3y) 5 (11, 33)<br />

QB > 5 (5, 2b)<br />

(3x 2 5(2), 3 2 5(3y)) 5 (11, 33)<br />

5 (0, 2) 2 (0, b)<br />

(3x 2 10, 3 2 15y) 5 (11, 33)<br />

5 (0, 2 2<br />

So since @QA > b)<br />

3 x 2 10 5 11<br />

@ 5 @QB > @<br />

3 2 15y 5 33<br />

,<br />

So x 5 7 and y 522.<br />

( 2b) 2 1 5 2 5 (2 2 b) 2<br />

b. 22(x, x 1 y) 2 3(6, y) 5 (6, 4)<br />

b 2 1 25 5 4 2 4b 1 b 2<br />

( 22x 2 18, 22x 2 5y) 5 (6, 4)<br />

4b 5221<br />

22x 2 18 5 6<br />

b 52 21<br />

22x 2 5y 5 4<br />

4<br />

To solve for x, use<br />

So Qa0, 2 21<br />

22x 2 18 5 6<br />

4 b.<br />

x 5212<br />

6-20 <strong>Chapter</strong> 6: Introduction to Vectors<br />

x


16. is in the direction opposite to and<br />

QP > 5 OP > 2 OQ > PQ > ,<br />

5 (11, 19) 2 (2, 221)<br />

5 (9, 40)<br />

@QP > @ 5 "9 2 1 40 2<br />

5 "1681<br />

5 41<br />

A unit vector in the direction of QP ><br />

is<br />

u > 5 1<br />

41 QP><br />

5 a 9 41 , 40<br />

Indeed, is obviously in the same direction as<br />

(since u > u > 41 b<br />

is a positive scalar multiple of QP > QP ><br />

), and<br />

notice that<br />

0 u > 0 5 Å<br />

a 9<br />

41 b 2<br />

1 a 40<br />

41 b 2<br />

81 1 1600<br />

5 Å 1681<br />

5 1<br />

17. a. O, P, and R can be thought of as the vertices<br />

of a triangle.<br />

PR > 5 OR > 2 OP ><br />

5 (28, 21) 2 (27, 24)<br />

@PR > 5 (21, 225)<br />

@ 2 5 (21) 2 1 (225) 2<br />

@OR > 5 626<br />

@ 2 5 (28) 2 1 (21) 2<br />

@OP > 5 65<br />

@ 2 5 (27) 2 1 24 2<br />

5 625<br />

By the cosine law, the angle, , between and<br />

OP > u OR ><br />

satisfies<br />

cos u5 @OR> @ 2 1 @OP > @ 2 2 @PR > 2<br />

@<br />

2 @OR > @ ? @OP > @<br />

65 1 625 2 626<br />

5<br />

2!65 ? !625<br />

65 1 625 2 626<br />

u5cos 21 a<br />

2!65 ? !625 b<br />

8 80.9°<br />

So the angle between OR ><br />

and is about .<br />

b. We found the vector in part a.,<br />

so RP > and<br />

@RP > 52PR > PR > OP ><br />

80.86°<br />

5 (21, 225)<br />

@ 2 5 @PR > 5 (1, 25)<br />

2<br />

@<br />

5 626<br />

Also, by the parallelogram law of vector addition,<br />

OQ > 5 OR > 1 OP ><br />

5 (28, 21) 1 (27, 24)<br />

@OQ > 5 (215, 23)<br />

@ 2 5 (215) 2 1 23 2<br />

5 754<br />

Placing RP > 5 (1, 25) and OQ > 5 (215, 23) with<br />

their tails at the origin, a triangle is formed by<br />

joining the heads of these two vectors. The third<br />

side of this triangle is the vector<br />

v > 5 RP > 2 OQ ><br />

5 (1, 25) 2 (215, 23)<br />

0 v > 5 (16, 2)<br />

0 2 5 16 2 1 2 2<br />

5 260<br />

Now by reasoning similar to part a., the cosine law<br />

implies that the angle, u, between RP ><br />

and OQ ><br />

satisfies<br />

cos u5 @RP> @ 2 1 @OQ > @ 2 2 @v > 2<br />

@<br />

2 @RP > @ ? @OQ > @<br />

626 1 754 2 260<br />

5<br />

2!626 ? !754<br />

u5cos 21 626 1 754 2 260<br />

a<br />

2!626 ? !754 b<br />

8 35.4°<br />

So the angle between RP ><br />

and OQ ><br />

is about 35.40° .<br />

However, since we are discussing the diagonals of<br />

parallelogram OPQR here, it would also have been<br />

appropriate to report the supplement of this angle,<br />

or about 180°235.40° 5 144.60°, as the angle<br />

between these vectors.<br />

6.7 Operations with Vectors in R 3 ,<br />

pp. 332–333<br />

OA > 521i > 1 2j > 1 4k ><br />

1. a.<br />

b. @OA > @<br />

2. OB > 5 "(21) 2 1 2 2 1 4 2 5 "21 8 4.58<br />

@OB > 5 (3, 4, 24)<br />

@ 5 "3 2 1 4 2 1 (24) 2 5 "41 8 6.40<br />

3. a > 1 1 3 b> 2 c > 5 (1, 3, 23) 1 (21, 2, 4)<br />

2 (0, 8, 1)<br />

5 (1 1 (21) 2 0, 3 1 2 2 8,<br />

(23) 1 4 2 1)<br />

5 (0, 23, 0)<br />

` a > 1 1 3 b> 2 c >` 5 "0 2 1 (23) 2 1 0 2<br />

QP > 6-21<br />

5 3<br />

Calculus and Vectors <strong>Solutions</strong> Manual


16. is in the direction opposite to and<br />

QP > 5 OP > 2 OQ > PQ > ,<br />

5 (11, 19) 2 (2, 221)<br />

5 (9, 40)<br />

@QP > @ 5 "9 2 1 40 2<br />

5 "1681<br />

5 41<br />

A unit vector in the direction of QP ><br />

is<br />

u > 5 1<br />

41 QP><br />

5 a 9 41 , 40<br />

Indeed, is obviously in the same direction as<br />

(since u > u > 41 b<br />

is a positive scalar multiple of QP > QP ><br />

), and<br />

notice that<br />

0 u > 0 5 Å<br />

a 9<br />

41 b 2<br />

1 a 40<br />

41 b 2<br />

81 1 1600<br />

5 Å 1681<br />

5 1<br />

17. a. O, P, and R can be thought of as the vertices<br />

of a triangle.<br />

PR > 5 OR > 2 OP ><br />

5 (28, 21) 2 (27, 24)<br />

@PR > 5 (21, 225)<br />

@ 2 5 (21) 2 1 (225) 2<br />

@OR > 5 626<br />

@ 2 5 (28) 2 1 (21) 2<br />

@OP > 5 65<br />

@ 2 5 (27) 2 1 24 2<br />

5 625<br />

By the cosine law, the angle, , between and<br />

OP > u OR ><br />

satisfies<br />

cos u5 @OR> @ 2 1 @OP > @ 2 2 @PR > 2<br />

@<br />

2 @OR > @ ? @OP > @<br />

65 1 625 2 626<br />

5<br />

2!65 ? !625<br />

65 1 625 2 626<br />

u5cos 21 a<br />

2!65 ? !625 b<br />

8 80.9°<br />

So the angle between OR ><br />

and is about .<br />

b. We found the vector in part a.,<br />

so RP > and<br />

@RP > 52PR > PR > OP ><br />

80.86°<br />

5 (21, 225)<br />

@ 2 5 @PR > 5 (1, 25)<br />

2<br />

@<br />

5 626<br />

Also, by the parallelogram law of vector addition,<br />

OQ > 5 OR > 1 OP ><br />

5 (28, 21) 1 (27, 24)<br />

@OQ > 5 (215, 23)<br />

@ 2 5 (215) 2 1 23 2<br />

5 754<br />

Placing RP > 5 (1, 25) and OQ > 5 (215, 23) with<br />

their tails at the origin, a triangle is formed by<br />

joining the heads of these two vectors. The third<br />

side of this triangle is the vector<br />

v > 5 RP > 2 OQ ><br />

5 (1, 25) 2 (215, 23)<br />

0 v > 5 (16, 2)<br />

0 2 5 16 2 1 2 2<br />

5 260<br />

Now by reasoning similar to part a., the cosine law<br />

implies that the angle, u, between RP ><br />

and OQ ><br />

satisfies<br />

cos u5 @RP> @ 2 1 @OQ > @ 2 2 @v > 2<br />

@<br />

2 @RP > @ ? @OQ > @<br />

626 1 754 2 260<br />

5<br />

2!626 ? !754<br />

u5cos 21 626 1 754 2 260<br />

a<br />

2!626 ? !754 b<br />

8 35.4°<br />

So the angle between RP ><br />

and OQ ><br />

is about 35.40° .<br />

However, since we are discussing the diagonals of<br />

parallelogram OPQR here, it would also have been<br />

appropriate to report the supplement of this angle,<br />

or about 180°235.40° 5 144.60°, as the angle<br />

between these vectors.<br />

6.7 Operations with Vectors in R 3 ,<br />

pp. 332–333<br />

OA > 521i > 1 2j > 1 4k ><br />

1. a.<br />

b. @OA > @<br />

2. OB > 5 "(21) 2 1 2 2 1 4 2 5 "21 8 4.58<br />

@OB > 5 (3, 4, 24)<br />

@ 5 "3 2 1 4 2 1 (24) 2 5 "41 8 6.40<br />

3. a > 1 1 3 b> 2 c > 5 (1, 3, 23) 1 (21, 2, 4)<br />

2 (0, 8, 1)<br />

5 (1 1 (21) 2 0, 3 1 2 2 8,<br />

(23) 1 4 2 1)<br />

5 (0, 23, 0)<br />

` a > 1 1 3 b> 2 c >` 5 "0 2 1 (23) 2 1 0 2<br />

QP > 6-21<br />

5 3<br />

Calculus and Vectors <strong>Solutions</strong> Manual


4. a. OP > 5 OA > 1 OB ><br />

5 ((23) 1 2, 4 1 2, 12 1 (21))<br />

5 (21, 6, 11)<br />

b.<br />

c.<br />

c.<br />

@OA > @ 5 "(23) 2 1 4 2 1 12 2 5 13<br />

@OB > @ 5 "2 2 1 2 2 1 (21) 2 5 3<br />

@OP > @ 5 "(21) 2 1 6 2 1 11 2 8 12.57<br />

AB > 5 OB > 2 OA ><br />

5 (2, 2, 21) 2 (23, 4, 12)<br />

5 (2 2 (23), 2 2 4, (21) 2 12)<br />

5 (5, 22, 213)<br />

@AB > @<br />

AB > 5 "5 2 1 (22) 2 1 (213) 2 5 "198 8 14.07<br />

represents the vector from the tip of to the tip<br />

of OB > OA ><br />

It is the difference between the two vectors.<br />

5. a. x > .<br />

2 2y > 2 z ><br />

5 (1, 4, 21) 2 2(1, 3, 22) 2 (22, 1, 0)<br />

5 (1, 4, 21) 2 (2, 6, 24) 2 (22, 1, 0)<br />

5 (1 2 2 2 (22), 4 2 6 2 1, 21 2 (24) 2 0)<br />

5 (1, 23, 3)<br />

b. 22x > 2 3y > 1 z ><br />

522(1, 4, 21) 2 3(1, 3, 22) 1 (22, 1, 0)<br />

5 (22, 28, 2) 2 (3, 9, 26) 1 (22, 1, 0)<br />

5 (22 2 3 2 2, 28 2 9 1 1, 2 1 6 1 0)<br />

5 (27, 216, 8)<br />

1<br />

2 x> 2 y > 1 3z ><br />

5 1 (1, 4, 21) 2 (1, 3, 22) 1 3(22, 1, 0)<br />

2<br />

5 a 1 2 , 2, 21 b 2 (1, 3, 22) 1 (26, 3, 0)<br />

2<br />

5 a 1 2 2 1 1 (26), 2 2 3 1 3, 21 2 (22) 1 0b<br />

2<br />

5 a2 13<br />

d. 3x > 2 , 2, 3<br />

1 5y > 2 b<br />

1 3z ><br />

5 3(1, 4, 21) 1 5(1, 3, 22) 1 3(22, 1, 0)<br />

5 (3, 12, 23) 1 (5, 15, 210) 1 (26, 3, 0)<br />

5 (3 1 5 2 6, 12 1 15 1 3, 23 2 10 1 0)<br />

5 (2, 30, 213)<br />

6. a. p > 1 q > 5 (2i > 2 j > 1<br />

5 (2<br />

b. p > 2 q > 5 i > 2<br />

5 (2i > 2 2j > 1)i > k > ) 1 (2i > 2<br />

1<br />

2 j > 1<br />

1<br />

5 (2<br />

5 3i > 1 1)i > k > 2k > (21 2 1)j > j > 1 k > )<br />

1 (1 1 1)k ><br />

) 2 (2i > 2<br />

1 0j > 1 (21<br />

1 0k > 1 1)j > j > 1 k > )<br />

1 (1 2 1)k ><br />

c. 2p > 2 5q > 5 2(2i ><br />

5 (4i > 2 j ><br />

2 2j > 1 k > )<br />

5 (4<br />

d. 22p > 5<br />

1<br />

5 (24i > 5q > 9i > 1 5)i > 1 2k > 2 5(2i > 2<br />

) 2 (25i > j > 1<br />

1 3j > 1 (22<br />

2<br />

1 2j > 522(2i > 3k > 1 5)j > 2 5j > k > )<br />

1 5k > )<br />

1 (2 2 5)k ><br />

5 (24<br />

529i > 2 5)i > 2 2k > 2 j > 1<br />

) 1 (25i > k > ) 1<br />

7. a. m > 2 3j > 1 (2<br />

2 n > 1 3k > 2 5)j > 2 5j > 5(2i ><br />

1 5k > 2 j > 1 k > )<br />

)<br />

1 (22 1 5)k ><br />

5 (2i ><br />

5 (2<br />

5<br />

0 m > 4i > 2<br />

2<br />

2<br />

b. m > n > j > (22))i > 2 k > )2<br />

2 3k > 1 (21)j > (22i > 1 j > 1 2k ><br />

1 (21 2 2)k > )<br />

0 5<br />

1 n > "4 2<br />

5 (2i > 1 (21) 2<br />

2 k > 1 (23) 2<br />

) 1 (22i ><br />

5 (2<br />

5 0i > 1<br />

1 j > (22))i ><br />

1 k > 1 j > 1 j > 5 "26<br />

1 2k > 8 5.10<br />

)<br />

1 (21 1 2)k ><br />

0 m > 1<br />

c. 2m > n > 0 5<br />

1 3n > "0 2 1<br />

5 2(2i > 1 2 1<br />

5 (4i > 2 k > 1 2 5 "2<br />

2 2k > ) 1 3(22i > 8 1.41<br />

) 1<br />

5 (4 1<br />

522i > (26))i > (26i > 1 j ><br />

1 3j > 1<br />

1 4k > 3j > 1 3j > 1 2k ><br />

1 6k > )<br />

)<br />

1 (22 1 6)k ><br />

0 2m > 1<br />

d. 25m > 3n > 0 5 "(22) 2<br />

525(2i > 2 k > 1 3 2 1<br />

) 5210i > 4 2 5<br />

1 5k > "29 8 5.39<br />

0 25m ><br />

8. x > 0 5<br />

1 x > 1 y > "(210) 2 1 (5) 2 5 "125 8 11.18<br />

2x > 2 y > 52i ><br />

5 3i > 1 2j ><br />

5 2i > 1 6j > 1 5k ><br />

1 8j > 2 7k ><br />

2 2k ><br />

Divide by 2 on both sides to get:<br />

x > 5 i > 1 4j > 2 k ><br />

Plug this equation into the first given equation:<br />

i > 1 4j > 2 k > 1 y > y > 52i ><br />

y > 52i > 1 2j ><br />

1 2j > 1 5k ><br />

1<br />

y > 5 (21<br />

522i > 2 1)i > 5k > 2 (i > 1<br />

9. a. The vectors OA > 2<br />

, OB > 2j > 1 (2<br />

1 6k > 2 4)j > 4j > 2 k > )<br />

1 (5 1 1)k ><br />

, and OC ><br />

represent the<br />

xy-plane, xz-plane, and yz-plane, respectively.<br />

They are also the vector from the origin to points<br />

(a, b, 0), (a, 0,c), and (0, b, c), respectively.<br />

b. OA ><br />

OB > 5 ai ><br />

OC > 5 ai > 1 bj ><br />

c. @OA > 5 0i > 1 0j > 1 0k ><br />

1 bj > 1 ck ><br />

1 ck ><br />

@ 5 "a 2 1 b 2<br />

@OB > @ 5 "a 2 1 c 2<br />

@OB > @ 5 "b 2 1 c 2<br />

6-22 <strong>Chapter</strong> 6: Introduction to Vectors


d. AB > AB > 5 (a, 0, c) 2 (a, b, 0) 5 (0, 2b, c)<br />

is a direction vector from A to B.<br />

10. a. @OA > @ 5 "(22) 2 1 (26) 2 1 3 2 5 "49 5 7<br />

b.<br />

c.<br />

d. @AB > @ 5 "5 2 1 2 2 1 9 2 5 "110 8 10.49<br />

e. BA > 5 OA > 2 OB ><br />

5 (22, 26, 3) 2 (3, 24, 12)<br />

5 (25, 22, 29)<br />

f.<br />

11.<br />

@OB > @<br />

AB > 5 "(3) 2<br />

5 OB > 1<br />

2 OA > (24) 2 1 12 2 5 "169 5 13<br />

5 (3, 24, 12) 2 (22, 26, 3)<br />

5 (3 2 (22), 24 2 (26), 12 2 3)<br />

5 (5, 2, 9)<br />

@BA > @ 5 "(25) 2 1 (22) 2 1 (29) 2<br />

5 "110 8 10.49<br />

z<br />

B(3, –1, 17)<br />

BC<br />

C(7, –3, 15)<br />

DC<br />

D(4, 1, 3)<br />

x<br />

AB<br />

A(0, 3, 5)<br />

AD<br />

In order to show that ABCD is a parallelogram, we<br />

must show that AB > 5 DC ><br />

or BC > 5 AD > . This will<br />

show they have the same direction, thus the opposite<br />

sides are parallel. By showing the vectors are<br />

equal they will have the same magnitude, implying<br />

the opposite sides having congruency.<br />

AB > 5 (3, 21, 17) 2 (0, 3, 5)<br />

5 (3 2 0, 21 2 3, 17 2 5)<br />

DC > 5 (3, 24, 12)<br />

5 (7, 23, 15) 2 (4, 1, 3)<br />

5 (7 2 4, 23 2 1, 15 2 3)<br />

5 (3,<br />

Thus AB > 24, 12)<br />

5 DC > . Do the calculations for the other<br />

pair as a check.<br />

BC > 5 (7, 23, 15) 2 (3, 21, 17)<br />

5 (7 2 3, 23 2 (21), 15 2 17)<br />

AD > 5 (4, 22, 22)<br />

5 (4, 1, 3) 2 (0, 3, 5)<br />

5 (4 2 0, 1 2 3, 3 2 5)<br />

5<br />

So BC > (4, 22,<br />

5 AD > 22)<br />

.<br />

We have shown AB > 5 DC ><br />

and BC > 5 AD > , so<br />

ABCD is a parallelogram.<br />

y<br />

12. 2x > 1 y > 2 2z ><br />

5 2(21, b, c) 1 (a, 22, c) 2 2(2a, 6, c)<br />

5 (22, 2b, 2c) 1 (a, 22, c) 2 (22a, 12, 2c)<br />

5 (22 1 a 1 2a, 2b 2 2 2 12, 2c 1 c 2 2c)<br />

5 (22 1 3a, 2b 2 14, c)<br />

5 (0, 0, 0)<br />

22 1 3a 5 0;<br />

3a 5 2; a 5 2 3<br />

2b 5 14; b 5 7<br />

c 5 0<br />

13. a.<br />

x<br />

2b 2 14 5 0; c 5 0<br />

z<br />

OB<br />

O<br />

OA<br />

OC<br />

b. V 1 5 (0, 0, 0), the origin<br />

V 2 5 end point of OA ><br />

V 3 5 end point of OB > 5 (22, 2, 5)<br />

V 4 5 end<br />

V 5 5 OA > point of<br />

1 OB > OC > 5 (0, 4, 1)<br />

5 (0, 5, 21)<br />

5 (22, 2, 5) 1 (0, 4, 1)<br />

5 (22 1 0, 2 1 4, 5 1 1)<br />

V 6 5 OA > 1 OC > 5 (22, 6, 6)<br />

5 (22, 2, 5) 1 (0, 5, 21)<br />

5 (22 1 0, 2 1 5, 5 2 1)<br />

V 7 5 OB > 1 OC > 5 (22, 7, 4)<br />

5 (0, 4, 1) 1 (0, 5, 21)<br />

5 (0 1 0, 4 1 5, 1 2 1)<br />

V 8 5 OA > 1 OB > 5 (0,<br />

1 OC > 9, 0)<br />

5 (22, 2, 5) 1 (0, 9, 0) (by V 7 )<br />

5 (22 1 0, 2 1 9, 5 1 0)<br />

5 (22, 11, 5)<br />

14. Any point on the x-axis has y-coordinate 0 and<br />

z-coordinate 0. The z-coordinate of each of A and B<br />

is 3, so the z-component of the distance from the<br />

desired point is the same for each of A and B. The<br />

y-component of the distance from the desired point<br />

will be 1 for each of A and B, 12 5 (21) 2 . So, the<br />

x-coordinate of the desired point has to be halfway<br />

between the x-coordinates of A and B. The desired<br />

point is (1, 0, 0).<br />

A<br />

B<br />

C<br />

y<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-23


15.<br />

a<br />

A<br />

a – b<br />

b<br />

To solve this problem, we must first consider the<br />

triangle formed by a > , b > , and a > 1 b > . We will use<br />

their magnitudes to solve for angle A, which will be<br />

1<br />

used to solve for in the triangle formed by<br />

1<br />

1<br />

b > , and 2 a > 2 b > 2 a > 1 b > 2 a > 2 b ><br />

,<br />

.<br />

Using the cosine law, we see that:<br />

cos (A) 5 @b> @ 2 1 @a > 1 b > @ 2 2 @a > 2<br />

@<br />

2 @b > @@a > 1 b > @<br />

25 1 49 2 9<br />

5<br />

5 13<br />

70<br />

14<br />

Now, consider the triangle formed by ,<br />

1<br />

and 2 a > 2 b > 1<br />

b > , 2 a > 1 b ><br />

. Using the cosine law again:<br />

cos (A) 5 @b> @ 2 1 ( 1 2 @a> 1 b > @ ) 2 2 ( 1 2 @a> 2 b > @ ) 2<br />

@b > @@a > 1 b > @<br />

149<br />

13<br />

14 5 4 2 (1 2 @a> 2 b > @) 2<br />

35<br />

@a > 2 b > @ 2 524a 65<br />

@a > 2 2 149<br />

4 b<br />

@a > 2 b > @ 2 5 19<br />

2 b > @ 5 "19 or 4.36<br />

6.8 Linear Combinations and Spanning<br />

Sets, pp. 340–341<br />

1. They are collinear, thus a linear combination is<br />

not applicable.<br />

2. It is not possible to use 0 ><br />

in a spanning set.<br />

Therefore, the remaining vectors only span R 2 .<br />

3. The set of vectors spanned by (0, 1) is m(0, 1).<br />

If we let m 521, then m(0, 1) 5 (0, 21).<br />

4. i > spans the set m(1, 0, 0). This is any vector<br />

along the x-axis. Examples: (2, 0, 0), (221, 0,<br />

5. As in question 2, it is not possible to use 0 > 0)<br />

in a<br />

spanning set.<br />

b<br />

a + b<br />

C<br />

B<br />

a<br />

6. 5(21, 2), (21, 1)6, 5(2, 24), (21, 1)6,<br />

5(21, 1), (23, 6)6 are all the possible spanning sets<br />

for R 2 with 2 vectors.<br />

7. a.<br />

5 4i > 2(2a ><br />

5 6i > 2 8j > 2 3b ><br />

2<br />

2<br />

4(2a > 20j > 6j > 1 c > )<br />

1<br />

524i > 1 b > 1 22k > 18k > 5 4a ><br />

1 2i > 2 6b ><br />

2 6j > 1 2c ><br />

1 4k<br />

2<br />

528i > 1 8j > c > )<br />

1<br />

(a > 2 c > 1 24j > 4j > 524a ><br />

2<br />

) 5 i > 2 20k > 12k > 1 4b ><br />

2 4i > 2 4c ><br />

1 12j > 2 8k<br />

2(2a > 5<br />

5 6i > 2 3b > j > 2 2j ><br />

2<br />

2<br />

5 14i > 20j > 1 c > 2k > 2 i > 1 3j > 2 2k ><br />

)24(2a ><br />

1<br />

2 43j > 22k > 1<br />

1 40k > 8i > 1 b ><br />

2 24j > 2 c > ) 1<br />

1 20k > (a ><br />

1 j > 2 c > )<br />

2 2k ><br />

1<br />

b.<br />

2 (2a> 2 4b > 2 8c > ) 5 a > 2 2b > 2 4c ><br />

5 i > 2<br />

523i > 2j > 2<br />

1 8j > 2j > 1 6k ><br />

2 2k > 2 4i > 1 12j > 2 8k ><br />

1<br />

3 (3a> 2 6b > 1 9c > ) 5 a > 2 2b > 1 3c ><br />

5 i > 2<br />

5 4i > 2j > 2<br />

2 15j > 2j > 1 6k ><br />

1 12k > 1 3i > 2 9j > 1 6k ><br />

1<br />

2 (2a> 2 4b > 2 8c > )2 1<br />

523i ><br />

527i > 1 8j > 2<br />

1 23j > 2k > 3 (3a> 2 6b > 1 9c > )<br />

2<br />

2 14k > 4i > 1 15j > 2 12k ><br />

8. 5(1, 0, 0), (0, 1, 0)6:<br />

(21, 2, 0) 521(1, 0, 0) 1 2(0, 1, 0)<br />

(3, 4, 0) 5 3(1, 0, 0) 1 4(0, 1, 0)<br />

5(1, 1, 0), (0, 1, 0)6<br />

(21, 2, 0) 521(1, 1, 0) 1 3(0, 1, 0)<br />

(3, 4, 0) 5 3(1, 1, 0) 1 (0, 1, 0)<br />

9. a. It is the set of vectors in the xy-plane.<br />

b. (22, 4, 0) 522(1, 0, 0) 1 4(0, 1, 0)<br />

c. By part a. the vector is not in the xy-plane.<br />

There is no combination that would produce a<br />

number other than 0 for the z-component.<br />

d. It would still only span the xy-plane. There<br />

would be no need for that vector.<br />

10. Looking at the x-component:<br />

2a 1 3c 5 5<br />

The y-component:<br />

6 1 21 5 b 1 c<br />

The z-component:<br />

2c 1 3c 5 15<br />

5c 5 15<br />

c 5 3<br />

6-24 <strong>Chapter</strong> 6: Introduction to Vectors


15.<br />

a<br />

A<br />

a – b<br />

b<br />

To solve this problem, we must first consider the<br />

triangle formed by a > , b > , and a > 1 b > . We will use<br />

their magnitudes to solve for angle A, which will be<br />

1<br />

used to solve for in the triangle formed by<br />

1<br />

1<br />

b > , and 2 a > 2 b > 2 a > 1 b > 2 a > 2 b ><br />

,<br />

.<br />

Using the cosine law, we see that:<br />

cos (A) 5 @b> @ 2 1 @a > 1 b > @ 2 2 @a > 2<br />

@<br />

2 @b > @@a > 1 b > @<br />

25 1 49 2 9<br />

5<br />

5 13<br />

70<br />

14<br />

Now, consider the triangle formed by ,<br />

1<br />

and 2 a > 2 b > 1<br />

b > , 2 a > 1 b ><br />

. Using the cosine law again:<br />

cos (A) 5 @b> @ 2 1 ( 1 2 @a> 1 b > @ ) 2 2 ( 1 2 @a> 2 b > @ ) 2<br />

@b > @@a > 1 b > @<br />

149<br />

13<br />

14 5 4 2 (1 2 @a> 2 b > @) 2<br />

35<br />

@a > 2 b > @ 2 524a 65<br />

@a > 2 2 149<br />

4 b<br />

@a > 2 b > @ 2 5 19<br />

2 b > @ 5 "19 or 4.36<br />

6.8 Linear Combinations and Spanning<br />

Sets, pp. 340–341<br />

1. They are collinear, thus a linear combination is<br />

not applicable.<br />

2. It is not possible to use 0 ><br />

in a spanning set.<br />

Therefore, the remaining vectors only span R 2 .<br />

3. The set of vectors spanned by (0, 1) is m(0, 1).<br />

If we let m 521, then m(0, 1) 5 (0, 21).<br />

4. i > spans the set m(1, 0, 0). This is any vector<br />

along the x-axis. Examples: (2, 0, 0), (221, 0,<br />

5. As in question 2, it is not possible to use 0 > 0)<br />

in a<br />

spanning set.<br />

b<br />

a + b<br />

C<br />

B<br />

a<br />

6. 5(21, 2), (21, 1)6, 5(2, 24), (21, 1)6,<br />

5(21, 1), (23, 6)6 are all the possible spanning sets<br />

for R 2 with 2 vectors.<br />

7. a.<br />

5 4i > 2(2a ><br />

5 6i > 2 8j > 2 3b ><br />

2<br />

2<br />

4(2a > 20j > 6j > 1 c > )<br />

1<br />

524i > 1 b > 1 22k > 18k > 5 4a ><br />

1 2i > 2 6b ><br />

2 6j > 1 2c ><br />

1 4k<br />

2<br />

528i > 1 8j > c > )<br />

1<br />

(a > 2 c > 1 24j > 4j > 524a ><br />

2<br />

) 5 i > 2 20k > 12k > 1 4b ><br />

2 4i > 2 4c ><br />

1 12j > 2 8k<br />

2(2a > 5<br />

5 6i > 2 3b > j > 2 2j ><br />

2<br />

2<br />

5 14i > 20j > 1 c > 2k > 2 i > 1 3j > 2 2k ><br />

)24(2a ><br />

1<br />

2 43j > 22k > 1<br />

1 40k > 8i > 1 b ><br />

2 24j > 2 c > ) 1<br />

1 20k > (a ><br />

1 j > 2 c > )<br />

2 2k ><br />

1<br />

b.<br />

2 (2a> 2 4b > 2 8c > ) 5 a > 2 2b > 2 4c ><br />

5 i > 2<br />

523i > 2j > 2<br />

1 8j > 2j > 1 6k ><br />

2 2k > 2 4i > 1 12j > 2 8k ><br />

1<br />

3 (3a> 2 6b > 1 9c > ) 5 a > 2 2b > 1 3c ><br />

5 i > 2<br />

5 4i > 2j > 2<br />

2 15j > 2j > 1 6k ><br />

1 12k > 1 3i > 2 9j > 1 6k ><br />

1<br />

2 (2a> 2 4b > 2 8c > )2 1<br />

523i ><br />

527i > 1 8j > 2<br />

1 23j > 2k > 3 (3a> 2 6b > 1 9c > )<br />

2<br />

2 14k > 4i > 1 15j > 2 12k ><br />

8. 5(1, 0, 0), (0, 1, 0)6:<br />

(21, 2, 0) 521(1, 0, 0) 1 2(0, 1, 0)<br />

(3, 4, 0) 5 3(1, 0, 0) 1 4(0, 1, 0)<br />

5(1, 1, 0), (0, 1, 0)6<br />

(21, 2, 0) 521(1, 1, 0) 1 3(0, 1, 0)<br />

(3, 4, 0) 5 3(1, 1, 0) 1 (0, 1, 0)<br />

9. a. It is the set of vectors in the xy-plane.<br />

b. (22, 4, 0) 522(1, 0, 0) 1 4(0, 1, 0)<br />

c. By part a. the vector is not in the xy-plane.<br />

There is no combination that would produce a<br />

number other than 0 for the z-component.<br />

d. It would still only span the xy-plane. There<br />

would be no need for that vector.<br />

10. Looking at the x-component:<br />

2a 1 3c 5 5<br />

The y-component:<br />

6 1 21 5 b 1 c<br />

The z-component:<br />

2c 1 3c 5 15<br />

5c 5 15<br />

c 5 3<br />

6-24 <strong>Chapter</strong> 6: Introduction to Vectors


Substituting this into the first and second equation:<br />

2a 1 9 5 5<br />

a 522<br />

27 5 b 1 3<br />

b 5 24<br />

11. (210, 234) 5 a(21, 3) 1 b(1, 5)<br />

Looking at the x-component:<br />

210 52a 1 b a 5 10 1 b<br />

Looking at the y-component:<br />

234 5 3a 1 5b<br />

Substituting in a:<br />

234 5 30 1 3b 1 5b<br />

b 528<br />

Substituting b into x-component equation:<br />

210 52a 1 (28)<br />

a 522<br />

(210, 234) 5 22(21, 3) 2 8(1, 5)<br />

12. a. a(2, 21) 1 b(21, 1) 5 (x, y)<br />

x 5 2a 2 b<br />

b 5 2a 2 x<br />

y 52a 1 b<br />

Substitute in b:<br />

y 52a 1 2a 2 x<br />

a 5 x 1 y<br />

Substitute this back into the first equation:<br />

b 5 2x 1 2y 2 x<br />

b 5 x 1 2y<br />

b. Using the formulas in part a:<br />

For (2, 23):<br />

a 5 x 1 y 5 2 2 3 521<br />

b 5 x 1 2y 5 2 2 6 524<br />

(2, 23) 521(2, 21) 2 4(21, 1)<br />

For (124, 25):<br />

a 5 124 2 5 5 119<br />

b 5 124 2 10 5 114<br />

(124, 25) 5 119(2, 21) 1 114(21, 1)<br />

For (4, 211)<br />

a 5 4 2 11 527<br />

b 5 4 2 22 5218<br />

(4, 211) 527(2, 21) 2 18(21, 1)<br />

13. Try: a(21, 2, 3) 1 b(4, 1, 22)<br />

5 (214, 21, 16)<br />

x components:<br />

2a 1 4b 5214<br />

a 5 14 1 4b<br />

y components:<br />

2a 1 b 521<br />

Substitute in a:<br />

28 1 8b 1 b 521<br />

b 523<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

Substitute this result into the x-components:<br />

a 5 14 2 3 5 11<br />

Check by substituting into z-components:<br />

3a 2 2b 5 16<br />

33 1 5 5 16<br />

Therefore:<br />

a(21, 2, 3) 1 b(4, 1, 22) 2 (214, 21, 16) for<br />

any a and b. They do not lie on the same plane.<br />

b. a(21, 3, 4) 1 b(0, 21, 1) 5 (23, 14, 7)<br />

x components:<br />

2a 523<br />

a 5 3<br />

y components:<br />

3a 2 b 5 14<br />

Substitute in a:<br />

9 2 b 5 14<br />

b 525<br />

Check with z components:<br />

4a 1 b 5 7<br />

12 2 5 5 7<br />

Since there exists an a and b to form a linear<br />

combination of 2 of the vectors to form the third,<br />

they lie on the same plane.<br />

3(21, 3, 4) 2 5(0,<br />

14. Let vector a > 21, 1) 5 (23, 14,<br />

and b > 7)<br />

5 (21, 3, 4) 5 (22, 3, 21)<br />

(vectors from the origin to points A and B,<br />

respectively). To determine x, we let c > (vector from<br />

origin to C) be a linear combination of a > and b > .<br />

a(21, 3, 4) 1 b(22, 3, 21) 5 (25, 6, x)<br />

x components:<br />

2a 2 2b 525<br />

a 5 5 2 2b<br />

y components:<br />

3a 1 3b 5 6<br />

Substitute in a:<br />

15 2 6b 1 3b 5 6<br />

b 5 3<br />

Substitute in b in x component equation:<br />

a 5 5 2 6 521<br />

z components:<br />

4a 2 b 5 x<br />

Substitute in a and b:<br />

x 524 2 3 527<br />

15. m 5 2, n 5 3. Non-parallel vectors cannot be<br />

equal, unless their magnitudes equal 0.<br />

16. Answers may vary. For example:<br />

Try linear combinations of the 2 vectors such that<br />

6-25


the z component equals 5. Then calculate what p<br />

and q would equal.<br />

21(4, 1, 7) 1 2(21, 1, 6) 5 (26, 1, 5)<br />

So p 526 and q 5 1<br />

5(4, 1, 5) 2 5(21, 1, 6) 5 (25, 0, 5)<br />

So p 5 25 and q 5 0<br />

(4, 1, 7) 2 1 (21, 1, 6) 5 a13<br />

3 3 , 2 3 , 5b<br />

So p 5 13 and q 5 2 3 3<br />

17. As in question 15, non-parallel vectors.<br />

Their magnitudes must be 0 again to make the<br />

equality true.<br />

m 2 1 2m 2 3 5 (m 2 1)(m 1 3)<br />

m 5 1, 23<br />

m 2 1 m 2 6 5 (m 2 2)(m 1 3)<br />

m 5 2, 23<br />

So, when m 523, their sum will be 0.<br />

Review Exercise, pp. 344–347<br />

1. a. false; Let then:<br />

@a > 1 b > @ 5 0 a > b > 52a ><br />

1 (2a > 2 0<br />

) 0<br />

5 0 0 0<br />

5<br />

b. true; @a > 0 , 0 a > 0<br />

1 b > @ and 0 a > 1 c > 0 both represent the<br />

lengths of the diagonal of a parallelogram, the first<br />

with sides and and the second with sides and<br />

c > a > b ><br />

; since both parallelograms have as a side and<br />

diagonals of equal length .<br />

c. true; Subtracting from both sides shows that<br />

b > 5 c > a > @b > @ 5 0 c > a > a ><br />

0<br />

d. true; Draw the parallelogram formed by and<br />

SW > . FW ><br />

and RS ><br />

RF ><br />

are the opposite sides of a parallelogram<br />

and must be equal.<br />

e. true; The distributive law for scalars<br />

f. false; Let and let Then,<br />

0 a > 0 5 and<br />

but @a > 0 2a > b > 52a ><br />

0 5 @b > @<br />

1 b > @ 5 0 a > 0 c > c ><br />

0<br />

1 (2 a > 5 @d > 5 d > 2 0.<br />

@<br />

)0 5 0<br />

@c > 1 d > @ 5 0 c > 1 c > 0 5 0 2c > 0<br />

so @a > 1 b > @ 2 @c > 1 d > @<br />

2. a. Substitute the given values of and into<br />

the expression<br />

2x > 2 3y ><br />

5 2(2a > 1 5z > 2x> 2 3y > 1 5z > x > , y > , z ><br />

2<br />

1 5(2a > 3b > 2<br />

2 3b > 4c > ) 2<br />

1 5c > 3 (22a > 1 3b > 1 3c > )<br />

)<br />

5 4a > 2<br />

1<br />

5 4a > 25c > 6b > 2 8c > 1 6a > 2 9b > 2 9c > 1 10a > 2 15b ><br />

1<br />

1 25c > 6a > 1 10a > 2 6b > 2 9b ><br />

2 15b > 2 8c > 2 9c ><br />

5 20a > 2 30b > 1 8c ><br />

b. Simplify the expression before substituting the<br />

given values of and<br />

3(22x > 2 4y > x > ,<br />

2 2(24x > 1 z > y > , z ><br />

5 26x > 2<br />

2<br />

1 10y > 12y > 5y > ) 2<br />

1<br />

5 26x > 2 2z > 1 3z > z > (2x > 2 y > 1 z > )<br />

)<br />

2 2x > 1 y > 2 z > 1 8x ><br />

2<br />

5 0x > z > 2 2x ><br />

2<br />

5 2y > 2 y > 2z > 1 8x > 2 12y > 1 y > 1 10y > 1 3z ><br />

1 0z ><br />

5 2a > 2<br />

3. a. XY > 3b > 2 3c ><br />

5 OY > 2 OX ><br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (24 2 (22), 4 2 1, 8 2 2)<br />

5 (22, 3, 6)<br />

@XY > @ 5 "(x 2 2 x 1 ) 2 1 (y 2 2 y 1 ) 2 1 (z 2 2 z 1 ) 2<br />

5 "(22) 2 1 (3) 2 1 (6) 2<br />

5 "4 1 9 1 36<br />

5 "49<br />

5 7<br />

b. The components of a unit vector in the same<br />

1<br />

direction as XY ><br />

are 7(22, 3, 6) 5 (2 2 7, 3 7, 6 7) .<br />

4. a. The position vector is equivalent to<br />

OP > 5 YX > OP ><br />

YX > .<br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (21 2 5, 2 2 5, 6 2 12)<br />

5 (26, 23, 26)<br />

b. @YX > @ 5 "(26) 2 1 (23) 2 1 (26) 2<br />

5 "81<br />

5 9<br />

The components of a unit vector in the same<br />

1<br />

direction as are<br />

5. 2MN > YX ><br />

5 NM ><br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (2 28, 3 2 1, 5 2 2)<br />

5 (26, 2, 3)<br />

9(26, 23, 26) 5 (2 2 3, 2 1 3, 2 2 3)<br />

6-26 <strong>Chapter</strong> 6: Introduction to Vectors


the z component equals 5. Then calculate what p<br />

and q would equal.<br />

21(4, 1, 7) 1 2(21, 1, 6) 5 (26, 1, 5)<br />

So p 526 and q 5 1<br />

5(4, 1, 5) 2 5(21, 1, 6) 5 (25, 0, 5)<br />

So p 5 25 and q 5 0<br />

(4, 1, 7) 2 1 (21, 1, 6) 5 a13<br />

3 3 , 2 3 , 5b<br />

So p 5 13 and q 5 2 3 3<br />

17. As in question 15, non-parallel vectors.<br />

Their magnitudes must be 0 again to make the<br />

equality true.<br />

m 2 1 2m 2 3 5 (m 2 1)(m 1 3)<br />

m 5 1, 23<br />

m 2 1 m 2 6 5 (m 2 2)(m 1 3)<br />

m 5 2, 23<br />

So, when m 523, their sum will be 0.<br />

Review Exercise, pp. 344–347<br />

1. a. false; Let then:<br />

@a > 1 b > @ 5 0 a > b > 52a ><br />

1 (2a > 2 0<br />

) 0<br />

5 0 0 0<br />

5<br />

b. true; @a > 0 , 0 a > 0<br />

1 b > @ and 0 a > 1 c > 0 both represent the<br />

lengths of the diagonal of a parallelogram, the first<br />

with sides and and the second with sides and<br />

c > a > b ><br />

; since both parallelograms have as a side and<br />

diagonals of equal length .<br />

c. true; Subtracting from both sides shows that<br />

b > 5 c > a > @b > @ 5 0 c > a > a ><br />

0<br />

d. true; Draw the parallelogram formed by and<br />

SW > . FW ><br />

and RS ><br />

RF ><br />

are the opposite sides of a parallelogram<br />

and must be equal.<br />

e. true; The distributive law for scalars<br />

f. false; Let and let Then,<br />

0 a > 0 5 and<br />

but @a > 0 2a > b > 52a ><br />

0 5 @b > @<br />

1 b > @ 5 0 a > 0 c > c ><br />

0<br />

1 (2 a > 5 @d > 5 d > 2 0.<br />

@<br />

)0 5 0<br />

@c > 1 d > @ 5 0 c > 1 c > 0 5 0 2c > 0<br />

so @a > 1 b > @ 2 @c > 1 d > @<br />

2. a. Substitute the given values of and into<br />

the expression<br />

2x > 2 3y ><br />

5 2(2a > 1 5z > 2x> 2 3y > 1 5z > x > , y > , z ><br />

2<br />

1 5(2a > 3b > 2<br />

2 3b > 4c > ) 2<br />

1 5c > 3 (22a > 1 3b > 1 3c > )<br />

)<br />

5 4a > 2<br />

1<br />

5 4a > 25c > 6b > 2 8c > 1 6a > 2 9b > 2 9c > 1 10a > 2 15b ><br />

1<br />

1 25c > 6a > 1 10a > 2 6b > 2 9b ><br />

2 15b > 2 8c > 2 9c ><br />

5 20a > 2 30b > 1 8c ><br />

b. Simplify the expression before substituting the<br />

given values of and<br />

3(22x > 2 4y > x > ,<br />

2 2(24x > 1 z > y > , z ><br />

5 26x > 2<br />

2<br />

1 10y > 12y > 5y > ) 2<br />

1<br />

5 26x > 2 2z > 1 3z > z > (2x > 2 y > 1 z > )<br />

)<br />

2 2x > 1 y > 2 z > 1 8x ><br />

2<br />

5 0x > z > 2 2x ><br />

2<br />

5 2y > 2 y > 2z > 1 8x > 2 12y > 1 y > 1 10y > 1 3z ><br />

1 0z ><br />

5 2a > 2<br />

3. a. XY > 3b > 2 3c ><br />

5 OY > 2 OX ><br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (24 2 (22), 4 2 1, 8 2 2)<br />

5 (22, 3, 6)<br />

@XY > @ 5 "(x 2 2 x 1 ) 2 1 (y 2 2 y 1 ) 2 1 (z 2 2 z 1 ) 2<br />

5 "(22) 2 1 (3) 2 1 (6) 2<br />

5 "4 1 9 1 36<br />

5 "49<br />

5 7<br />

b. The components of a unit vector in the same<br />

1<br />

direction as XY ><br />

are 7(22, 3, 6) 5 (2 2 7, 3 7, 6 7) .<br />

4. a. The position vector is equivalent to<br />

OP > 5 YX > OP ><br />

YX > .<br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (21 2 5, 2 2 5, 6 2 12)<br />

5 (26, 23, 26)<br />

b. @YX > @ 5 "(26) 2 1 (23) 2 1 (26) 2<br />

5 "81<br />

5 9<br />

The components of a unit vector in the same<br />

1<br />

direction as are<br />

5. 2MN > YX ><br />

5 NM ><br />

5 (x 2 , y 2 , z 2 ) 2 (x 1 , y 1 , z 1 )<br />

5 (x 2 2 x 1 , y 2 2 y 1 , z 2 2 z 1 )<br />

5 (2 28, 3 2 1, 5 2 2)<br />

5 (26, 2, 3)<br />

9(26, 23, 26) 5 (2 2 3, 2 1 3, 2 2 3)<br />

6-26 <strong>Chapter</strong> 6: Introduction to Vectors


@NM > @ 5 "(26) 2 1 (2) 2 1 (3) 2 6-27<br />

5 "49<br />

5 7<br />

The components of the unit vector with the opposite<br />

1<br />

direction to MN ><br />

are 7(26, 2, 3) 5 (2 6 7, 2 7, 3 7)<br />

6. a. The two diagonals can be found by calculating<br />

OA > 1 OB > and OA > 2 OB > .<br />

O<br />

OA > 1 OB > 5 (3, 2, 26) 1 (26, 6, 22)<br />

5 (3 126, 2 1 6, 26 122)<br />

OA > 2 OB > 5 (23, 8, 28)<br />

5 (3, 2, 26) 1 (26, 6, 22)<br />

5 (3 2 (26), 2 2 6, 26 2 (22))<br />

5 (9, 24, 24)<br />

b. To determine the angle between the sides of the<br />

parallelogram, calculate<br />

and<br />

@OA > 2 OB > @OA > @ , @OB > @ ,<br />

@ and apply<br />

the cosine law.<br />

@OA > @ 5 "(3) 2 1 (2) 2 1 (26) 2<br />

@OB > @ 5 "(26) 2 1 (6) 2 1 (22) 2<br />

@OA > 2 OB > @ 5 "(9) 2 1 (24) 2 1 (24) 2<br />

cos u5 @OA> @ 2 1 @OB > @ 2 2 @OA > 2 OB > @<br />

2<br />

2 @OA > @@OB > @<br />

cos u5 (7)2 1 (2"19) 2 2 ("113) 2<br />

2(7)(2"19)<br />

cos u 8 0.098<br />

u 8 84.4°<br />

7. a.<br />

5 "49<br />

5 7<br />

5 "76<br />

5 2"19<br />

A<br />

5 "113<br />

@AB > @ 5 "(2 2 (21)) 2 1 (0 2 1) 2 1 (3 2 1) 2<br />

5 "(3) 2 1 (21) 2 1 (2) 2<br />

5 "9 1 1 1 4<br />

5 "14<br />

OA – OB<br />

B<br />

OA + OB<br />

@BC > @ 5 "(3 2 2) 2 1 (3 2 0) 2 1 (24 2 3) 2<br />

5 "(1) 2 1 (3) 2 1 (27) 2<br />

5 "1 1 9 1 49<br />

5 "59<br />

0 CA > 0 5 "(21 2 3) 2 1 (1 2 3) 2 1 (1 2 (24)) 2<br />

5 "(24) 2 1 (22) 2 1 (5) 2<br />

5 "16 1 4 1 25<br />

5 "45<br />

Triangle ABC is a right triangle if and only if<br />

@AB > @ 2 1 @CA > @ 2 5 @BC > @ 2 .<br />

@AB > @ 2 1 @CA > @ 2 5 ("14) 2 1 ("45) 2<br />

5 14 1 45<br />

@BC > 5 59<br />

@ 2 5 ("59) 2<br />

5 59<br />

So triangle ABC is a right triangle.<br />

b. Area of a triangle For triangle ABC the<br />

longest side BC > 5 1 2bh.<br />

is the hypotenuse, so AB ><br />

and CA ><br />

are the base and height of the triangle.<br />

Area 5 1 2 ( 0 AB> 0 )(0 CA > 0 )<br />

5 1 2 "14"45<br />

5 1 2 "630<br />

5 3 or 12.5<br />

2 "70<br />

c. Perimeter of a triangle equals the sum of the sides.<br />

Perimeter 5 @AB > @ 1 @BC > @ 1 @CA > @<br />

5 "14 1 "59 1 "45<br />

8 18.13<br />

d. The fourth vertex D is the head of the diagonal<br />

vector from A. To find take .<br />

AB > AD ><br />

AB > 1 AC ><br />

5 (2 2 (21), 0 2 1, 3 2 1) 5 (3, 21, 2)<br />

AC > 5 (3 2 (21), 3 2 1, 24 2 1) 5 (4, 2, 25)<br />

AD > 5 AB > 1 AC ><br />

5 (3 1 4, 21 1 2, 2 1 (25))<br />

5 (7, 1, 23)<br />

So the fourth vertex is D(21 1 7, 1 1 1, 1 1 (23))<br />

or D(6, 2, 22).<br />

Calculus and Vectors <strong>Solutions</strong> Manual


8. a.<br />

b<br />

a – b<br />

a – b + c<br />

a<br />

c<br />

b. Since the vectors and are perpendicular,<br />

0 a > 1 b > 0 2 5 0 a > a ><br />

0 2 1 0 b > b ><br />

0 2 . So,<br />

0 a > 1 b > 0 2 5 (4) 2 1 (3) 2<br />

5 16 1 9<br />

0 a > 1 b > 5 25<br />

0 5 "25 5 5<br />

9. Express r > as a linear combination of p > and q > :<br />

Solve for a and b:<br />

r > 5 ap > 1 bq ><br />

( 21, 2) 5 a(211, 7) 1 b(23, 1)<br />

( 21, 2) 5 (211a, 7a) 1 (23b, b)<br />

( 21, 2) 5 (211a 2 3b, 7a 1 b)<br />

Solve the system of equations:<br />

21 5211a 2 3b<br />

2 5 7a 1 b<br />

Use the method of elimination:<br />

3(2) 5 3(7a 1 b)<br />

6 5 21a 1 3b<br />

1 21 5211a 2 3b<br />

5 5 10a<br />

a – b<br />

1<br />

2 5 a<br />

By substitution, b 52 3 2<br />

1<br />

Therefore<br />

Express q > 2(211, 7) 12 3 2(23, 1) 5 (21,<br />

as a linear combination of p > 2)<br />

and r > .<br />

Solve for a and b:<br />

q > 5 ap > 1 br ><br />

( 23, 1) 5 a(211, 7) 1 b(21, 2)<br />

( 23, 1) 5 (211a, 7a) 1 (2b, 2b)<br />

( 23, 1) 5 (211a 2 b, 7a 1 2b)<br />

Solve the system of equations:<br />

23 5211a 2 b<br />

1 5 7a 1 2b<br />

Use the method of elimination:<br />

2(23) 5 2(211a 2 b)<br />

26 5222a 2 2b<br />

1 1 5 7a 1 2b<br />

25 5215a<br />

1<br />

3 5 a<br />

By substitution, 2 2 3 5 b<br />

1<br />

Therefore<br />

Express p > 3(211, 7) 1 2 2 3(21, 2) 5 (23,<br />

as a linear combination of q > 1)<br />

and r > .<br />

Solve for a and b:<br />

p > 5 aq > 1 br ><br />

( 211, 7) 5 a(23, 1) 1 b(21, 2)<br />

( 211, 7) 5 (23a, a) 1 (2b, 2b)<br />

( 211, 7) 5 (23a 2 b, a 1 2b)<br />

Solve the system of equations:<br />

211 523a 2 b<br />

7 5 a 1 2b<br />

Use the method of elimination:<br />

2(211) 5 2(23a 2 b)<br />

222 526a 2 2b<br />

1 7 5 a 1 2b<br />

215 525a<br />

3 5 a<br />

By substitution, 2 5 b<br />

Therefore 3(23, 1) 1 2(21, 2) 5 (211, 7)<br />

10. a. Let P(x, y, z) be a point equidistant from A<br />

and B. Then @PA > @ 5 @PB > @ .<br />

(x 2 2) 2 1 (y 2 (21)) 2 1 (z 2 3) 2<br />

5 (x 2 1) 2 1 (y 2 2) 2 1 (z 2 (23)) 2<br />

x 2 2 4x 1 4 1 y 2 1 2y 1 1 1 z 2 2 6z 1 9<br />

5 x 2 2 2x 1 1 1 y 2 2 4y 1 4 1 z 2 1 6z 1 9<br />

22x 1 6y 2 12z 5 0<br />

x 2 3y 1 6z 5 0<br />

b. (0, 0, 0) and (1, 1 3, 0) clearly satisfy the equation<br />

and are equidistant from A and B.<br />

11. a.<br />

(224, 3, 25) 5 2(a, b, 4) 1 1 (6, 8, c) 2 3(7, c, 24)<br />

2<br />

(224, 3, 25) 5 (2a, 2b, 8) 1 a3, 4, c 2 b<br />

(224, 3, 25) 5 a2a 2 18, 2b 1 4 2 3c, c 2 1 20b<br />

Solve the equations:<br />

i. 224 5 2a 2 18<br />

26 5 2a<br />

23 5 a<br />

ii. 25 5 c 2 1 20<br />

5 5 c 2<br />

10 5 c<br />

iii. 3 5 2b 1 4 2 3c<br />

3 5 2b 1 4 2 3(18)<br />

3 5 2b 2 50<br />

53 5 2b<br />

26.5 5 b<br />

2 (21, 3c, 212)<br />

6-28 <strong>Chapter</strong> 6: Introduction to Vectors


. (3, 222, 54)<br />

5 2aa, a, 1 2 ab 1 (3b, 0, 25c) 1 2ac, 3 c, 0b<br />

2<br />

(3, 222, 54)<br />

5 (2a, 2a, a) 1 (3b, 0, 25c) 1 (2c, 3c, 0)<br />

(3, 222, 54) 5 (2a 1 3b 1 2c, 2a 1 3c, a 2 5c)<br />

Solve the system of equations:<br />

222 5 2a 1 3c<br />

54 5 a 2 5c<br />

Use the method of elimination:<br />

22(54) 522(a 2 5c)<br />

2108 522a 1 10c<br />

1222 5 2a 1 3c<br />

2130 5 13c<br />

210 5 c<br />

By substitution, 8 5 a<br />

Solve the equation:<br />

3 5 2a 1 3b 1 2c<br />

3 5 2(8) 1 3b 1 2(210)<br />

3 5 16 1 3b 2 20<br />

3 5 3b 2 4<br />

7 5 3b<br />

7<br />

3 5 b<br />

12. a. Find @AB > @ , @BC > @ ,<br />

Test<br />

5 "11<br />

5 "21<br />

5 "(3) 2 1 (21) 2<br />

5 "10<br />

@AB > @ ,<br />

@BC > @ ,<br />

@CA > @<br />

@CA > @<br />

@AB > @ 5 "(2 2 1) 2 1 (2 2 (21)) 2 1 (2 2 1) 2<br />

5 "(1) 2 1 (3) 2 1 (1) 2<br />

@BC > @ 5 "(4 2 2) 2 1 (22 2 2) 2 1 (1 2 2) 2<br />

5 "(2) 2 1 (24) 2 1 (21) 2<br />

@CA > @ 5 "(4 2 1) 2 1 (22 2 (21)) 2 1 (1 2 1) 2<br />

in the Pythagorean theorem:<br />

@AB > @ 2 1 @CA > @ 2 5 ("11) 2 1 ("10) 2<br />

5 11 1 10<br />

@BC > 5 21<br />

@ 2 5 ("21) 2<br />

5 21<br />

So triangle ABC is a right triangle.<br />

b. Yes, P(1, 2, 3), Q(2, 4, 6), and R(21, 22, 23)<br />

are collinear because:<br />

2P 5 (2, 4, 6)<br />

1Q 5 (2, 4, 6)<br />

22R 5 (2, 4, 6)<br />

13. a. Find 0 AB > 0, 0 BC > 0, 0 CA > 0<br />

@AB > @ 5 "(1 2 3) 2 1 (2 2 0) 2 1 (5 2 4) 2<br />

5 "9<br />

@BC > 5 3<br />

@ 5 "(2 2 1) 2 1 (1 2 2) 2 1 (3 2 5) 2<br />

@CA > @ 5 "(2 2 3) 2 1 (1 2 0) 2 1 (3 2 4) 2<br />

Test<br />

in the Pythagorean theorem:<br />

@BC > @ 2 1 @CA > @ 2 5 A"6B 2 1 A"3B 2<br />

5 6 1 3<br />

@AB > 5 9<br />

@ 2 5 (3) 2<br />

5 9<br />

So triangle ABC is a right triangle.<br />

b. Since triangle ABC is a right triangle,<br />

6<br />

cos/ABC 5 Å 3<br />

14. a. DA > , BC ><br />

and EB > , ED ><br />

b. DC > , AB ><br />

and CE > , EA ><br />

c. @AD > @ 2 1 @DC > @ 2 5 @AC > @ 2 ,<br />

But @AC > @ 2 5 @DB > 2<br />

@<br />

Therefore, @AD > @ 2 1 @DC > @ 2 5 @DB > 2<br />

@<br />

15. a. C(3, 0, 5); P(3, 4, 5); E(0, 4, 5); F(0, 4, 0)<br />

b. DB > 5 (3 2 0, 4 2 0, 0 2 5)<br />

5 (3, 4, 25)<br />

CF > 5 (0 2 3, 4 2 0, 0 2 5)<br />

5 (23, 4, 25)<br />

c. D<br />

P<br />

O<br />

5 "(22) 2 1 (2) 2 1 (1) 2<br />

5 "(1) 2 1 (21) 2 1 (22) 2<br />

5 "6<br />

5 "(21) 2 1 (1) 2 1 (21) 2<br />

5 "3<br />

@AB > @ ,<br />

@BC > @ ,<br />

X<br />

u<br />

@CA > @<br />

B<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-29


0 OD > 0 5 5<br />

0 DP > 0 5 5 by the Pythagorean theorem<br />

Thus ODPB is a square and cos u50,<br />

so the angle<br />

between the vectors is 90°.<br />

d. E<br />

P<br />

O<br />

u<br />

5 180° 2 2acos 21 a 3<br />

!50 bb<br />

8 50.2°<br />

16. a.<br />

d + e<br />

150° 30° d<br />

e<br />

Use the cosine law to evaluate 0 d > 1 e > 0<br />

0 d > 1 e > 0 2 5 0 d > 0 2 1 0 e > 0 2 2 20 d > 00e > 0 cos u<br />

5 (3) 2 1 (5) 3 2 2(3)(5) cos 150°<br />

5 9 1 25 2 30 2"3<br />

2<br />

0 d > 1 e > 8 59.98<br />

0 8 "59.98<br />

8 7.74<br />

b.<br />

d d – e<br />

e<br />

Use the cosine law to evaluate 0 d > 2 e > 0<br />

0 d > 2 e > 0 2 5 0 d > 0 2 1 0 e > 0 2 2 20 d > 00e > 0 cos u<br />

5 (3) 2 1 (5) 3 2 2(3)(5) cos 30°<br />

5 9 1 25 2 30 "3<br />

2<br />

0 d > 2 e > 8 8.02<br />

0 8 "8.02<br />

c. 0 e > 2 d > 8 2.83<br />

0 5 0 2 (d > 2 e > ) 0 5 0 d > 2 e > 0 8 2.83<br />

17. a.<br />

u<br />

A<br />

@OA > @ 5 3, @OP > @ 5 "5<br />

u5180° 2 2(m/POA)<br />

A:<br />

400 km/h<br />

Let represent the air speed of the airplane and let<br />

W > A ><br />

represent the velocity of the wind. In one hour,<br />

the plane will travel 0 A > 1 W > 0 kilometers. Because<br />

A > and W ><br />

make a right angle, use the Pythagorean<br />

theorem:<br />

0 A > 1 W > 0 2 5 0 A > 0 2 1 0 W > 2<br />

0<br />

0 A > 1 W > 0 5 "170000<br />

8 412.3 km<br />

So in 3 hours, the plane will travel<br />

3(412.3)km 8 1236.9 km<br />

b.<br />

5 170 000<br />

tan u5 0 W> 0<br />

0 A > 0<br />

5 100<br />

400<br />

5 (400) 2 1 (100) 2<br />

u5tan 21 a 1 4 b<br />

8 14.0°<br />

The direction of the airplane is S14.0°<br />

W.<br />

18. a. Any pair of nonzero, noncollinear vectors<br />

will span R 2 . To show that (2, 3) and (3, 5) are<br />

noncollinear, show that there does not exist any<br />

number k such that k(2, 3) 5 (3, 5) . Solve the<br />

system of equations:<br />

2k 5 3<br />

3k 5 5<br />

Solving both equations gives two different values<br />

3 5<br />

for k, 2 and 3, so (2, 3) and (3, 5) are noncollinear<br />

and thus span R 2<br />

b. (323, 795) 5 m(2, 3) 1 n(3, 5)<br />

(323, 795) 5 (2m, 3m) 1 (3n, 5n)<br />

(323, 795) 5 (2m 1 3n, 3m 1 5n)<br />

Solve the system of equations:<br />

323 5 2m 1 3n<br />

795 5 3m 1 5n<br />

Use the method of elimination:<br />

23(323) 523(2m 1 3n)<br />

2(795) 5 2(3m 1 5n)<br />

2969 526m 2 9n<br />

1 1590 5 6m 1 10n<br />

621 5 n<br />

By substitution, m 52770.<br />

W: 100 km/h<br />

6-30 <strong>Chapter</strong> 6: Introduction to Vectors


19. a. Find a and b such that<br />

(5, 9, 14) 5 a(22, 3, 1) 1 b(3, 1, 4)<br />

(5, 9, 14) 5 (22a, 3a, a) 1 (3b, b, 4b)<br />

(5, 9, 14) 5 (22a 1 3b, 3a 1 b, a 1 4b)<br />

i. 5 522a 1 3b<br />

ii. 9 5 3a 1 b<br />

iii. 14 5 a 1 4b<br />

Use the method of elimination with i. and iii.<br />

2(14) 5 2(a 1 4b)<br />

28 5 2a 1 8b<br />

1 5 522a 1 3b<br />

33 5 11b<br />

3 5 b<br />

By substitution, .<br />

a > a 5 2<br />

lies in the plane determined by b ><br />

and c > because it<br />

can be written as a linear combination of and<br />

b. If vector a > is in the span of b ><br />

and c > b ><br />

then can<br />

be written as a linear combination of b > a > c > .<br />

,<br />

and c > . Find<br />

m and n such that<br />

(213, 36, 23) 5 m(22, 3, 1) 1 n(3, 1, 4)<br />

5 (22m, 3m, m) 1 (3n, n, 4n)<br />

5 (22m 1 3n, 3m 1 n, m 1 4n)<br />

Solve the system of equations:<br />

213 522m 1 3n<br />

36 5 3m 1 n<br />

23 5 m 1 4n<br />

Use the method of elimination:<br />

2(23) 5 2(m 1 4n)<br />

46 5 2m 1 8n<br />

1 213 522m 1 3n<br />

33 5 11n<br />

3 5 n<br />

By substitution, .<br />

So, vector a > m 5 11<br />

is in the span of b ><br />

and c > .<br />

20. a.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 0, 4)<br />

(4, 4, 4)<br />

b.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 4, 4)<br />

(4, 0, 4)<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

x (4, 0, 0) (4, 4, 0)<br />

PO > 5 (4, 4, 4) so,<br />

OP > 52PO > 52(4, 4, 4) 5 (24, 24, 24)<br />

c.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 0, 4)<br />

(4, 4, 4)<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

x (4, 0, 0) (4, 4, 0)<br />

The vector PQ ><br />

from P(4, 4, 4) to Q(0, 4, 0) can be<br />

written as PQ > 5 (24, 0, 24) .<br />

d.<br />

z<br />

(0, 0, 4)<br />

(0, 4, 4)<br />

(4, 0, 4)<br />

(4, 4, 4)<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

x (4, 0, 0) (4, 4, 0)<br />

x<br />

(0, 0, 0)<br />

y<br />

(0, 4, 0)<br />

(4, 0, 0) (4, 4, 0)<br />

The vector with the coordinates (4, 4, 0).<br />

21. 0 2(a ><br />

5 0 2a > 1 b ><br />

5 0 4a > 1 2b > 2 c > )<br />

2 3b > 2 2c > 2 (a ><br />

1 c > 2 a > 1 2b ><br />

2 2b > ) 1<br />

1 3a > 3(a > 2<br />

2 3b > b > 1<br />

1 3c > c > ) 0<br />

0<br />

0<br />

5 0 4(1, 1, 21) 2 3(2, 21, 3) 1 (2, 0, 13) 0<br />

5 0 (4, 4, 24) 1 (26, 3, 29) 1 (2, 0, 13) 0<br />

5 0 (0, 7, 0) 0<br />

5 7<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-31


22.<br />

a. @AB > @ 5 10 because it is the diameter of the circle.<br />

5 2"5 or 4.47<br />

5 "80 or 8.94<br />

b. If A, B, and C are vertices of a right triangle, then<br />

5 100<br />

So, triangle ABC is a right triangle.<br />

23. a. FL ><br />

b. MK > 5<br />

5 JK > FG > 1<br />

2 JM > GH ><br />

5 a > 1 HL ><br />

2 b > 5 a > 1 b > 1 c ><br />

c.<br />

d.<br />

e.<br />

24.<br />

5 "20<br />

A(–3, 4)<br />

@BC > @ 5 "(5 2 3) 2 1 (0 2 (24)) 2<br />

@CA > @ 5 "(5 2 (23)) 2 1 (0 2 4) 2<br />

@BC > @ 2 1 @CA > @ 2 5 @AB > @<br />

2<br />

@BC > @ 2 1 @CA > @ 2 5 A2"5B 2 1 A"80B 2<br />

b<br />

y<br />

5 "(2) 2 1 (4) 2<br />

5 "(8) 2 1 (24) 2<br />

B(3, –4)<br />

5 20 1 80<br />

@AB > 5 100<br />

@ 2 5 10 2<br />

HJ ><br />

IH > 5 HG ><br />

IK > 1 KJ > 1 GF ><br />

2 IH > 5 FG > 1 FJ ><br />

5 HK > 1 GF > 52b > 2 a > 1 c ><br />

5 IJ > 5<br />

5 b > 0<br />

2 c ><br />

a<br />

C(5, 0)<br />

25. a. "0a > 0 2 1 0 b > 2<br />

0 by the Pythagorean theorem<br />

b. "0a > 0 2 1 0 b > 2<br />

0 by the Pythagorean theorem<br />

c. "40 a > 0 2 1 90 b > 0<br />

2 by the Pythagorean theorem<br />

26. Case 1 If b ><br />

and c > are collinear, then is<br />

also collinear with both and But is perpendicular<br />

to b > and c > , so a > b > c > . a > 2b > 1 4c ><br />

is perpendicular to 2b > 1 4c > .<br />

x<br />

Case 2 If and are not collinear, then by spanning<br />

sets, b > b ><br />

and c > c ><br />

span a plane in and is in<br />

that plane. If a > R 3 ,<br />

is perpendicular to b > 2b > 1<br />

and c > 4c ><br />

, then it is<br />

perpendicular to the plane and all vectors in the<br />

plane. So, a > is perpendicular to 2b > 1 4c > .<br />

<strong>Chapter</strong> 6 Test, p. 348<br />

1. Let P be the tail of and let Q be the head of<br />

The vector sums 3a > a ><br />

1 (b > 1 c > and 3(a > 1 b > c ><br />

) 1 c > .<br />

)4<br />

4<br />

can be depicted as in the diagram below, using the<br />

triangle law of addition. We see that<br />

PQ > 5 a > 1 (b > 1 c > ) 5 (a > 1 b > ) 1 c > . This is the<br />

associative property for vector addition.<br />

P<br />

a<br />

(a + b)<br />

PQ= (a + b) + c = a + (b + c)<br />

2. a. AB ><br />

b. @AB > 5 (6 2 (22), 7 2 3, 3 2 (25)) 5 (8, 4, 8)<br />

@<br />

c. BA > 5 "8 2 1<br />

5 (21)AB > 4 2 1 8 2 5 12<br />

5 (28, 24, 28);<br />

@BA > @ 5 @AB > @ 5 12; unit vector in direction of<br />

@BA > @ 5 1<br />

@BA > BA ><br />

@<br />

5 1 (28, 24, 28)<br />

12<br />

5 a2 2 3 2 1 3 , 22 3 b<br />

b<br />

(b + c)<br />

3. Let x > 5 PQ > , y > 5 QR > , and as in the<br />

diagram below. Note that 0 RS > 2y > 5<br />

0 5 0 2y > QS > ,<br />

0 5 6 and that<br />

triangle PQR and triangle PRS share angle u.<br />

c<br />

Q<br />

6-32 <strong>Chapter</strong> 6: Introduction to Vectors


22.<br />

a. @AB > @ 5 10 because it is the diameter of the circle.<br />

5 2"5 or 4.47<br />

5 "80 or 8.94<br />

b. If A, B, and C are vertices of a right triangle, then<br />

5 100<br />

So, triangle ABC is a right triangle.<br />

23. a. FL ><br />

b. MK > 5<br />

5 JK > FG > 1<br />

2 JM > GH ><br />

5 a > 1 HL ><br />

2 b > 5 a > 1 b > 1 c ><br />

c.<br />

d.<br />

e.<br />

24.<br />

5 "20<br />

A(–3, 4)<br />

@BC > @ 5 "(5 2 3) 2 1 (0 2 (24)) 2<br />

@CA > @ 5 "(5 2 (23)) 2 1 (0 2 4) 2<br />

@BC > @ 2 1 @CA > @ 2 5 @AB > @<br />

2<br />

@BC > @ 2 1 @CA > @ 2 5 A2"5B 2 1 A"80B 2<br />

b<br />

y<br />

5 "(2) 2 1 (4) 2<br />

5 "(8) 2 1 (24) 2<br />

B(3, –4)<br />

5 20 1 80<br />

@AB > 5 100<br />

@ 2 5 10 2<br />

HJ ><br />

IH > 5 HG ><br />

IK > 1 KJ > 1 GF ><br />

2 IH > 5 FG > 1 FJ ><br />

5 HK > 1 GF > 52b > 2 a > 1 c ><br />

5 IJ > 5<br />

5 b > 0<br />

2 c ><br />

a<br />

C(5, 0)<br />

25. a. "0a > 0 2 1 0 b > 2<br />

0 by the Pythagorean theorem<br />

b. "0a > 0 2 1 0 b > 2<br />

0 by the Pythagorean theorem<br />

c. "40 a > 0 2 1 90 b > 0<br />

2 by the Pythagorean theorem<br />

26. Case 1 If b ><br />

and c > are collinear, then is<br />

also collinear with both and But is perpendicular<br />

to b > and c > , so a > b > c > . a > 2b > 1 4c ><br />

is perpendicular to 2b > 1 4c > .<br />

x<br />

Case 2 If and are not collinear, then by spanning<br />

sets, b > b ><br />

and c > c ><br />

span a plane in and is in<br />

that plane. If a > R 3 ,<br />

is perpendicular to b > 2b > 1<br />

and c > 4c ><br />

, then it is<br />

perpendicular to the plane and all vectors in the<br />

plane. So, a > is perpendicular to 2b > 1 4c > .<br />

<strong>Chapter</strong> 6 Test, p. 348<br />

1. Let P be the tail of and let Q be the head of<br />

The vector sums 3a > a ><br />

1 (b > 1 c > and 3(a > 1 b > c ><br />

) 1 c > .<br />

)4<br />

4<br />

can be depicted as in the diagram below, using the<br />

triangle law of addition. We see that<br />

PQ > 5 a > 1 (b > 1 c > ) 5 (a > 1 b > ) 1 c > . This is the<br />

associative property for vector addition.<br />

P<br />

a<br />

(a + b)<br />

PQ= (a + b) + c = a + (b + c)<br />

2. a. AB ><br />

b. @AB > 5 (6 2 (22), 7 2 3, 3 2 (25)) 5 (8, 4, 8)<br />

@<br />

c. BA > 5 "8 2 1<br />

5 (21)AB > 4 2 1 8 2 5 12<br />

5 (28, 24, 28);<br />

@BA > @ 5 @AB > @ 5 12; unit vector in direction of<br />

@BA > @ 5 1<br />

@BA > BA ><br />

@<br />

5 1 (28, 24, 28)<br />

12<br />

5 a2 2 3 2 1 3 , 22 3 b<br />

b<br />

(b + c)<br />

3. Let x > 5 PQ > , y > 5 QR > , and as in the<br />

diagram below. Note that 0 RS > 2y > 5<br />

0 5 0 2y > QS > ,<br />

0 5 6 and that<br />

triangle PQR and triangle PRS share angle u.<br />

c<br />

Q<br />

6-32 <strong>Chapter</strong> 6: Introduction to Vectors


P<br />

x + y<br />

x – y<br />

x<br />

By the cosine law:<br />

cos u5 0 y> 0<br />

2 1 0 x > 1 y > 0<br />

2 2 0 x > 2<br />

0<br />

and<br />

20 y > 00x > 1 y > 0<br />

cos u5 0 2y> 0<br />

2 1 0 x > 1 y > 0<br />

2 2 0 x > 2 y > 2<br />

0<br />

20 2y > 00x > 1 y > .<br />

0<br />

Hence,<br />

0 y > 0 2 1 0 x > 1 y > 0 2 2 0 x > 2<br />

0<br />

20 y > 00x > 1 y > 0<br />

5 0 2y> 0<br />

2 1 0 x > 1 y > 0<br />

2 2 0 x > 2 y > 2<br />

0<br />

20 2y > 00x > 1 y > 0<br />

2(0 y > 0 2<br />

5<br />

1 20 x > 2<br />

0 x > 0 2y > 1 0 x > 1<br />

0 2<br />

2 y > 1 0 x > y > 0 2<br />

1<br />

0 2 5 20 y > y > 2 0 x > 0 2<br />

0 2 2<br />

0 2 2 0 x > 0 x > )<br />

2<br />

1 y > y > 2<br />

0<br />

2<br />

0 0<br />

1 "20 x > 2<br />

0 x > 2 y > 0 5 "20 y > 0 2 2 0 x > 1 y > 2<br />

0 0<br />

0 x > 2 y > 0 5 "2(3) 2 2 ("17) 2 1 "2(3) 2<br />

0 x > 2 y > 0 5 "19<br />

u<br />

R<br />

y<br />

Q<br />

S<br />

– y<br />

4. a. We have 3x > 2 2y > 5 a > and 5x > 2 3y > 5 b > .<br />

Multiplying the first equation by and the second<br />

equation by 2 yields:<br />

and<br />

10x > Adding these equations, we have:<br />

x > 2<br />

5 2b > 6y > 5<br />

2 3a > 2b > 29x > 1 6y > 23<br />

523a ><br />

.<br />

Substituting this into the first equation<br />

yields:<br />

Simplifying, we<br />

have: y > 3(2b > .<br />

2 3a > )<br />

5 3b > 2 5a > 2 2y > 5 a > .<br />

.<br />

b. First, conduct scalar multiplication on the third<br />

vector, yielding:<br />

(2, 21, c) 1 (a, b, 1) 2 (6, 3a, 12) 5 (23, 1, 2c).<br />

Now, each of the three components corresponds to<br />

an equation. First, 2 1 a 2 6 523, which implies<br />

a 5 1. Second, 21 1 b 2 3a 5 1. Substituting<br />

a 5 1 and simplifying yields b 5 5. Third,<br />

c 1 1 so<br />

5. a. a > 2 12 5<br />

and b > 2c, c 5211.<br />

span R 2 , because any vector (x, y) in<br />

R 2 can be written as a linear combination of a > and b > .<br />

These two vectors are not multiples of each other.<br />

b. First, conduct scalar multiplication on the vectors,<br />

yielding: (22p, 3p) 1 (3q, 2q) 5 (13, 29).<br />

Now, each component corresponds to an equation.<br />

First, 22p 1 3q 5 13. Second, 3p 2 q 529.<br />

Multiplying the second equation by 3 and adding<br />

the result to the first equation yields: 7p 5214,<br />

which implies p 522. Substituting this into the<br />

first equation and simplifying yields<br />

6. a. a > 5 mb > 1 nc > q 5 3.<br />

(1, 12, 229) 5 m(3, 1, 4) 1 n(1, 2, 23)<br />

(1, 12, 229) 5 (3m, m, 4m) 1 (n, 2n, 23n)<br />

Each of the three components corresponds to an<br />

equation. First, 1 5 3m 1 n. Second, 12 5 m 1 2n.<br />

Third, 229 5 4m 2 3n. Multiplying the first<br />

equation by 22 and adding the result to the second<br />

equation yields m 522. Substituting m 522 into<br />

the first equation yields n 5 7. Since m 522 and<br />

n also solves the third component’s equation,<br />

a > 5 7<br />

5 mb > 1 nc > for m 522 and n 5 7. Hence, can<br />

be written as a linear combination of and<br />

b. r > 5 mp > 1 nq > b > c > a ><br />

.<br />

(16, 11, 224) 5 m(22, 3, 4) 1 n(4, 1, 26)<br />

(16, 11, 224) 5 (22m, 3m, 4m) 1 (4n, n, 26n)<br />

Each of the three components corresponds to an<br />

equation. First, 16 522m 1 4n. Second,<br />

11 5 3m 1 n. Third, 224 5 4m 2 6n. Multiplying<br />

the first equation by 2 and adding the result to the<br />

third equation yields n 5 4. Substituting n 5 4 into<br />

the first equation yields m 5 0. We have that n 5 4<br />

and m 5 0 is the unique solution to the first and<br />

third equations, but n 5 4 and m 5 0 does not<br />

solve the second equation. Hence, this system of<br />

equations has no solution, and cannot be written<br />

as a linear combination of and In other words,<br />

r > p > r > q > .<br />

does not lie in the plane determined by and<br />

7. x > and y ><br />

p > q > .<br />

have magnitudes of 1 and 2, respectively,<br />

and have an angle of 120° between them, as depicted<br />

in the picture below.<br />

y<br />

120˚<br />

x<br />

Calculus and Vectors <strong>Solutions</strong> Manual<br />

6-33


Since 60° is the complement of 120° 3x > 1 2y > can<br />

be depicted as below.<br />

3x + 2y<br />

u<br />

3x<br />

60˚<br />

2y<br />

By the cosine law:<br />

0 3x ><br />

0 3x > 1 2y ><br />

0 3x > 1 2y > 0 2 5 0 3x ><br />

1 2y > 0 2 5 90 x > 0 2 1 0 2y ><br />

0 2 1 40 y > 0 2 2 20 3x ><br />

0 2 2 60 x > 002y ><br />

00y > 0 cos 60<br />

0<br />

0 2 5 9 1 16 2 12<br />

0 3x > 1 2y > 0 5 "13 or 3.61<br />

The direction of 3x > 1 2y > is u, the angle from x > .<br />

This can be computed from the sine law:<br />

0 3x > 1 2y > 0<br />

5 0 2y> 0<br />

sin 60 sin u<br />

sin u5 0 2y> 0 sin 60<br />

0 3x > 1 2y > 0<br />

u5sin 21 (4) sin 60<br />

a b<br />

"13<br />

relative to x<br />

8. DE > u 8<br />

DE > 5 CE > 73.9°<br />

5 b > 2<br />

2 a > CD ><br />

Also,<br />

BA ><br />

BA > 5 CA ><br />

5 2b > 2 CB ><br />

2 2a ><br />

Thus,<br />

DE > 5 1 2 BA><br />

u5sin 21 a 0 2y> 0 sin 60<br />

0 3x > 1 2y > b<br />

0<br />

6-34 <strong>Chapter</strong> 6: Introduction to Vectors

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