Chapter 6 Solutions.pdf
Chapter 6 Solutions.pdf
Chapter 6 Solutions.pdf
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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