Chapter 7 Solution.pdf - TeacherWeb
Chapter 7 Solution.pdf - TeacherWeb
Chapter 7 Solution.pdf - TeacherWeb
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7. Arms 90 cm apart will yield a resultant with asmaller magnitude than at 30 cm apart. A resultantwith a smaller magnitude means less force tocounter your weight, hence a harder chin-up.8. Using the cosine law, the resultant has a magnitude,r,of5 6 2 1 8 2 2 2(6)(8)a2 1 2 b5 36 1 64 1 485 148r 5 "1488 12.17 NUsing the sine law, the resultant’s angle, u, can befound bysin u sin 120°58 12.17"32sin u5812.17u5sin 21 812.178 34.7° from the 6 N force toward the 8 Nforce. The equilibrant, then, would be 12.17 N at180° 2 34.7° 5 145.3° from the 6 N force awayfrom the 8 N force.9.10 N"32Now we look at x and We know> 1 x 2 .x 1 5 @ f 1 @ sin 15>x 2 5 @ f 2 @ sin 75x 1 1 x 2 5 10> >So @ f1 @ sin 15 1 @ f2 @ sin 75 5 10>Substituting then solving for f 2 yields> cos 75>@ f2 @ @ f 2 @ sin 75 5 10cos 15 sin 15 1> cos 75@ f 2 @ a sin 15 1 sin 75b 5 10cos 15>@ f 2 @ (1.035) 5 10>@ f 2 @ 5 9.66 N>Now we solve for f 1 :> > cos 75@ f 1 @ 5 @ f 2 @cos 15> cos 75@ f 1 @ 5 (9.66)cos 15>@ f 1 @ 5 (9.66)(0.268)@ f 1>@ 5 2.59 NSo the force 15° from the 10 N force is 9.66 N andthe force perpendicular to it is 2.59 N.10. The force of the block is(10 kg)(9.8 N>kg) 5 98 N. The component of thisforce parallel to the ramp is(98) sin 30° 5 (98)A 1 2B 5 49 N, directed down theramp. So the force preventing this block frommoving would be 49 N directed up the ramp.11. a.r 2 5 @ f 1>@ 2 1 @ f 2>@ 2 2 2 @ f 1>@@f2>@ cos 120°7-3f 1f 27 N13 N>f 1 5 force 15° from the 10 N force> >f 2 5 force perpendicular to f1>x 1 5 component of f 1 parallel to the 10 N force>x 2 5 component of f 2 parallel to the 10 N force> >We know that the components of f 1 and f 2perpendicular to the 10 N force must be equal, so wecan write> >@ f1 @ cos 15 5 @ f 2 @ cos 75> > cos 75@ f 1 @ 5 @ f 2 @cos 15Calculus and Vectors <strong>Solution</strong>s Manualb. Using the cosine law for the angle, u, we have13 2 5 8 2 1 7 2 2 2(8)(7) cos u169 5 64 1 49 2 112 cos u56 52112 cos ucos u5 256112u5cos5 120212128 N
T 2 (1.12) 5 196T 2 8 175.73 Ncos 45°T 1 5 (175.73)cos 30°8 143.48 NThus the tension in the 45° rope is 175.73 N and thetension in the 30° rope is 143.48 N.17.40 cmThus the tension in the 24 cm string is 39.2 N andthe tension in the 32 cm string is 29.4 N.18.xresultant35°u2x24 cm 32 cm5 kgFirst, use the Cosine Law to find the angles thestrings make at the point of suspension. Let u 1 be theangle made by the 32 cm string and u 2 be the anglemade by the 24 cm string.24 2 5 32 2 1 40 2 2 2(32)(40) cos u 122048 522560 cos u 1212048u 1 5 cos25608 36.9°32 2 5 24 2 1 40 2 2 2(24)(40) cos u 221152 521920 cos u 2211152u 2 5 cos19208 53.1°A keen eye could have recognized this triangle as a3-4-5 right triangle and simply used the Pythagoreantheorem as well. Now we set up the same system ofequations as in problem 16, with T 1 being the tensionin the 32 cm string and T 2 being the tension in the24 cm string, and the force of the mass being(5 kg)(9.8 N>kg) 5 49 N.cos 53.1°T 1 5 T 2cos 36.9°T 1 sin 36.9° 1 T 2 sin 53.1° 5 49cos 53.1°aT 2cos 36.9° b sin 36.9° 1 T 2 sin 53.1° 5 49cos 53.1°T 2 aa b sin 36.9° 1 sin 53.1°b 5 49cos 36.9°T 2 (1.25) 5 49T 2 8 39.2 Ncos 53.1°T 1 5 (39.2)cos 36.9°8 29.4 NCalculus and Vectors <strong>Solution</strong>s ManualT 1 cos 36.9° 5 T 2 cos 53.1°WNES(Port means left and starboard means right.) We arelooking for the resultant of these two force vectorsthat are 35° apart. We don’t know the exact valueof the force, so we will call it x. So the small tugis pulling with a force of x and the large tug ispulling with a force of 2x. To find the magnitudeof the resultant, r, in terms of x, we use the cosinelaw.r 2 5 x 2 1 (2x) 2 2 2(x)(2x) cos 145°5 x 2 1 4x 2 2 4x 2 cos 145°8 5x 2 2 4x 2 (20.8192)8 5x 2 1 3.2768x 28 8.2768x 2r 8 "8.2768x 28 2.8769xNow we use the cosine law again to find the angle,u, made by the resultant.x 2 5 r 2 1 (2x) 2 2 2(2.8769x)(2x) cos ux 2 5 8.2768x 2 1 4x 2 2 11.5076x 2 cos ux 2 5 12.2768x 2 2 11.5076x 2 cos u211.2768x 2 5211.5076x 2 cos ucos u5 11.276811.5076u5cos 21 a 11.276811.5076 b8 11.5° from the large tug toward thesmall tug, for a net of 8.5° to the starboard side.7-5
19.8N10N5Na. First we will find the resultant of the 5 N and8 N forces. Use the Pythagorean theorem to find themagnitude, m.m 2 5 5 2 1 8 25 25 1 645 89m 5 "89 8 9.4Next we use the Pythagorean theorem again to findthe magnitude, M, of the resultant of this net forceand the 10 N force.M 2 5 m 2 1 10 25 89 1 1005 189M 5 "189 8 13.75Since the equilibrant is equal in magnitude to theresultant, we have the magnitude of the equilibrantequal to approximately 13.75 N.b. To find each angle, use the definition of cosinewith respect each force as a leg and the resultant asthe hypotenuse. Let u 5N be the angle from the5 N force to the resultant, u 8N be the angle from the8 N force to the resultant, and u 10N be the anglefrom the 10 N force to the resultant.Let the sign of the resultant be negative, since it isin a direction away from the head of each of thegiven forces.5cos u 5N 5213.75u 5N 5 cos 21 5a213.75 b8 111.3°8cos u 8N 5213.75u 8N 5 cos 21 8a213.75 b8 125.6°cos u 10N 5 10213.75u 10N 5 cos 21 10a213.75 b8 136.7°20. We know that the resultant of these two forces isequal in magnitude and angle to the>diagonal>lineof the parallelogram formed with f and as legs> > 1 f 2and has diagonal length @ f 1 1 f2 @ . We also knowfrom the cosine law that> >@ f 1 1 f2 @ 2 >5 @ f 1 @ 2 >1 @ f 2 @ 2 > >2 2 @ f 1 @@f2@ cos fwhere f is the supplement to u in our parallelogram.Since we know f5180 2u, thencos f5cos (180 2u) 52cos u.Thus>we>have@ f 1 1 f2 @ 2 >5 @ f 1 @ 2 >1 @ f 2 @ 2 > >2 2 @ f 1 @@f2@ cos f5 @ f 1>@ 2 1 @ f 2>@ 2 1 2 @ f 1>@@f2>@ cos u@ f 1>1 f2>@ 5 " @ f1>@ 2 1 @ f 2>@ 2 1 2 @ f 1>@@f2>@ cos u7.2 Velocity, pp. 367–3701. a. Both the woman and the train’s velocities arein the same direction, so we add them.80 km>h 1 4 km>h 5 84 km>hb. The woman’s velocity is directed opposite that oftrain, so we subtract her velocity from the train’s.80 km>h 2 4 km>h 5 76 km>h. The resultant is inthe same direction as the train’s movement.2. a. The velocity of the wind is directed opposite thatof the airplane, so we subtract the wind’s velocityfrom the airplane’s.600 km>h 2 100 km>h 5 500 km>h north.b. Both the wind and the airplane’s velocities are inthe same direction, so we add them.600 km>h 1 100 km>h 5 700 km>h north.3. We use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, u, made.m 2 5 300 2 1 50 25 90 000 1 25005 92 500m 5 "92 5008 304.14 km>htan u5 503002150u5tan3008 9.5°. The resultant is 304.14 km>h, W 9.5° S.7-6 <strong>Chapter</strong> 7: Applications of Vectors
4. Adam must swim at an angle, u, upstream so asto counter the 1 km>h velocity of the stream. This isequivalent to Adam swimming along the hypotenuseof a right traingle with 1 km>h leg and a 2 km>hhypotenuse. So the angle is found using the definitionof cosine.cos u5 1 2u5cos 21 1 25 60° upstream5. a. 2 m>s forwardb. 20 m>s 1 2 m>s 5 22 m>s in the direction of the car6. Since the two velocities are at right angles wecan use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, u, made.m 2 5 12 2 1 5 25 144 1 255 169m 5 "1695 13 m>ssin u5 513u5sin 21 5138 22.6° from the direction of the boat towardthe direction of the current. This results in a net of22.6° 1 15° 5 37.6° , or N 37.6° W.7. a. First we find the components of the resultantdirected north and directed west. The componentdirected north is the velocity of the airplane, 800,minus 100 sin 45°, since the wind forms a 45°angle south of west. The western component ofthe resultant is simply 100 cos 45°. So we use thePythagorean theorem to find the magnitude, m, ofthe resultant and the definition of sine to find theangle, u, of the resultant.m 2 5 (800 2 100 sin 45°) 2 1 (100 cos 45°) 28 (729.29) 2 1 (71.71) 28 536 863.8082m 8 732.71 km>hUse the sine law to determine the direction.sin u sin 45°5100 732.71u 8 5.5°The direction is N 5.5° W.Calculus and Vectors <strong>Solution</strong>s Manualb. The airplane is travelling at approximately732.71 km>h, so in 1 hour the airplane will travelabout 732.71 km.8. a. First we find the velocity of the airplane. Weuse the Pythagorean theorem to find the magnitude,m, of the resultant.m 2 5 450 2 1 100 25 202 500 1 10 0005 212 500m 5 "212 5008 461 km>hSo in 3 hours, the airplane will travel about(461 km>h)(3 h) 5 1383 km.b. To find the angle, u, the airplane travels, we usethe definition of sine.sin u5 10046121100u5sin4618 12.5° east of north.9. a. To find the angle, u, at which to fly is theequivalent of the angle of a right triangle with 44 asthe opposite leg and 244 as the hypotenuse. So weuse the definition of sine to find this angle.sin u5 442442144u5sin2448 10.4° south of west.b. By the Pythagorean Theorem, the resultant groundspeed of the airplane is "(244 2 2 44 2 ) 5 240 km>h.Since time 5 distance>rate, the duration of theflight is simply (480 km)>(240 km>h) 5 2 h.10. a. Since Judy is swimming perpendicular tothe flow of the river, her resultant velocity is simplythe hypotenuse of a right triangle with 3 and 4 asbases, which is a 3-4-5 right triangle. Thus, Judy’sresultant velocity is 5 km>h. The direction isdetermined by tan u5 4 3. u 8 53.1° downstreamb. Judy’s distance traveled down the river would be the“4” leg of the 3-4-5 triangle formed by the vectors, butscaled down so that 1m (the width of the river) isequivalent to the “3” leg. So her distance traveled is43 8 1.33 km. This makes her about 0.67 km fromHelen’s cottage.c. While in the river, Judy is swimming at55 km>h for a distance of 3 km. Sincetime 5 distance>rate, her time taken is53 km5 km>h 5 1 3 hours 5 20 minutes.7-7
11.vu205 km/ha. and b. Here, 205 km>h directed 30° north of eastis the resultant of 212 km>h directed east, and thewind speed, v, directed at some angle. This problemis more easily approached finding the wind speed,v, first. So we will do that using the cosine law.v 2 5 205 2 1 212 2 2 2(205)(212) cos 30°5 42 025 1 44 944 2 86 920 cos 30°5 86 969 2 75 2755 11 694v 5 "11 6948 108 km>hNow to find the wind’s direction, we simply find theangle supplementary to the lesser angle, u, formedby the parallelogram of these three velocities. Wecan use the sine law for this.sin u sin 30°5205 108sin 30°sin u5205a108 bu5sin 21 sin 30°a205a108 bb8 71.6°Thus, the direction of v is the angle supplementaryto u in the parallelogram:180° 2 71.6° 5 108.4° 5 18.4° west of north.12.430°212 km/h5Since her swimming speed is a maximum of 4 km>h,this is her maximum resultant magnitude, which isalso the hypotenuse of the triangle formed by her andthe river’s velocity vector. Since one of these legs is5 km>h, we have a triangle with a leg larger than itshypotenuse, which is impossible.13. a. First we need to find Mary’s resultantvelocity, v. Since this resultant is the diagonal of theparallelogram formed by hers and the river’svelocity, we can use the cosine law with the angle, u,of the parallelogram adjacent 30°.v 2 5 3 2 1 4 2 2 2(3)(4) cos 150°5 9 1 16 2 24 cos 150°5 25 1 20.85 45.8v 5 "45.88 6.8 m>sSo in 10 seconds, Mary travels about(6.8 m>s)(10 s) 5 68 m.b. Since Mary is travelling at 3 m>s at an angle of 30°,to find the component of her velocity, v, perpendicularto the current, we use the definition of sine.v 5 3 sin 305 3a 1 2 b5 1.5 m>s perpendicular to the current.So since time 5 distance>rate, the time taken is(150 m)>(1.5 m>s) 5 100 s.14. a. So we have a 5.5 m>s vector and a 4 m>svector with a resultant vector that is directed 45°south of west. Letting u be the angle between the4 km>h vector and the resultant, we can constructa parallelogram using these three vectors and asubsequent triangle with u opposite the 5.5 m>svector and 45° opposite the 4 m>s vector. We nowuse the sine law to find u.sin u sin 45°55.5 4sin 45°sin u55.5a b4u5sin 21 sin 45°a5.5a bb48 76.5° from the resultant.Since the resultant is 45° west of south, Dave’sdirection is 76.5° 1 45° 5 121.5° west of south,which is equivalent to about 180° 2 121.5° 5 58.5°upstream.b. First, we find the magnitude, m, of Dave’s 4 m>svelocity in the direction perpendicular to the river.This is done using the definition of sine.m 5 4 sin 58.5°8 3.41 m>s perpendicular to the river.Since time is distance>rate, we have(200 m)>(3.41 m>s) 8 58.6 s.15. Let b represent the speed of the steamboat and crepresent the speed of the current. On the waydownstream, the effective speed is b 1 c, andupstream is b 2 c. The distance upstream anddownstream is the same, so 5(b 1 c) 5 7(b 2 c).So, b 5 6c. This means that the speed of the boatis 6 times the speed of the current. So, (6c 1 c) ? 57-8 <strong>Chapter</strong> 7: Applications of Vectors
or 35c is the distance. This means that it would takea raft 35 hours moving with the speed of the currentto get from A to B.7.3 The Dot Product of Two GeometricVectors, pp. 377–3781. a > This meansor @b > ? b > 5 0 a > 0 @b > @ cos u50.0 a > 0 5 0,@ 5 0, or cos u50.To be guaranteed that thetwo vectors are perpendicular, the vectors must benonzero.2. a > ? b > is a scalar, and a dot product is only definedfor vectors, so (a > ? b > ) ? c > is meaningless.3. Answers may vary. Leta > ? b > but4. a > ? b > b >5 because5. Since a > b > ? c >? a >and b > 5 b > ? c > a > a > 5 i > ,52c > b > 5 j > , c > 52i > .5 0, 5 0,.c > 5are unit vectors, 0 a > a >0 5 @b > @ 5 1 andsince they are pointing in opposite directions thenu5180° soTherefore6. a. p > ? q > cos5 0 p > u00q > 521. a > ? b > 521.0cos u5 (4)(8) cos (60°)5 (32)(.5)b. x > ? y > 55 0 x > 1600y > 0cos u5 (2)(4) cos (150°)c.d.e.f.a > ? b > 8 26.935 0 a > 0 @b > @ cos u5 (0)(8) cos (100°)p > ? q > 5 05 0 p > 00q > 0cos u5 (1)(1) cos (180°)5 (1)(21)m > ? n > 5215 0 m > 00n > 0cos u5 (2)(5) cos (90°)5 (10)(0)u > ? v > 5 05 0 u > 00v > 0cos u5 (4)(8) cos 145°x > 8? y > 226.25 0 x > 00y > 0cos u7. a.5 (8)a2 "32 b12"3 5 (8)(3) cos u"32 5 cos uu530°b. m > ? n > 5 0 m > 00n > 0cos u(6) 5 (6)(6) cos u16 5 cos uc. p > ? q > u 8 80°5 0 p > 00q > 0cos u3 5 (5)(1) cos u35 5 cos ud. p > u? q > 8 53°5 0 p > 00q > 0cos u23 5 (5)(1) cos ue.f.8.2 3 5 5 cos ua > u? b > 8 127°5 0 a > 0 @b > @ cos u10.5 5 (7)(3) cos u12 5 cos uu > u560°? v > 5 0 u > 00v > 0cos u250 5 (10)(10) cos u2 1 2 5 cos ua > u5120°? b > 5 0 a > 0 @b > @ cos u5 (7.5)(6) cos (180° 2 120°)5 (45)a 1 2 b5 22.5Note: u is the angle between the two vectors whenthey are tail to tail, so9. a. (a > 1 5b > ) ? (2a > u 22 3b > 120°.) 5 a > ? 2a > 21 5b >5 20 a > ? 2a > a > ? 3b >202 2 15 @b > 5b > ? 3b >2@2 3a >5 20 a > ? b > 1 10a >02 2 15 @b > ? b >2@3x > ? (x > 2 3y > ) 2 (x > 2 3y > 1 7a > ?) ? (23x > b >1 y > )b.5 30 x > 021 305 60 x > y > 2 3x > ? 3y > 1 30 x > 0 2 2 x > ? y > 2 (23y > ? 23x > )205 60 x > 0 2 2 9x > ?0 2 2 19x > y > 2? y > x > ? y >1 30 y > 2 9x > ? y > 1 30 y > 202010. so the dot product of any vector with0 > @0 > @ 5 0is 0.Calculus and Vectors <strong>Solution</strong>s Manual7-9
11.0 a > (a > 20 2 2 a > 5b >? b > ) ? (a >2 5b > 2 b >?0 a > a > ) 5 @a >1 5 @b > 2 5b > @@a > 2 b > @ cos (90°)@ 2 5 002 1 5 @b > @ 2 5 6a > ? b >12. a.b.13. a.b. b > ? c > 5 @b > @Therefore 0 a > 0 c > 0cos (90°) 502 5 @b > @ 2 1 0 c > 002 .This is just what the Pythagorean theorem says,where and are the legs of the right triangle.14. (u > b >5 u > 1?15 0 u > v > u > v > c >11? w > u > w >? v > ) ? (u >10 2 1 0 v > w > 1? u > u > 1? w > v > 10 2 1 0 w > 1 w > 1? v > v > w > )? u >0 2 1 20 u > 100v > w > 1 v >? w > ? v >0cos (90°)1 20 u > 00w > 0cos (90°) 1 20 v > 00w > 0cos (90°)5 (1) 2 1 (2) 2 1 (3) 25 1415. 0 u > 1 v > 0 2 1 0 u > 2 v > 205 (u > 1 v > ) ? (u > 1 v > ) 1 (u > 2 v > ) ? (u > 2 v > )5 0 u > 0 2 1 2u > ? v > 1 0 v > 0 2 1 0 u > 0 2 2 2u > ? v > 1 0 v > 205 20 u > 0 2 1 20 v > 2016. (a >5 0 a > 1 b > )02 1 a > ? (a > 1? b > 1 a > b > 1 c > )? c > 1 b > ? a > 1 @b > @ 2 1 b > ? c >5 1 1 20 a > 0 @b > @ cos (60°) 1 0 a > 00c > 0cos (60°) 1 11 @b > @ 0 c > 0cos (120°)5 2 1 2a 1 2 b 1 1 2 2 1 25 317.a > a >? (a > 1 b >1 b > 1 c >1 c > 5 0 >) 1 b > ? (a > 1 b > 1 c > )10 a > c > ? (a >02 1 a > 1 b > 1? b > 1 a > c > )? c > 5 01 b > ? a > 1 @b > @ 2 1 b > ? c >1 c > ? a > 1 c > ? b > 1 0 c > 0 2 5 0a > ? b > 5 1 6 ( 0 a> 0 2 1 5 @b > @ 2 )(a > 1 b > ) ? (a > 1 b > ) 5 a > 5? a > 11 a > ? b >15 0 a > b > ? a >02 1 a > 1 b >? b > ? b >1 a > ? b >1(a > 1 b > ) ? (a > 2 b > ) 5 a > 5 0? a > a > @b > 2@02 12 a > 2a >? b > ? b > 11 b > @b > 2? a > @25 0 a > b > ? b >0 a > 02 5 a > ?5 (b > a > 5 0 a > 0 2 2 a >02 2 @b > ? b > 1 a > ? b > 2 @b > 2@2@5 @b > 1 c > ) ?@ 2 1 2b > (b >? c > 1 c > )1 0 c > 201 1 4 1 9 1 2(a > ? b > 1 a > ? c > 1 b > ? c > ) 5 018. d >c > b > 5 b >? a > 5 d > 2 c > a > ? b > 1 a > ? c > 1 b > ? c > 52715 ((b > c >? a > ) a > ) ? a >c > ? a > 5 (b > ? a > )(a > ? a > ) because b > ? a > is a scalarc > c > ? a > becausec > ? a > 5 (b > ? a > )0d > ? a > 5 (d > 1? a > 5 d > ? a > c > a > 20) ?1 c > a >? a > 0 a > 0 5 15 07.4 The Dot Product for AlgebraicVectors, pp. 385–3871. a > ? b > 5 0(21)b 1 1 b 2 5 0b 2 5 b 1Any vector of the form (c, c) is perpendicularto a > . Therefore there are infinitely many vectorsperpendicular to a > . Answers may vary. For example:(1, 1), (2, 2), (3, 3).2. a. a > ? b > 5 (22)(1) 1 (1)(2)5 0b. a > ? b > u590°5 (2)(4) 1 (3)(3) 1 (21)(217)5 8 1 9 1 175 34 . 0cos u .02(a > ? b > 1 a > ? c > 1 b > ? c > ) 5214u is acutec. a > ? b > 5 (1)(3) 1 (22)(22) 1 (5)(22)5 3 1 4 2 10523 , 0cos u ,0u is obtuse3. Any vector in the xy-plane is of the forma > 5 (a Let b > 1 , a 2 , 0). 5 (0, 0, 1).a > ? b > 5 (0)(a 1 ) 1 (0)(a 2 ) 1 (0)(1)5 0Therefore (0, 0, 1) is perpendicular to every vectorin the xy-plane.Any vector in the xz-plane is of the formc > 5 (c Let d > 1 , 0, c 3 ). 5 (0, 1, 0).c > ? d > 5 (0)(c 1 ) 1 (0)(1) 1 (0)(c 3 )5 0Therefore (0, 1, 0) is perpendicular to every vectorin the xz-plane.7-10 <strong>Chapter</strong> 7: Applications of Vectors
Any vector in the yz-plane is of the forme > 5 (0, e Let f > 2 , e 3 ). 5 (1, 0, 0) .e > ? f > 5 (1)(0) 1 (0)(e 2 ) 1 (0)(e 3 )5 0Therefore (1, 0, 0) is perpendicular to every vectorin the yz-plane.4. a. (1, 2, 21) ? (4, 3, 10) 5 4 1 6 2 105 0( 24, 25, 26) ? a5, 23, 2 5 b 5220 1 15 1 565 0b. If any of the vectors were collinear then onewould be a scalar multiple of the other. Comparingthe signs of the individual components of eachvector eliminates (1, 2, 21) and (5, 23, 2 5 6). All ofthe components of (24, 25, 26) have the samesign and the same is true for (4, 3, 10), but (4, 3, 10)is not a scalar multiple of (24, 25, 26). Thereforenone of the vectors are collinear.5. a. Using the strategy of Example 5 yields(x, y) ? (1, 22) 5 0 and (x, y) ? (1, 1) 5 0x 2 2y 5 0 and x 1 y 5 03y 5 0Therefore the only result is x 5 y 5 0, or (0, 0).This is because (1, 22) and (1, 1) both lie on thexy-plane and are not collinear, so any vector that isperpendicular to both vectors must be in R 3 whichdoes not exist in R 2 .b. If we select any two vectors that are not collinearin R 2 , then any vector that is perpendicular to bothcannot be in R 2 and must be in R 3 This is notpossible since R 3 does not exist in.R6. a. cos u5 a> ? b >2 .0 a > 0 @b > @(5)(21) 1 (3)(22)5"25 1 9"1 1 45 211"(34)(5)5 211"170u 8 148°b. cos u5 a> ? b >0 a > 0 @b > @(21)(6) 1 (4)(22)5"1 1 16"36 1 45 214"680u 8 123°Calculus and Vectors <strong>Solution</strong>s Manualc. cos u5 a> ? b >0 a > 0 @b > @(2)(2) 1 (2)(1) 1 (1)(22)5"4 1 4 1 1"4 1 1 1 45 4(3)(3)d. cos u5 a> ? b >0 a > 0 @b > @(2)(25) 1 (3)(0) 1 (26)(12)5"4 1 9 1 36"25 1 1445 282(7)(13)5 28291u 8 154°7. a.a > ? b > 5 0 a >(21)(26k) 1 (2)(21) 1 (23)(k) 5 0 a > 0 @b > @0 @b > cos u@b.(1)(0) 1 (1)(k) 5 "1 1 1"k 2 cos (45°)1k 5 "2 0 k 0"28. a.b.5 4 9u 8 64°–2 –1a > ? b > 5 0 a > 0 @b > @ cos u2(0, 1)1(1, 0) x0 1 2–1–2k 5 0 k 0k $ 0yy2(0, 1)1(1, 0) x–2 –1 0–1–21 26 k 2 2 2 3k 5 03 k 5 2k 5 2 3cos (90°)7-11
The diagonals are (1, 0) 1 (0, 1) 5 (1, 1) and(1, 0) 2 (0, 1) 5 (1, 21) or(1, 0) 1 (0, 1) 5 (1, 1) and(0, 1) 2 (1, 0) 5 (21, 0) .c. (1, 1) ? (1, 21)5 1 2 15 0or (1, 1) ? (21, 1)521 1 15 09. a. cos u5 a> ? b >0 a > 0 @b > @5 (1 2 "2)(1) 1 ("2 2 1)(1)0 a > 0 @b > @5 0u590°b. cos u5 a> ? b >0 a > 0 @b > @"2 2 1 1 "2 1 1 1 "25"(2 2 2"2 1 1) 1 (2 1 2"2 1 1) 1 2"1 1 1 1 15 3"2"8"35 "32u530°10. a. i. a > 5 kb >8 5 12kk 5 2 3p 5 4a 2 3 bp 5 8 32 5 2 3 qq 5 3ii. Answers may vary. For example:a > ? b > 5 02q 1 4p 1 96 5 0q 522p 2 48Let p 5 1q 5250b. In part a., the values are unique because bothvectors have their third component specified, andthe ratios must be the same for each component b > .In part b. the values are not unique; any value ofp could have been chosen, each resulting in adifferent value of q.11. AB > 5 (2, 6), BC > 5 (25, 25), CA > 5 (3, 21)cos (180° 2u A ) 5 AB> ? CA >@AB > @@CA > @5 6 2 6@AB > @@CA > @5 0180° 2u A 5 90°u A 5 90°cos (180° 2u B ) 5 AB> ? BC >@AB > @@BC > @210 2 305"4 1 36"25 1 252405"(40)(50)452 Å 5180° 2u B 8 153.4°u B 8 26.6°u C 5 180° 2u A 2u Bu C 8 63.4°12. a. O 5 (0, 0, 0), A 5 (7, 0, 0), B 5 (7, 4, 0),C 5 (0, 4, 0), D 5 (7, 0, 5), E 5 (0, 4, 5),F 5 (0, 0, 5)b.AE > ? BF > 5 @AE > @@BF > @ cos u(27, 4, 5) ? (27, 24, 5) 5 "49 1 16 1 253 "49 1 16 1 25 cos u49 2 16 1 25 5 90 cos u5890 5 cos uu 8 50°13. a. Answers may vary. For example:( x, y, z) ? (21, 3, 0) 5 02x 1 3y 5 0x 5 3y( x, y, z) ? (1, 25, 2) 5 0x 2 5y 1 2z 5 022y 1 2z 5 0y 5 zLet y 5 1.(3, 1, 1) is perpendicular to (21, 3, 0) and(1, 25, 2).b. Answers may vary. For example:( x, y, z) ? (1, 3, 24) 5 0x 1 3y 2 4z 5 0x 5 4z 2 3y( x, y, z) ? (21, 22, 3) 5 07-12 <strong>Chapter</strong> 7: Applications of Vectors
2x 2 2y 1 3z 5 03y 2 4z 2 2y 1 3z 5 0y 5 zLet y 5 1.(1, 1, 1) is perpendicular to (1, 3, 24) and(21, 22, 3).14. (p, p, 1) ? (p, 22, 23) 5 0p 2 2 2p 2 3 5 0p 5 2 6 "22 2 4(23)2p 5 1 6 2p 5 3 or 2115. a. (23, p, 21) ? (1, 24, q) 5 023 2 4p 2 q 5 03 1 4p 1 q 5 0b. 3 1 4p 2 3 5 0p 5 016. Answers may vary. For example: Note thats > 522r > , so they are collinear. Therefore anyvector that is perpendicular to is alsoperpendicular to r > s >.( x, y, z) ? (1, 2, 21) 5 0x 1 2y 2 z 5 0Let x 5 z 5 1.(1, 0, 1) is perpendicular to (1, 2, 21) and(22, 24, 2).Let x 5 y 5 1.(1, 1, 3) is perpendicular to (1, 2 2 1) and(22,17. x > 24,? y > 2).5 0 x > 00y > 0cos u(24, p, 22) ? (22, 3, 6)5 "16 1 p 2 1 4"4 1 9 1 36 cos u9p 2 2 24p 1 16 5 49(20 1 p 2 )a 4 21 b 29p 2 2 24p 1 16 5 3209 1 1665p 2 2 216p 2 176 5 0p 5 216 6 "(2216)2 2 4(65)(2176)2(65)p 5 4 or8 1 3p 2 12 5 "20 1 p 2 (7) cos u2(3p 2 4) 2 5 a7"20 1 p 2 cos ub2 44659 p218. a. a > ? b > 523 1 35 0Therefore, since the two diagonals are perpendicular,all the sides must be the same length.Calculus and Vectors <strong>Solution</strong>s Manualb. AB > 5 1 2 (a> 1 b > )5 (1, 2, 21)BC > 5 1 2 (a> 2 b > )5 (2, 1, 1)@AB > @ 5 @BC > @ 5 "6c. AB > ? BC > 5 @AB > @@BC > @ cos u 12 1 2 2 1 5 6 cos u 112 5 cos u 1u 1 5 60°2u 1 1 2u 2 5 360°u 2 5 120°19. a. AB > 5 (3, 4, 212), DA > 5 (24, 2 2 q, 25)AB > ? DA > 5 0212 1 8 2 4q 1 60 5 021 2 q 1 15 5 0q 5 14DA > 5 CB >( 24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)x 5 6, y 5 18, z 524The coordinates of vertex C are (6, 18, 24).b. AC > ? BD > 5 @AC > @@BD > @ cos u(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 493 "1 1 64 1 289 cos u7 1 128 2 119 5 354 cos u16354 5 cos uu 8 87.4°20. The two vectors representing the body diagonalsare (0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1) and(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)( 21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u1 2 1 1 1 5 3 cos u13 5 cos uu 8 70.5°a5180° 2ua 8 109.5°Mid-<strong>Chapter</strong> Review, pp. 388–3891. a. a > ? b > 5 (3)(2) cos (60°)5 (6) 1 25 37-13
14. a. 45 sin (150°) 5 500 sin uu 8 N 2.6° Eb. v 5 500 cos (2.6°) 2 45 cos (30°)8 460.5 km>ht 8 1000460.5t 8 2.17 hours15.a > ? x > 5 02x 1 1 2x 2 1 5x 3 5 0b > x 1? x > 5 2x 2 1 5x 35 0x 1 1 3x 2 1 5x 3 5 02x 2 1 5x 3 1 3x 2 1 5x 3 5 0x 2 1 2x 3 5 0choose x 3 5 1x 2 522x 1 5 1x > 5 1 (1, 22, 1)!6x> or a2 1 !6 , 25 a 1 !6 , 2 1 !6 ,2 2 !6 , 1!6 b!6 b16. a. v 5 4 1 3 cos (45°)8 6.12 ms >d 8 (6.12)(10)8 61.2 mb. w 5 3 sin (45°)8 2.12 ms >t 8 1802.12t 8 84.9 seconds17. a.0 x > 0 2 2 x > (x > 1? y > y > )1 y > ? (x >? x > 2 y >2 0 y > ) 5 0( x > 1 y > ) ? (x > 0 x > 0 2 5 02 y > 0 2 5 0 y > 20) 5 0 when x > and y > havethe same length.b. Vectors and determine a parallelogram. Theirsum a > 1 b > a > b >is one diagonal of the parallelogramformed, with its tail in the same location as the tailsof a > and b > . Their difference a > 2 b >is the otherdiagonal of the parallelogram.18. @F > @ 5 350 cos (40°)8 268.12 N7.5 Scalar and Vector Projections,pp. 398–4001. a. Scalar projection of a > on b >is wherea > 5 (2, 3) and is the positive x-axis (X, 0).a > ? b > 5 (2X) 1 (3 3 0)5 2X 1 05 2X@b > @ 5 "X 2 1 0 2a > ? b > 5 X@b > 5 2X @ X5 2;The vector projection is the scalar projectionmultiplied by where is the x-axis divided bythe magnitude of the x-axis which is equal toThe scalar projection of 2 multiplied by equalsa > i > i > .2i > .>?bb. Scalar projection of a > on b >is wherea > 5 (2, 3) and>b@ b > @a > ? b > 5 (2 3 0) 1 (3Y)5 0 1 3Y@b > 5 3Y@ 5 "0 2 1 Y 25 Y>bis now the positive y-axis (0, Y).a > @b > @@b > 5 3Y @ Y5 3;The vector projection is the scalar projectionmultiplied by where is the y-axis divided@ b > @b >b >>b@ b > @>b@ b > @by the magnitude of the y-axis which is equal toThe scalar projection of 3 multiplied by j > j > .equals 3j > .2. Using the formula would cause a division by 0.Generally the 0 > has any direction and 0 magnitude.You can not project onto nothing.3. You are projecting a > onto the tail of b >whichis a point with magnitude 0. Therefore it is theprojections of onto the tail of are also 0and 0 > b > a > 0 > ;.@b > @a > >?b@b > @7-16 <strong>Chapter</strong> 7: Applications of Vectors
4. Answers may vary. For example:q > 5 AB > p > 5 AE > ,pEDA C q BScalar projection p > onVector projection p > q >onScalar projection q > q > 5 @AC > @onVector projection on5. When a > q > p > 5 AC > ;;p > 5 @AD > @5 AD > ;5 (21, 2, 5) and b > 5 (1, 0, 0) thena > ? b > 5 (21 3 1 1 2 3 0 1 5 3 0)@b > 521@ 5 "1 2 1 0 2 1 0 25 1a > ? b >Therefore the scalar projection is@b > 5 21@ 1521;The vector equation isUnder the same approach, whenand b > 5 (0, 1, 0), thena > ? b > 5 (21 3 0 1 2 3 1 1 5 3 0)@b > 5 2@ 5 "0 2 1 1 1 0 25 1a > ? b >Therefore the scalar projection is@b > 5 2 @ 1The vector equation is21 3 b>@b > @2 3 b>@b > @521 3521;a > 5 (21, 2, 5)5 2 35 2,(0, 1, 0)1(1, 0, 0)1The same is also true whenandb > a > 5 2;5 (21, 2, 5)thena > 5? b > (0, 0, 1)5 (21 3 0 1 2 3 0 1 5 3 1)@b > 5 5@ 5 "0 2 1 0 2 1 1 25 1a > ? b >Therefore the scalar projection is@b > 5 5 @ 15 5,The vector equation is 5 3 b> (0, 0, 1)@b > 5 5 3@15 5;Calculus and Vectors <strong>Solution</strong>s ManualWithout having to use formulae, a projection of(21, 2, 5) on i > , j > , or is the same as a projectionof on i > k >(21, 0, 0) , (0, 2, 0) on j > , and (0, 0, 5) onwhich intuitively yields the same result.6. a. p > ? q > 5 (3 324) 1 (6 3 5)1 (222 3220)5212 1 30 1 4400 q > 5 4580 5 "(24) 2 1 5 2 1 (220) 25 "16 1 25 1 4005 "4415 21p > ? q >Therefore the scalar projection is0 q > 0The vector equation 5 45821 3 q>0 q > 05 45821b. Direction angles for p > where p > 5 (a, b, c)ainclude a, b, and g. cos a5"a 2 1 b 2 1 c 235"3 2 1 6 2 1 (222) 235"9 1 36 1 4845 3"529Therefore a5cos 21 a 38 82.5°;bcos b5"a 2 1 b 2 1 c 265"3 2 1 6 2 1 (222) 265"9 1 36 1 4845 6"5295 623 ,Therefore b5cos 21 a 623 b8 74.9°;5 458 (24, 5, 20).44123 b 5 323 ,5 45821 ,(24, 5, 220).21k >7-17
ccos g5"a 2 1 b 2 1 c 22225"3 2 1 6 2 1 (222) 22225"9 1 36 1 4845 222"5295 22223 ,Therefore g5cos 21 a 22223 b7. a. x > ? y > 8 163.0°5 (1 3 1) 1 (1 321)5 1 1 (21)0 y > 5 00 5 "1 2 1 (21) 25 "2x > ? y >The scalar projection is0 y > 5 0 0 "25 0;The vector projection is 0 3 y>b. x > ? y > 0 y > 5 0 >05 (2 3 1) 1 (2"3 3 0)5 20 y > 0 5 "1 2 1 0 25 1x > ? y >The scalar projection is0 y > 5 2 0 15 2;The vector projection is 2 3 y>0 y > 5 2 30c. x > ? y > 5 2i >5 (2 325) 1 (5 3 12)5210 1 600 y > 5 500 5 "(25) 2 1 12 25 "25 1 1445 "1695 13x > ? y >The scalar projection is0 y > 0The vector projection is5 5013 .5013 3 y>0 y > 0(1, 0)15 50 (25, 12)313 135 50 (25, 12)1698. a. The scalar projection of a > on the x-axis(X, 0, 0) isa > ? (X, 0, 0)0 (X, 0, 0) 0a > ? (X, 0, 0)0 (X, 0, 0) 0(21 3 X) 1 (2 3 0) 1 (4 3 0)5"X 2 1 0 2 1 0 25 2XX521;The vector projection of a > on the x-axis is(X, 0, 0)(X, 0, 0)21 3521 3"X 2 1 0 2 21 052i > XThe scalar projection of on the y-axis isa > a > ;(0, Y, 0)? (0, Y, 0) (21 3 0) 1 (2 3 Y) 1 (4 3 0)50 (0, Y, 0) 0"0 2 1 Y 2 1 0 25 2Y Y5 2The vector projection of a > on the y-axis is(0, Y, 0)(0, Y, 0)2 35 2 3"0 2 1 Y 2 21 05 2j > YThe scalar projection of on the z-axis isa > a > ;(0, 0, Z)? (0, 0, Z) (21 3 0) 1 (2 3 0) 1 (4 3 Z)50 (0, 0, Z) 0"0 2 1 0 2 1 Z 25 4Z Z5 4;The vector projection of a > on the z-axis is(0, 0, Z)(0, 0, Z)4 35 4 3"0 2 1 0 2 21 Z5 4k > Z.b. The scalar projection of m a > on the x-axis(X, isma > 0, 0)? (X, 0, 0) (2m 3 X) 1 (2m 3 0)50 (X, 0, 0) 0 "X 2 1 0 2 1 0 2(4m 3 0)1"X 2 1 0 2 1 0 25 2mXX52mThe vector projection of ma > on the x-axis is(X, 0, 0)(X, 0, 0)2m 352m 3"X 2 1 0 2 21 052mi > X;7-18 <strong>Chapter</strong> 7: Applications of Vectors
The scalar projection of on the y-axis isma > (0, Y, 0)? (0, Y, 0) (2m 3 0) 1 (2m 3 Y)50 (0, Y, 0) 0 "0 2 1 Y 2 1 0 2(4m 3 0)1"0 2 1 y 2 1 0 25 2mYY5 2m;The vector projection of ma > on the y-axis is(0, Y, 0)(0, Y, 0)2m 35 2m 3"0 2 1 Y 2 21 05 2mj > YThe scalar projection of on the z-axis isma > ma > ;(0, 0, Z)? (0, 0, Z) (2m 3 0) 1 (2m 3 0)50 (0, 0, Z) 0 "0 2 1 0 2 1 Z 2(4m 3 Z)1"0 2 1 0 2 1 Z 25 4mZZ5 4m;The vector projection of ma > on the z-axis is(0, 0, Z)(0, 0, Z)4m 35 4m 3"0 2 1 0 2 21 Z5 4mk > Z.9. a.a11. a. AB > 5 Point B 2 Point A5 (21, 3, 4) 2 (1, 2, 2)5 (22, 1, 2)The scalar projection of on the x-axis isa > AB >(X, 0, 0)? (X, 0, 0) (22 3 X) 1 (1 3 0) 1 (2 3 0)50 (X, 0, 0) 0"X 2 1 0 2 1 0 25 22XX522;The vector projection of AB >on the x-axis is(X, 0, 0)(X, 0, 0)22 3522 3"X 2 1 0 2 21 0522i > XThe scalar projection of on the y-axis isa > AB > ;(0, Y, 0)? (0, Y, 0) (22 3 0) 1 (1 3 Y) 1 (2 3 0)50 (0, Y, 0) 0"0 2 1 Y 2 1 0 25 Y Y5 1;The vector projection of AB >on the y-axis is(0, Y, 0)(0, Y, 0)1 35 1 3"0 2 1 Y 2 21 05 j > Y;The scalar projection of on the z-axis isa > AB >(0, 0, Z)? (0, 0, Z) (22 3 0) 1 (1 3 0) 1 (2 3 Z)50 (0, 0, Z) 0"0 2 1 0 2 1 Z 2ma > 7-19a > projected onto itself will yield itself. The scalarprojection will be the magnitude of itself.b. Using the formula for the scalar projection0 a > 0cos u50 a >5 0 a > 0cos 05 0 a > 0 (1)0.The vector projection is the scalar projectionmultiplied by10. a.a >0 a > ,0B –a O a A(2a > ) ? a >b.0 a > 2 0 a > 2050 0 a >520 a > 000 a > 0 3 a>0 a > 05 a > .So the vector projection is 2 0 a > 0a 0 a> 0b 5 2a > .0 a > 05 2Z Z5 2;The vector projection of AB >on the z-axis is(0, 0, Z)(0, 0, Z)2 35 2 3"0 2 1 0 2 21 Z5 2k > Zb. The angle made with the y-axis is bbcos b5"a 2 1 b 2 1 c 215"(22) 2 1 1 2 1 2 215"4 1 1 1 45 1 "95 1 3 ,Calculus and Vectors <strong>Solution</strong>s Manual
Therefore b5cos 21 a 1 3 b12. a.b.@BD > @@BD > @B8 70.5°Bbuc. In an isosceles triangle, CD is a median and aright bisector of BA. Therefore and have thesame magnitude projected on c > a > b >.d. Yes, not only do they have the same magnitude,but they are in the same direction as well whichmakes them have equivalent vector projections.13. a. Use the formula for the scalar projection of onb > 5 0 a > a >0cos u5 10 cos 135°527.07And the formula for the scalar projection of ona > 5 @b > b >@ cos u5 12 cos 135°528.49b. b12aCD135°10aQ OPOQ > is the vector projection of onOP > b >is the vector projection of a > a >on b >14. a. AB > 5 Point B 2 Point A5 (1, 3, 3) 2 (22, 1, 4)5 (3, 2, 21)The scalar projection of on isAB > ? OD >AB >OD >(3 321) 1 (2 3 2) 1 (21 3 2)@ OD > 5@"(21) 2 1 2 2 1 2 2(23) 1 4 1 (22)5"1 1 4 1 4Cabu u AD ccbuAb.52 1BC > 35 Point C 2 Point B5 (26, 7, 5) 2 (1, 3, 3)5 (27, 4, 2)The scalar projection of on isBC > ? OD >BC >OD >(27 321) 1 (4 3 2) 1 (2 3 2)@ OD > 5@"(21) 2 1 2 2 1 2 25 7 1 8 1 4"1 1 4 1 4AB > ? OD >@OD > @5 21"95 19"95 1931 BC> ? OD >@OD > @AC > 5 65 Point C 2 Point A5 (26, 7, 5) 2 (22, 1, 4)5 (24, 6, 1)The scalar projection of on isAC > ? OD >AC >OD >(24 321) 1 (6 3 2) 1 (1 3 2)@OD > 5@"(21) 2 1 2 2 1 2 25 4 1 12 1 2"1 1 4 1 45 18"952 1 3 1 1935 1835 1835 6c. Same lengths and both are in the direction of OD > .Add to get one vector.15. a. 1 5 cos 2 a1cos 2 b1cos 2 g22ab5 a"a 2 1 b 2 1 c 2b 1 a"a 2 1 b 2 1 c 2b2c1 a"a 2 1 b 2 1 c 2ba 25a 2 1 b 2 1 c 2 1 b 2a 2 1 b 2 1 c 2c 21a 2 1 b 2 1 c 27-20 <strong>Chapter</strong> 7: Applications of Vectors
5 a2 1 b 2 1 c 25 1b. a590°, b530°, g560°cos a5cos 90°5 0,x 5 0cos b5cos 30°5 "32 ,"3y is a multiple of 2 .cos g5cos 60°5 1 2 ,18. Answers may vary. For example:zB (0, c, d)yxA (a, b, 0)a 2 1 b 2 1 c 2 7-211z is a multiple of2 .Answers include Q0, "32 , 1 2R, Q0, "3, 1R, etc.c. If two angles add to 90° , then all three will add to180° .16. a. a5b5gcos a5cos b5cos gcos 2 a5cos 2 b5cos 2 g1 5 cos 2 a1cos 2 b1cos 2 g1 5 3 cos 2 x13 5 cos2 x1Å 3 5 cos x1x 5 cos 21 Å 3x 8 54.7°1b. For obtuse, use cos x 52 Å 3 .x 5 cos 21 1a2 Å 3 bx 8 125.3°17. cos 2 x 1 sin 2 x 5 1cos 2 x 5 1 2 sin 2 x1 5 cos 2 a1cos 2 b1cos 2 g1 5 (1 2 sin 2 a) 1 (1 2 sin 2 b) 1 (1 2 sin 2 g)1 5 3 2 (sin 2 a1sin 2 b1sin 2 g)sin 2 a1sin 2 b1sin 2 g527.6 The Cross Product of TwoVectors, pp. 407–4081. a.xa x bzaba > 3 b >is perpendicular to a > . Thus, their dot productmust equal 0. The same applies to the second case.za3bba1byab. is still in the same plane formed by andb > a > x1 b >thus a > 1 b >is perpendicular to a > 3 b > a >,making thedot product 0 again.c. Once again, is still in the same planeformed by and thus is perpendicular toa > 3 making the dot product 0 again.2. a > b > a > a > 2b > b >, a > 2 b >3 b >produces a vector, not a scalar. Thus, theequality is meaningless.3. a. It’s possible because there is a vector crossedwith a vector, then dotted with another vector,producing a scalar.b. This is meaningless because a > ? b >produces ascalar. This results in a scalar crossed with a vector,which is meaningless.yCalculus and Vectors <strong>Solution</strong>s Manual
c. This is possible. produces a vector, andc > 1 d > a > 3 b >also produces a vector. The result is a vectordotted with a vector producing a scalar.d. This is possible. produces a scalar, andc > 3 d > a > ? b >produces a vector. The product of a scalarand vector produces a vector.e. This is possible. produces a vector, andc > 3 d > a > 3 b >produces a vector. The cross product of avector and vector produces a vector.f. This is possible. a > 3 b >produces a vector. Whenadded to another vector, it produces another vector.4. a. (2, 23, 5) 3 (0, 21, 4)5 (23(4) 2 5(21), 5(0) 2 2(4),2(21) 2 (23)(0))5 (27, 28, 22)(2, 23, 5) ? (27, 28, 22) 5 0(0, 21, 4) ? (27, 28, 22) 5 0b. (2, 21, 3) 3 (3, 21, 2)5 (21(2) 2 3(21), 3(3) 2 2(2),2(21) 2 (21)(3))5 (1, 5, 1)(2, 21, 3) ? (1, 5, 1) 5 0(3, 21, 2) ? (1, 5, 1) 5 0c. (5, 21, 1) 3 (2, 4, 7)5 (21(7) 2 1(4), 1(2) 2 5(7),5(4) 2 (21)(2))5 (211, 233, 22)(5, 21, 1) ? (211, 233, 22) 5 0(2, 4, 7) ? (211, 233, 22) 5 0d. (1, 2, 9) 3 (22, 3, 4)5 (2(4) 2 9(3), 9(22) 2 1(4),1(3) 2 2(22))5 (219, 222, 7)(1, 2, 9) ? (219, 222, 7) 5 0(22, 3, 4) ? (219, 222, 7) 5 0e. (22, 3, 3) 3 (1, 21, 0)5 (3(0) 2 3(21), 3(1) 2 (22)(0),22(21) 2 3(1))5 (3, 3, 21)(22, 3, 3) ? (3, 3, 21) 5 0(1, 21, 0) ? (3, 3, 21) 5 0f. (5, 1, 6) 3 (21, 2, 4)5 (1(4) 2 6(2), 6(21) 2 5(4),5(2) 2 1(21))5 (28, 226, 11)(5, 1, 6) ? (28, 226, 11) 5 0(21, 2, 4) ? (28, 226, 11) 5 05. (21, 3, 5) 3 (0, a, 1)5 (3(1) 2 5(a), 5(0) 2 (21)(1),21(a) 2 3(0))If we look at the x component, we know that:3(1) 2 5(a) 52225(a) 5256. a. a > 3 b > a 5 15 (1(1) 2 1(5), 1(0) 2 0(1),0(5) 2 0(1))5 (24, 0, 0)b. Vectors of the form (0, b, c) are in theyz-plane. Thus, the only vectors perpendicular to theyz-plane are those of the form (a, 0, 0) because theyare parallel to the x-axis.7. a. (1, 2, 1) 3 (2, 4, 2)5 (2(2) 2 1(4), 1(2) 2 1(2), 1(4) 2 2(2))5 (0, 0, 0)b. (a, b, c) 3 (ka, kb, kc)5 (b(kc) 2 c(kb), c(ka) 2 a(kc),a(kb) 2 b(ka))Using the commutative law of multiplication wecan rearrange this:5 (bck 2 bck, ack 2 ack, abk 2 abk)5 (0,8. a. p > 0, 0)3 (q > 1 r > ) 5 (1, 22, 4) 3 3(1, 2, 7)1 (21, 1, 0)45 (1, 22, 4) 3 (1 2 1, 2 1 1, 7 1 0)5 (1, 22, 4) 3 (0, 3, 7)5 (22(7) 2 4(3), 4(0) 2 1(7),1(3) 1 2(0))p > 53 q > (226,1 p > 27,3 r > 3)5 (22(7) 2 4(2),b.p > 3 (q > 1 r > 5 (226, 27, 3)) 5 (4, 1, 2) 3 3(3, 1, 21)1 (0, 1, 2)45 (4, 1, 2) 3 (3, 1 1 1, 21 1 2)5 (4, 1, 2) 3 (3, 2, 1)5 (1(1) 2 2(2), 3(2) 2 4(1),4(2) 2 1(3))p > 3 q > 1 p > 3 r > 5 (23, 2, 5)5 (1(21) 2 2(1), 2(3) 2 4(21),4(1) 2 1(3)) 1 (1(2) 2 2(1),9. a.4(1) 2 1(7), 1(2) 1 2(1))1 (22(0) 2 4(1),4(21) 2 1(0), 1(1) 1 2(21))5 (222, 23, 4) 1 (24, 24, 21)2(0) 2 4(2), 4(1) 2 1(0))5 (23, 10, 1) 1 (0, 28, 4)i > 3 j > 5 (23, 2, 5)5 (1, 0, 0) 3 (0, 1, 0)5 (0 2 0, 0 2 0, 1 2 0)5 (0,5 k > 0, 1)7-22 <strong>Chapter</strong> 7: Applications of Vectors
.10.2j > 3 i > 5 (0, 21, 0) 3 (1, 0, 0)5 (0 2 0, 0 2 0, 0 2 (21))5 (0,j > 3 k > 5 k > 0, 1)5 (0, 1, 0) 3 (0, 0, 1)5 (1 2 0, 0 2 0, 0 2 0)5 (1,2k > 3 j > 5 i > 0, 0)5 (0, 0, 21) 3 (0, 1, 0)5 (0 2 (21), 0 2 0, 0 2 0)5 (1,c. k > 3 i > 5 i > 0, 0)5 (0, 0, 1) 3 (1, 0, 0)5 (0 2 0, 1 2 0, 0 2 0)5 (0,2i > 3 k > 5 j > 1, 0)5 (21, 0, 0) 3 (0, 0, 1)b.(by part a.)c. All the vectors are in the xy-plane. Thus, the crossproduct in part b. is between vectors parallel to thez-axis and so parallel to each other. The crossproduct of parallel vectors is12. Let x > 0 > .y > 5 (1, 0, 1)5 (1, 1, 1)Then5 (0 2 0, 0 2 (21), 0 2 0)5 (0,5 j > 1, 0)k(a 2 b 3 2 a 3 b 2 , a 3 b 1 2 a 1 b 3 , a 1 b 2 2 a 2 b 1 )? (a 1 , a 2 , a 3 )5 k(a 1 a 2 b 3 2 a 1 a 3 b 2 1 a 2 a 3 b 1 2 a 2 a 1 b 31 a 3 a 1 b 2 2 a 3 a 2 b 1 )5 k(0)a > 5 0is perpendicular to11. a. a > 3 b > k(a > 3 b > ).5 (2, 0, 0) 3 (0, 3, 0)5 (0 2 0, 0 2 0, 6 2 0)c > 3 d > 5 (0, 0, 6)5 (2, 3, 0) 3 (4, 3, 0)5 (0 2 0, 0 2 0, 6 2 12)(a > 3 b > 5 (0,) 3 (c > 0,3 d > 26)) 5 (0, 0, 6) 3 (0, 0, 26)5 (0 2 0, 0 2 0, 0 2 0)5 (0, 0, 0)z >x > 53 y > (1, 2, 3)5 (0 2 1, 1 2 1, 1 2 0)5 (21, 0, 1)(x > 3 y > ) 3 z > 5 (0 2 2, 1 2 (23), 23 2 0)5 (22, 4, 23)y > 3 z > 5 (3 2 2, 1 2 3, 2 2 1)x > 3 (y > 3 z > 5 (1, 22, 1)) 5 (0 1 2, 1 2 1, 22 2 0)Thus (x >13. (a > 32 b > y > 5 (2,) 3 z > 0, 22)) 3 (a > 2 x >1 b > 3 (y > 3 z > ).)By the distributive property of cross product:5 (a > 2 b > ) 3 a > 1 (a > 2 b > ) 3 b >By the distributive property again:5 a > 3 a > 2 b > 3 a > 1 a > 3 b > 2 b > 3A vector crossed with itself equals thus:52b >5 a > 35 a > 3 b > a > 135 2a > b > 2 b > a > 3323 b > (2a > a > b > 0 > b >,3 b > )7.7 Applications of the Dot Productand Cross Product, pp. 414–4151. By pushing as far away from the hinge aspossible, 0 r > 0 is increased making the cross productbigger. By pushing at right angles, sine is its largestvalue, 1, making the cross product larger.2. a. a > 3 b > 5 (1, 2, 1) 3 (2, 4, 2)5 (2(2) 2 1(4), 1(2)2 1(2), 1(4) 2 2(2))@a > 3 b > 5 (0, 0, 0)@ 5 0b. This makes sense because the vectors lie on thesame line. Thus, the parallelogram would just be aline making its area 0.3. a.b.f > ? s > 5 3 ? 150 5 450 Jy x40 m50° 392 NThe axes are tilted to illustrate the force of gravitycan be split up into components to find the part inthe direction of the motion. Let x be the componentof force going in the motion’s direction.cos (50°) 5x392x 5 (392) cos (50°)Now we have our force, so:(392) cos 50° N ? 40 m 8 10 078.91 JCalculus and Vectors <strong>Solution</strong>s Manual7-23
c.140 N20°250 mFirst find the x component of the force:(140) cos (20°) 5 xCalculate work:140 cos 20° N ? 250 m 8 32 889.24 Jd.100 N45°500 mFirst calculate the x component of the force:x 5 (100) cos (45°)Calculate work:100 cos4. a. i > 45°3 j > ? 5005 k > m 5 35 355.34 JThe square formed by the 2 vectors has an area of 1.The 2 vectors are on the xy-plane, thus, the crossproduct must be by the right hand rule.b. 2i > 3 j > 52k > k>Once again, the area is 1, making the possible vectorhave a magnitude of 1. Also, the 2 vectors are on thexy-plane again so the cross product must lie on thez axis. However, because of the right hand rule, theproduct must be this time.c. i > 3 k > 52j > 2k >The square has an area of 1, so the magnitude of thevector produced must be 1. The 2 vectors are on thexz-plane. The new vector must be on the y axismaking it – because of the right hand rule.d. 2i > 3 k > j >52j >The square has an area of 1. The 2 vectors are onthe xz-plane. So the new vector must be j >becauseof the right hand rule.5. a. a > 3 b > 5 (1, 1, 0) 3 (1, 0, 1)5 (1 2 0, 0 2 1, 0 2 1)@a > 3 b > 5 (1, 21, 21)@ 5 "1 1 1 1 1 5 "3So the area of the parallelogram is square units.b. a > 3 b > "35 (1, 22, 3) 3 (1, 2, 4)5 (28 2 6, 3 2 4, 2 1 2)@a > 3 b > 5 (214, 21, 4)@ 5 "196 1 1 1 16 5 "213So the area of the parallelogram is "213 square units.6. p > 3 q > 5 (a, 1, 21) 3 (1, 1, 2)5 (2 1 1, 22a 2 1, a 2 1)0 p > 3 q > 5 (3, 2a 1 1, a 2 1)0 5 "9 1 (2a 1 1) 2 1 (a 2 1) 2 5 "359 1 (2a 1 1) 2 1 (a 2 1) 2 5 359 1 4a 2 1 4a 1 1 1 a 2 2 2a 1 1 5 355a 2 1 2a 2 24 5 0a 5 22 6 "22 2 4(5)(224)2(5)22 6 225102125 2,57. a.BABAAC CAs we see from the picture, the area of the triangleABC is just half the area of the parallelogramdetermined by vectors AB >and AC > . Thus, we use themagnitude of the cross product to calculate the area.AB >AC > 5 (1 1 2, 0 2 1, 1 2 3) 5 (3, 21, 22)AB > 5 (2 1@AB > 3 AC > 2, 3 2 1, 2 2 3) 5 (4, 2, 21)3 AC > 5 (1 1 4, 23 1 8, 6 1 4) 5 (5, 5, 10)@ 5 "25 1 25 1 100 5 5"6Since triangle ABC is half the area of the5"6parallelogram, its area is 2 square units.b. This is just a different way of describing the first5"6triangle, thus the area is 2 square units.c. Any two sides of a triangle can be used tocalculate its area.8. @r > 3 f > @ 5 ( 0 r > 0sin (u)) @ f > @5 (0.14) sin (45°) ? 108 0.99 J9.AOA BNOCOB BWe know that the area of a parallelogram is equal toits height multiplied with its base. Its height is BNand its base is AC > 5 OB >as can be seen from thepicture. We can calculate the area using the givenvectors, then use the area to find BN.OA > 3 OB > 5 (8 2 4, 12 2 16, 4 2 6)5 (4, 24, 22)@ OA > 3 OB > @ 5 "16 1 16 1 4 5 "36 5 67-24 <strong>Chapter</strong> 7: Applications of Vectors
Now we need to calculate @OB > @ to know the lengthof the base.AC > 5 @OB > @ 5 "9 1 1 1 16 5 "26Substituting these results into the equation for area:@OB > @ ? BN 5 6"26 BN 5 6BN 5 6 or about 1.18"2610. a.p > 3 q > 5 (26 2 3, 6 2 3, 1 1 4)( p > 3 q > ) 3 r > 5 (29, 3, 5)5 (0 2 5, 5 1 0, 29 2 3)5 (25, 5, 212)a(1, 22, 3) 1 b(2, 1, 3) 5 (25, 5, 212)Looking at x components:a 1 2b 525;a 525 2 2by components:22a 1 b 5 5Substitute in a:10 1 4b 1 b 5 55b 525b 521Substitute b back into the x components:a 525 1 2;a 523Check in z components:3a 1 3b 521229b. p > 2q > ? r > 3 5212( p > ? r > 5 1 2 2 1 0 521? r > 5)q > 2 12 (q > 1 1? r > 0)p > 5 3521(2, 1, 3) 2 3(1, 22, 3)5 (2, 21, 23) 2 (3, 26, 9)5 (22 2 3, 21 1 6, 23 2 9)5 (25, 5, 212)Review Exercise, pp. 418–4211. a. a > 3 b > 5 (2 2 0, 21 1 1, 0 1 2)b. b > 3 c > 5 (2, 0, 2)5 (0 2 4, 25 1 5, 24 2 0)5 (24, 0, 24)c. 16d. The cross products are parallel, so the originalvectors are in the same plane.2. a.b. @b > 0 a > 0 5 "2 2 1 (21) 2 1 2 2 5 3c. a > @ 52 b > "6 2 1 3 2 1 (22) 2 5 75 (2 2 6, 21 2 3, 2 1 2)@a > 2 b > 5 (24, 24, 4)@ 5 "(24) 2 1 (24) 2 1 4 2 5 4"3d. a > 1 b > 5 (2 1 6, 21 1 3, 2 2 2)@a >e. a > 1 b > 5 (8, 2, 0)? b > @ 5 "8 2 1 2 2 1 0 2 5 2"17f. a > 5 2(6)2 2b > 2 1(3) 1 2(22) 5 55 (2 2 12, 21 2 6, 2 1 4)a > ? (a > 2 2b > 5 (210, 27, 6)) 5 2(210)3. a. If then will be twice thus collinear.b. x > 3 y > y > 2 1(27) 1x > 2(6) 521a 5 6,,5 (3, a, 9) ? (a, 12, 18) 5 03 a 1 12a 1 162 5 015a 52162a 5 2544. cos (u) 5 a> ? b > 5a > ? b > 0 a > 0 @b > @0 a > 5 4(23) 1 5(6) 1 20(22) 5 458@b > 0 5 "4 2 1 5 2 1 20 2 5 21@ 5 "(23) 2 1 6 2 1 22 2 5 23u5cos 21 a 458483 bu 8 18.52°5. a.y42OB OA x–4 –2 0 2 4–2–4b. We can use the dot product of the 2 diagonals tocalculate the angle. The diagonals are the vectorsOA > andOA > 1 OB >OA > 1 OB > OA > 2 OB > .2 OB > 5 (5 2 1, 1 1 4) 5 (4, 5)5 (5 1 1, 1 2cos (u) 5 (OA> 1 OB > 4) 5 (6,) ? (OA > 23)2 OB > )@OA > 1 OB > @@OA > 2 OB > @(OA > 1 OB > ) ? (OA > 2 OB > ) 5 4(6) 1 5(23) 5 9@OA > 1 OB > @ 5 "4 2 1 5 2 5 "41@OA > 2 OB > @ 5 "6 2 1 (23) 2 5 3"59u5cos 21 a3"205 bu 8 77.9°Calculus and Vectors <strong>Solution</strong>s Manual7-25
6.T130°The vertical components of the tensions must equalthe downward force:T 1 sin (30°) 1 T 2 sin (45°) 5 98 N12 T 1 1 1 "2 T 2 5 98T 1 5 196 2 "2T 2The horizontal components:T 1 cos (30°) 1 T 2 cos (45°) 5 0 N"32 T 1 2 1 "2 T 2 5 0Substitute in T 1 :98"3 2 "62 T 2 5298"32"6 2 "2T2 2 5298"3T 2 8 87.86NSubstitute this back in to get T 1 :T 1 8 71.74N7.x50 km/hx 5 "50 2 1 300 2 8 304.14tan 21 a 50300 b 8 9.46°The resultant velocity is 304.14 km>h, W 9.46° N.8. a.zx98 NyT 245°300 km/hxxyyb. x > 3 y > 5 (215 2 35, 25 2 15, 21 2 3)0 x > 3 y > 5 (250, 220, 18)0 5 "50 2 1 20 2 1 18 2 5 "3224 8 56.789. (0, 3, 25) 3 (2, 3, 1)5 (3 1 15, 210 2 0, 0 2 6) 5 (18, 210, 26)The cross product is perpendicular to the givenvectors, but its magnitude is"18 2 1 (210) 2 1 (26) 2 , or 2"115. A unit vectorperpendicular to the given vectors isa 9!115 , 2 5!115 , 2 3!115 b.10. a. cos (a) 5 AB> ? AC >AB > @AB > @@AC > @AB > AC > 5 (0, 23, 4) 2 (2, 3, 7) 5 (22, 26, 23)? AC > 5 (5, 2, 24) 2 (2, 3, 7) 5 (3, 21, 211)@AB > 522(3) 2 6(21) 2 3(211) 5 33@@AC > 5 "(22) 2 1 (26) 2 1 (23) 2 5 7@ 5 "3 2 1 (21) 2 1a5cos 21 AB> ? AC > (211) 2 5 "131@AB > @@AC > @335 cos 217"1318 65.68°b5cos 21 BA> ? BC >BA >BA > BC > 52AB > @BA > @@BC > @5 (2, 6, 3)? BC > 5 (5 2 0, 2 1 3, 24 2 4, ) 5 (5, 5, 28)@BA > 5 2(5) 1 6(5) 1 3(28) 5 16@@BC > 5 "2 2 1 6 2 1 3 2 5 7@ 5 "5 2 1 5 2 1 (28) 2 5 "144b5cos 21 167"1148 77.64°g5180 2a2b 8 36.68°So b 8 77.64° is the largest angle.b. The area is half the magnitude of the crossproduct of AB >and AC > .12 AB> 3 AC > 5 1 0 (63, 231, 20) 0 8 36.50211. The triangle formed by the two strings and theceiling is similar to a 3-4-5 right triangle, with the30 cm and 40 cm strings as legs. So the angleadjacent to the 30 cm leg satisfiescos u 5 3 57-26 <strong>Chapter</strong> 7: Applications of Vectors
The angle adjacent to the 40 cm leg satisfiescos f 5 4 5Also,sin u 5 4 and sin f 5 3 55 .Let T 1be the tension in the 30 cm string, and T 2bethe tension in the 40 cm string. ThenT 1 cos u2T 2 cos f503T 15 2 T 425 5 0Also,T 1 sin u1T 2 sin f5(10)(9.8) 5 984T 15 2 T 325 5 985 78.4 NSo the tension in the 30 cm string is 78.4 N and thetension in the 40 cm string is 58.8 N.12. a.30 NT 1 5 4 3 T 2a 4 3 T 2b 4 5 1 T 325 5 9853 T 2 5 98T 2 5 58.8 N54 N42 N25 NT 1 5 4 3 (58.8)b. The east- and west-pulling forces result in a forceof 5 N west. The north- and south-pulling forcesresult in a force of 12 N north. The 5 N west and12 N north forces result in a force pulling in thenorth-westerly direction with a force of"5 2 1 12 2 5 13 N,by using the Pythagorean theorem. To find the exactdirection of this force, use the definition of sine.Calculus and Vectors <strong>Solution</strong>s ManualIf u is the angle west of north, thensin u 5 513u 8 22.6°So the resultant is 13 N in a directionN22.6°W. The equilibrant is 13 N in a directionS22.6°E.13. a. Let D be the origin, then:A 5 (2, 0, 0), B 5 (2, 4, 0), C 5 (0, 4, 0),D 5 (0, 0, 0), E 5 (2, 0, 3), F 5 (2, 4, 3),G 5 (0, 4, 3) H 5 (0, 0, 3)b. AF >AF > AC > 5 (0, 4, 3)? AC > 5 (22, 4, 0)@AF > 5 0 1 16 1 0 5 16@@AC > 5 "0 2 1 4 2 1 3 2 5 5@ 5 "(22) 2cos (u) 5 AF> ? AC > 1 4 2 1 0 2 5 2"5@AF > @@AC > @u5cos 21 a 1610"5 bu 8 44.31°c. Scalar projection 5 @AF > @ cos (u)By part b.:5 (5) cos (44.31°)8 3.5814. a > ? b > 5 0 a > 0 @b > @ cos (u) 5 cos (u)cos (u) 52 1 (cosine law)(2a > 2 5b > 25213a > ) ?5213a > ? b > (b > 1? b > 1 6a > 3a > )? a > 2 5b > ? b >1 15213 cos (u) 1 15 7.515. a. The angle to the bank, u, will satisfysin (90° 2u) 5 2 390° 2u8 41.8°u 8 48.2°b. By the Pythagorean theorem, Kayla’s netswimming speed will be"3 2 2 2 2 5 "5 km>h.So since distance 5 rate 3 time, it will take hert 5 0.3"58 0.13 h8 8 min 3 secto swim across.7-27
c. Such a situation would have resulted in a righttriangle where one of the legs is longer than thehypotenuse, which is impossible.16. a. The diagonals are OA > 1 OB >andOA > 2 OB > .OA > 1 OB > 5 (3 2 6, 2 1 6, 26 2 2)OA > 2 OB > 5 (23, 8, 28)5 (3 1 6, 2 2 6, 26 1 2)b. OA > 5? OB > (9, 24, 24)@OA > 5 3(26) 1 2(6) 2 6(22) 5 6@ 5 "3 2 1 2 2 1 (26) 2 5 7@OB > @ 5 "(26) 2 1cos (u) 5 OA> ? OB > 6 2 1 (22) 2 5 2"19@OA > @@OB > @6u5cos 21 5 a14"19 b8 84.36°17. a. The z value is double, so if andthe vector will be collinear.b. If p > and q > q >a 5 4b 524,are perpendicular, then their dotproduct will equal 0.p > ? q > 5 2a 2 2b 2 18 5 0c. Let a 5 9, and b then we have a vectorperpendicular to p > 5 0,. Now it must be divided by itsmagnitude to make it a unit vector:0 p > 0 5 "81 1 0 1 324 5 9"5So the unit vector is:a 1 !5 , 0, 218. a. m > !5 b? n >0 m > 5 2"3 2 2"3 1 3 5 30 n > 0 5 "3 1 4 1 9 5 40 5 "4 1cos (u) 5 m> ? n > 3 1 1 5 2"20 m > 00n > 0u5cos 21 a 38"2 b8 74.62°b. Scalar projection 5 0 n > 0cos (u)5 2"2 cos (74.62°)8 0.75c. Scalar projection multiplied with the unit vectorin the direction of m > :5 (0.75) m>0 m > 0("3, 22, 23)5 (0.75)45 (0.1875)("3, 22, 23)d. m > ? k > 523u5cos 21 a 234 b8 138.59°19. a. If the dot product is 0, then the vectors areperpendicular:(1, 0, 0) ? (0, 0, 21) 5 0 1 0 1 0 5 0(1, 0, 0) ? (0, 1, 0) 5 0 1 0 1 0 5 0(0, 0, 21) ? (0, 1, 0) 5 0 1 0 1 0 5 0 specialb. a 1 !2 , 1 21, 0b ? a!2 !3 , 1!3 , 1!3 b52 1 !6 1 1 !6 1 05 0a 1 !2 , 1, 0b ? (0, 0, 21) 5 0 1 0 1 0 5 0!2a 21!3 , 1!3 , 1b ? (0, 0, 21)!35 0 1 0 12 1p > 3 q > !3 52 1 !3not special20. a.5 (22(1) 2 1(21), 1(2) 2 1(1), 1(21) 1 2(2))5 (21,b. p > p > 2 q > 1, 3)1 q > 5 (21, 21, 0)5 (3, 23, 2)(p > 2 q > ) 3 (p > 1 q > ) 5 (22 2 0, 0 1 2, 3 2 (23))c. p > 3 r > 5 (22, 2, 6)5 (4 2 1, 0 1 2, 1 2 0)5 (3, 2, 1)( p > 3d. p > r > )3 q > ? r > 5 0 1 2 2 2 5 05 (22 1 1, 2 2 1, 21 1 4)5 (21, 1, 3)21. Since the angle between the two vectors is 60°,the angle formed when they are placed head-to-tailis 120°. So the resultant, along with these twovectors, forms an isosceles triangle with top angle120° and two equal angles 30°. By the cosine law,the two equal forces satisfy20 2 5 2F 2 2 2F 2 cos 120°F 2 5 4003F 5 20"3822. a > 11.553 b > N5 (2 2 0, 25 2 3, 0 2 10)5 (2, 28, 210)7-28 <strong>Chapter</strong> 7: Applications of Vectors
23. First we need to determine the dot product ofandx > ? y > y > :5 0 x > 00y > 0cos u5 (10) cos (60°)(x > 52 2y > 5) ? (x > 1 3y > )By the distributive property:5 x > ? x > 1 3x > ? y > 2 2x > ? y > 2 6y > ? y >5 4 1 15 2 10 2 1505214124. 0 (2, 2, 1) 0 5 "2 2 1 2 2 1 1 2 5 3Since the magnitude of the scalar projection is 4,the scalar projection itself has value 4 or 24.If it is 4, we get(1, m, 0)?(2, 2, 1)5 432 1 2m 5 12m 5 5If it is 24, we get(1, m, 0)?(2, 2, 1)52432 1 2m 5 212m 5 27So the two possible values for m are 5 and25. a > ?0 a > j > 27.5230 5 "144 1 9 1 16 5 13u5cos 21 a 2313 b8 103.34°26. a.b. CF > C 5 (3, 0, 5), F 5 (0, 4, 0)5 (0, 4, 0) 2 (3, 0, 5) 5 (23, 4, 25)c. @CF > @OP > 5 "9 1 16 1 25 5 5"25CF > @OP > (3, 4, 5)@? OP > 5 "9 1 16 1 25 5 5"2529 1 16 2 25 5218u5cos 21 a 21850 b8 111.1°27.d 130°50°ea. Using properties of parallelograms, we know thatthe other angle is 130 ° (Angles must add up to360 ° , opposite angles are congruent).Using the cosine law,@d > 1 e > @ 2 5 3 2 1 5 2 2 2(3)(5) cos 130°@d > 1 e > @ 8 7.30b. Using the cosine law,@d > 2 e > @ 2 5 3 2 1 5 2 2 2(3)(5) cos 50°@d > 2 e > @ 8 3.84c. is the vector in the opposite direction ofd > e > 22 e > d >, but with the same magnitude. So:@e > 2 d > @ 5 @d > 2(i > e > @ 81 j > 3.84) ? (i > )28. a. Scalar:@i > 5 1@Vector:(i > 1 j > ) ? (j > )b. Scalar:@ j > 5 1@Vector:1a i>1a j>@i > b 5 i >@@ j > b 5 j >@(i > 1 j > ) ? (k > 1 j > )c. Scalar:@k > 1 j > 5 1 @ "21Vector:"2 ? (k> 1 j > )@k > 1 j > 5 1 @ 2 (k> 1 j > )29. a. If its magnitude is 1, it’s a unit vector:0 a > 0 5 " 1 not a unit vector@b > 4 1 1 9 1 136 2 1@ 5 " 1 3 1 1 3 1 1 3 5 1, unit vector0 c > 0 5 " 1 4 1 1 2 1 1 4 5 1, unit vector@d > @ 5 not a unit vectorb. a >"1 1 1 1 1 2 1,is. When dotted with d > , it equals 0.30. 25 ?31. a.b. a > a > sin? b > (30°) ? 0.6 5 7.50 J5 6 2 5 2 1 5 0with the x-axis:0 a > 0 5 "4 1 25 1 1 5 "30cos (a) 5 2a > "30with the y-axis:cos (b) 5 5a > "30with the z-axis:cos (g) 5 21@b > "30@b > 5 "9 1 1 1 1 5 "11with the x-axis:cos (a) 5 3"11x > 7-29Calculus and Vectors <strong>Solution</strong>s Manual
7-30with the y-axis:cos (b) 5 21b > "11with the z-axis:cos (g) 5 1"11c. m > 1 ? m > 2 5 6!330 2 5!330 2 1!330 5 032. Need to show that the magnitudes of thediagonals are equal to show that it is a rectangle.@3i > 1 3j > 1 10k > @ 5 "9 1 9 1 100 5 "118@2i > 1 9j > 2 6k > @ 5 "1 1 81 1 36 5 "11833. a. Direction cosine for x-axis:cos (30°) 5 "32We know the identitycos 2 a1cos 2 b1cos 2 g51.Since a530g, and b5g, we get2 cos 2 b51 2 3cos b5cos g56 142"2cos a5 "32So there are two possibilities, depending uponwhether b5gis acute or obtuse.b. If g is acute, thencos g5 12"2g 8 69.3°If Á is obtuse, thencos g5 12"2g 8 110.7°34. a > ? b > 5 0 a > 0 @b > @ cos (u) 5 1(a >ma > 2 3b >? a > ) ?1 a > (ma >? b > 1 b > 2)2 3ma > 5 0? b > 2 3b > ? b > 5 0m 1 1 2 2 3 2 m 2 6 2 5 035.2 1 2 m 5 5 2m 525a > a > ? b >@a > 1 b > 5 0 2 20 1 12 5281 b > 5 (21, 21, 28)@ 5 "1 1 1 1 64 5 "66a >@a > 2 b >2 b > 5 (1, 9, 24)@ 5 "1 1 81 1 16 5 "9814 @a> 1 b > @ 2 2 136.0 c > c > 5 b > 4 @a> 2 b > @ 2 5 660 2 5 @b > 2 a >4 2 984 5282 a > 2@5 (b >5 b > 25 0 a > ? b > a > ) ?2 a > (b >02 1 @b > ? b > 2 a >1@ 2 2 2a > a > )?? b > a > 2 a > ? b >5 0 a > 0 2 1 @b > @ 2 2 20 a > 0 @b > @ cos u37. AB >@AB > 5 (2, 0, 4)@@AC > 5 "4 1 0 1 16 5 2"5@@AC > 5 (1, 0, 2)@BC > 5 "1 1 0 1 4 5 "5@BC > 5 (21, 0, 22)@ 5 "1 1 0 1 4 5 "5cos A 5 AB> ? AC >@ AB > @@AC > @5 10105 1But this means that angle A 5 0° , so that thistriangle is degenerate. For completeness, though,notice that BC > 52AC > and AB > 5 2 AC >. Thismeans that point C sits at the midpoint of the linesegment joining A and B. So angleC 5 180° and angle B 5 0° . Socos B 5 1;cos C 521.The area of triangle ABC is, of course, 0.<strong>Chapter</strong> 7 Test, p. 4221. a. We use the diagram to calculate a > 3 b > , notinga 1 521, a 2 5 1, a 3 5 1 and b 1 5 2, b 2 5 1,b 3 523.a > b >1 1xx 5 1(23) 2 1(1) 5241 23yy 5 1(2) 2 (21)(23) 52121 2zz 521(1) 2 1(2) 5231 1So, a > 3 b > 5 (24, 21, 23)<strong>Chapter</strong> 7: Applications of Vectors
. We use the diagram again:c >1 1xx 5 1(27) 2 (23)(1) 52423 2 7yy 523(5) 2 (2)(27) 5212 5zz 5 2(1) 2 1(5) 5231 1So,c. a > b > 3? (b > c > 53 c > (24, 21, 23)) 5 (21, 1, 1) ? (24, 21, 23)5 (21)(24) 1 (1)(21)1 (1)(23)5 0d. We could use the diagram method again, or, wenote that for any vectorsso letting y > 5 x > x > ,we have from the lastequation. Since a > 3 b > x > y > , x >5 from the first twoparts of the problem, (a > b > 3 x > 3 y > 52(y > 3 x > ),,33 b > c > 5 0) 3 (b > 3 c > ) 5 0.2. a. To find the scalar and vector projections ofon we need to calculate anda > b > ,? b > a > ? b >@b > @ 5 "b > a >? b >5 (1, 21, 1) ? (2, 21, 22)5 (1)(2) 1 (21)(21) 1 (1)(22)@b > 5 1@ 5 "2 2 1 (21) 2 1 (22) 2So, @b > 5 3@ 5 3The scalar projection of on isa > >a > b >? b> 5 1 and@b @3,the vector projection of a > on b >is>a a> ? b> 2 bb > 5 1@b @9(2, 21, 22).b. We find the direction cosines for b > :cos (a) 5 b 1@b > 5 2 @ 3a 8 48.2°.cos (b) 5 b 2@b > 5 21@ 3b 8 109.5°.cos (g) 5 b 3@b > 5 22@ 3g 8 131.8°.c. The area of the parallelogram is the magnitude ofthe cross product.a > b >21 21xx 5 (21)(22) 2 1(21) 5 31 22yy 5 1(2) 2 (1)(22) 5 41 2zz 5 (1)(21) 2 (21)(2) 5 121So, a > 21and thus,@a > 3 b >3 b > 5 (3, 4, 1)@ 5 "3 2 1 4 2 1 1 25 "26So the area of the parallelogram formed by a > andis "26 or 5.10 square units.3. We first draw a diagram documenting thesituation:E40 N60°50 N40 N60°120°50 NIn triangle DEF, we use the cosine law:@R > @ 5 "40 2 1 50 2 2 2(40)(50) cos (120°)@R > @ 8 78.10We now use the sine law to find /EDF:sin /EDF sin /DEF@EF > 5@ @R > @sin /EDF sin 120°850 78.10sin /EDF 8 0.5544/EDF 8 33.7°The equilibrant force is equal in magnitude andopposite in direction to the resultant force, so bothforces have a magnitude of 78.10 N. The resultantmakes an angle 33.7°to the 40 N force and 26.3°tothe 50 N force. The equilibrant makes an angle 146.3°to the 40 N force and 153.7°to the 50 N force.DRGb >Fb > 7-31Calculus and Vectors <strong>Solution</strong>s Manual
4. We find the resultant velocity of the airplane.2.5 m/sF1000 km/hPosition diagramESince the airplane’s velocity is perpendicular to thewind, the resultant’s magnitude is given by thePythagorean theorem:@R > @@R > 5 "1000 2 1 100 2@ 8 1004.99The angle is determined using the tangent ratio:tan /EDF 5 1001000/EDF 8 5.7°Thus, the resultant velocity is 1004.99 km>h,N 5.7°W (or W 84.3°N).5. a. The canoeist will travel 200 m across thestream, so the total time he will paddle is:t 5dr canoeistt 5 200 m2.5 m>st 5 80 sThe current is flowing 1.2 m>s downstream, so thedistance that the canoeist travels downstream is:d 5 r current 3 td 5 (1.2 m>s)(80 s)d 5 96 mSo, the canoeist will drift 96 m south.b. In order to arrive directly across stream, thecanoeist must take into account the change in hisvelocity caused by the current. That is, he mustinitially paddle upstream in a direction such thatthe resultant velocity is directed straight acrossthe stream. The resultant velocity:E1.2 m/sRFG DVector diagramSince the resultant velocity is perpendicular to thecurrent, the direction in which the canoeist shouldhead is determined by the sine ratio.sin /EDF 5 1.22.5/EDF 8 28.7°The canoeist should head 28.7° upstream.6. The area of the triangle is exactly:A DABC 5 1 2 @AB> 3 BC > @AB > 5 (2, 1, 3) 2 (21, 3, 5)BC > 5 (3, 22, 22)5 (21, 1, 4) 2 (2, 1, 3)AB > 5 (23,BC > 0, 1)22 x 0x 5 (22)(1) 2 (22)(0) 52222 y 1y 5 (22)(23) 2 (3)(1) 5 33 z 23z 5 (3)(0) 2 (22)(23) 52622 0So, AB > 3 BC > 5 (22, 3, 26) and@AB > 3 BC > @ 5 "(22) 2 1 3 2 1 (26) 2So, A DABC 5 1 2 @AB> 3 BC > @ 5 7 2 .The area of the triangle is 3.50 square units.7.T 125 kgThe system is in equilibrium (i.e. it is not moving),so we know that the horizontal components of T > 1and are equal:T > 24585 "495 7708@T > 1@ sin (45°) 5 @T > 2@ sin (70°)@T > 2@ 5T2sin (45°)sin (70°) @T> 1@DRF7-32 <strong>Chapter</strong> 7: Applications of Vectors
Also, the vertical component of T > 1 1 T > 2 must equalthe gravitational force on the block:@T > 1@ cos 45° 1 @T > 2@ cos 70° 5 (25 kg)(9.8 m>s 2 )Substituting in for , we find that:@T > T > 21@ cos 45° 1sin 45°@T> 1@ cos 70° 5 (25 kg) (9.8 m>s 2 )sin 70°@T > 1@ acos 45° 1So, we can now find@T > sin (45°)2@ 5sin (70°) @T> 1@@T > 2@ 8sin (45°)(254.0 N)sin (70°)@T > 2@ 8 191.1 Nsin 45°cos 70°b 5 245 Nsin 70°@T > 1@ (0.9645) 8 245 N@T > 1@ 8 254.0 NThe direction of the tensions are indicated in thediagram.8. a. We explicitly calculate both sides of theequation. The left side is:x > ? y > 5 (3, 3, 1) ? (21, 2, 23)5 (3)(21) 1 (3)(2) 1 (1)(23)5 0We perform a few computations before computingthe right side:x > 1 y > 5 (3, 3, 1) 1 (21, 2, 23)5 (2, 5, 22)0 x > 1 y > 0 2 5 (x > 1 y > ) ? (x > 1 y > )5 2 2 1 5 2 1 (22) 2x > 2 y > 5 335 (3, 3, 1) 2 (21, 2, 23)0 x > 2 y > 5 (4,0 2 5 (x > 1, 4)2 y > ) ? (x > 2 y > )5 4 2 1 1 2 1 4 25 33Thus, the right side is14 0 x> 1 y > 02 2 1 4 0 x> 2 y > 02 5 1 4 (33) 2 1 4 (33)5 0So, the equation holds for these vectors.b. We now verify that the formula holds in general.We will compute the right side of the equation, butwe first perform some intermediary computations:0 x > 1 y > 0 2 5 (x >5 (x > 10 x > 2 y > 5 (x > ? x > y > ) ? (x >0 2 5 (x > ? x > ) 1 (x > 1?)5 (x > 2? x > y > 1 2(x > y > y > ))) ? (x > ? y > 1 (y > ?) 115 (x > (2y > (x > 2 y > ) 1 (y > x > )? y > 1 (y > ? y > )))? x > ? 2y > ? 2y > ) 1 (2y > ? x > )) 2 2(x > )? y > ) 1 (y > ? y > )So, the right side of the equation is:14 0 x> 1 y > 02 2 1 4 0 x> 2 y > 02 5 1 4 (4(x> ? y > ))5 x > ? y >Thus, the equation holds for arbitrary vectors.Calculus and Vectors <strong>Solution</strong>s Manual7-33
Review of Prerequisite Skills, p. 3501. The velocity relative to the ground has amagnitude equivalent to the hypotenuse of atriangle with sides 800 and 100. So, by thePythagorean theorem we can find the magnitude of8005 640 000 1 10 0005 650 000v 5 "650 0008 806 km>h100Applications of VectorsCHAPTER 7the velocity.v 2 5 800 2 1 100 2m 2 5 5 2 1 12 2 2 2(5)(12) cos 1355 25 1 144 2 120a 2 !2 b25 169 1 84.855 253.85m 5 "253.858 15.93 units12 2 5 15.93 2 1 5 2 2 2(15.93)(5) cos u144 5 253.76 1 25 2 159.3 cos u2134.76 52159.3 cos ucos u5 134.76159.3u5cos 21 a 134.76159.3 b8 32.2°So the displacement is 15.93 units, W 32.2° N.3.zutan u 5 100800u5tan 21 a 100800 bu 8 7.1°The velocity of the airplane relative to the ground isabout 806 km>h N 7.1° E.2.12displacementu5The angle between the two displacements is 135°.The magnitude, m, and the angle, u, of thedisplacement can be found using the cosine law.x4. a. (3, 22, 7)l 5 magnitude5 "3 2 1 (22) 2 1 7 25 "9 1 4 1 495 "628 7.87b. (29, 3, 14)l 5 magnitude5 "(29) 2 1 3 2 1 14 25 !81 1 9 1 1965 !2868 16.91C(–2, 0, 1)B(–3, 2, 0)yA(0, 1, 0)D(0, 2, –3)Calculus and Vectors <strong>Solution</strong>s Manual7-1
c. (1, 1, 0)l 5 magnitude5 "28 1.41d. (2, 0, 29)l 5 magnitude5 "858 9.225. a. A(x, y, 0)In the xy-plane at the point (x, y).b. B(x, 0, z)In the xz-plane at the point (x, z).c. C(0, y, z)In the yz-plane at the point (y, z).6. a. (26,5 (26i > 0)5 (26i > 1 0j > 1 7(1,5 i > 2 7j > 1 0j > ) 1 7(i > 21)) 1 (7i > 2 j > )2 7j > )b. (4, 21,5 (4i >5 6i > 2 j > 3) 22 2j > 1 3k > (22, 1, 3)) 2 (22i > 1 j > 1 3k > )c. 2(21,5 2(2i > 1, 3)5 (22i > 1 j > 1 3(22,1528i > 1 2j > 3k > 3, 21))1 11j > 1 6k > 1 3(22i >1 3k > ) 1 (26i > 1 3j >1 9j > 2 k > )2 3k > )d.5 "1 2 1 1 2 1 0 25 "2 2 1 0 2 1 (29) 25 "4 1 0 1 812 1 2 (4, 26, 8) 1 3 (4, 26, 8)252 1 2 (4i> 2 6j > 1 8k > ) 1 3 2 (4i> 2 6j > 1 8k > )5 (22i >5 4i > 12 6j > 3j > 21 8k > 4k > ) 1 (6i > 2 9j > 1 12k > )7. a. a >5 (3i > 1 b >5 i > 1b. a > 1 3j > 2j > 22 k > k > ) 1 (22i > 1 j > )25 (3i > b >5 (3i > 1 2j >5 5i > 1c. 2a > 1 j > 2j > 2 k >22 k > k > ) 2 (22i >) 1 (2i > 12 j > j > ))25 2(3i > 3b >5 (6i > 1 2j >5 12i > 1 4j > 2 k >1 j > 2 2k > ) 2 3(22i >2 2k > ) 1 (6i > 12 3j > j > ))7.1 Vectors as Forces, pp. 362–3641. a. 10 N is a melon, 50 N is a chair, 100 N is acomputerb. Answers will vary.2. a.b. 180°3.The forces should be placed in a line along thesame direction.4. For three forces to be in equilibrium, they mustform a triangle, which is a planar figure.5.equilibrant30 N10 N20 Nf 2f 110 N20 Nresultanta. The resultant is equivalent in magnitude to thehypotenuse, h, of the triangle with 5 and 12 as sidesand is directed northeast at an angle of sin 21 12 h .Thus, the resultant is "5 2 1 12 2 5 13 N at an angle21 12of sin 13 5 N 22.6° E. The equlibrant is equal inmagnitude and opposite in direction of the resultant.Thus, the equilibrant is 13 N at an angle of S 22.6° W.b. The resultant is "9 2 1 12 2 5 15 N at an angle of21 12sin 15 5 S 36.9° W. The equilibrant, then, is 15 Nat N 36.9° E.6. For three forces to form equilibrium, they mustbe able to form a triangle or a balanced line, soa. Yes, since 3 1 4 . 7 these can form a triangle.b. Yes, since 9 1 40 . 41 these can form a triangle.c. No, since "5 1 6 , 9 these cannot form atriangle.d. Yes, since 9 1 10 5 19, placing the 9 N and 10 Nforce in a line directly opposing the 19 N forceachieves equilibrium.7-2 <strong>Chapter</strong> 7: Applications of Vectors
c. (1, 1, 0)l 5 magnitude5 "28 1.41d. (2, 0, 29)l 5 magnitude5 "858 9.225. a. A(x, y, 0)In the xy-plane at the point (x, y).b. B(x, 0, z)In the xz-plane at the point (x, z).c. C(0, y, z)In the yz-plane at the point (y, z).6. a. (26,5 (26i > 0)5 (26i > 1 0j > 1 7(1,5 i > 2 7j > 1 0j > ) 1 7(i > 21)) 1 (7i > 2 j > )2 7j > )b. (4, 21,5 (4i >5 6i > 2 j > 3) 22 2j > 1 3k > (22, 1, 3)) 2 (22i > 1 j > 1 3k > )c. 2(21,5 2(2i > 1, 3)5 (22i > 1 j > 1 3(22,1528i > 1 2j > 3k > 3, 21))1 11j > 1 6k > 1 3(22i >1 3k > ) 1 (26i > 1 3j >1 9j > 2 k > )2 3k > )d.5 "1 2 1 1 2 1 0 25 "2 2 1 0 2 1 (29) 25 "4 1 0 1 812 1 2 (4, 26, 8) 1 3 (4, 26, 8)252 1 2 (4i> 2 6j > 1 8k > ) 1 3 2 (4i> 2 6j > 1 8k > )5 (22i >5 4i > 12 6j > 3j > 21 8k > 4k > ) 1 (6i > 2 9j > 1 12k > )7. a. a >5 (3i > 1 b >5 i > 1b. a > 1 3j > 2j > 22 k > k > ) 1 (22i > 1 j > )25 (3i > b >5 (3i > 1 2j >5 5i > 1c. 2a > 1 j > 2j > 2 k >22 k > k > ) 2 (22i >) 1 (2i > 12 j > j > ))25 2(3i > 3b >5 (6i > 1 2j >5 12i > 1 4j > 2 k >1 j > 2 2k > ) 2 3(22i >2 2k > ) 1 (6i > 12 3j > j > ))7.1 Vectors as Forces, pp. 362–3641. a. 10 N is a melon, 50 N is a chair, 100 N is acomputerb. Answers will vary.2. a.b. 180°3.The forces should be placed in a line along thesame direction.4. For three forces to be in equilibrium, they mustform a triangle, which is a planar figure.5.equilibrant30 N10 N20 Nf 2f 110 N20 Nresultanta. The resultant is equivalent in magnitude to thehypotenuse, h, of the triangle with 5 and 12 as sidesand is directed northeast at an angle of sin 21 12 h .Thus, the resultant is "5 2 1 12 2 5 13 N at an angle21 12of sin 13 5 N 22.6° E. The equlibrant is equal inmagnitude and opposite in direction of the resultant.Thus, the equilibrant is 13 N at an angle of S 22.6° W.b. The resultant is "9 2 1 12 2 5 15 N at an angle of21 12sin 15 5 S 36.9° W. The equilibrant, then, is 15 Nat N 36.9° E.6. For three forces to form equilibrium, they mustbe able to form a triangle or a balanced line, soa. Yes, since 3 1 4 . 7 these can form a triangle.b. Yes, since 9 1 40 . 41 these can form a triangle.c. No, since "5 1 6 , 9 these cannot form atriangle.d. Yes, since 9 1 10 5 19, placing the 9 N and 10 Nforce in a line directly opposing the 19 N forceachieves equilibrium.7-2 <strong>Chapter</strong> 7: Applications of Vectors
7. Arms 90 cm apart will yield a resultant with asmaller magnitude than at 30 cm apart. A resultantwith a smaller magnitude means less force tocounter your weight, hence a harder chin-up.8. Using the cosine law, the resultant has a magnitude,r,of5 6 2 1 8 2 2 2(6)(8)a2 1 2 b5 36 1 64 1 485 148r 5 "1488 12.17 NUsing the sine law, the resultant’s angle, u, can befound bysin u sin 120°58 12.17"32sin u5812.17u5sin 21 812.178 34.7° from the 6 N force toward the 8 Nforce. The equilibrant, then, would be 12.17 N at180° 2 34.7° 5 145.3° from the 6 N force awayfrom the 8 N force.9.10 N"32Now we look at x and We know> 1 x 2 .x 1 5 @ f 1 @ sin 15>x 2 5 @ f 2 @ sin 75x 1 1 x 2 5 10> >So @ f1 @ sin 15 1 @ f2 @ sin 75 5 10>Substituting then solving for f 2 yields> cos 75>@ f2 @ @ f 2 @ sin 75 5 10cos 15 sin 15 1> cos 75@ f 2 @ a sin 15 1 sin 75b 5 10cos 15>@ f 2 @ (1.035) 5 10>@ f 2 @ 5 9.66 N>Now we solve for f 1 :> > cos 75@ f 1 @ 5 @ f 2 @cos 15> cos 75@ f 1 @ 5 (9.66)cos 15>@ f 1 @ 5 (9.66)(0.268)@ f 1>@ 5 2.59 NSo the force 15° from the 10 N force is 9.66 N andthe force perpendicular to it is 2.59 N.10. The force of the block is(10 kg)(9.8 N>kg) 5 98 N. The component of thisforce parallel to the ramp is(98) sin 30° 5 (98)A 1 2B 5 49 N, directed down theramp. So the force preventing this block frommoving would be 49 N directed up the ramp.11. a.r 2 5 @ f 1>@ 2 1 @ f 2>@ 2 2 2 @ f 1>@@f2>@ cos 120°7-3f 1f 27 N13 N>f 1 5 force 15° from the 10 N force> >f 2 5 force perpendicular to f1>x 1 5 component of f 1 parallel to the 10 N force>x 2 5 component of f 2 parallel to the 10 N force> >We know that the components of f 1 and f 2perpendicular to the 10 N force must be equal, so wecan write> >@ f1 @ cos 15 5 @ f 2 @ cos 75> > cos 75@ f 1 @ 5 @ f 2 @cos 15Calculus and Vectors <strong>Solution</strong>s Manualb. Using the cosine law for the angle, u, we have13 2 5 8 2 1 7 2 2 2(8)(7) cos u169 5 64 1 49 2 112 cos u56 52112 cos ucos u5 256112u5cos5 120212128 N
This is the angle between the vectors when placedhead to tail. So the angle between the vectors whenplaced tail to tail is 180° 2 120° 5 60°.12. The 10 N force and the 5 N force result in a 5 Nforce east. The 9 N force and the 14 N force resultin a 5 N force south. The resultant of these is nowequivalent to the hypotenuse of the right trianglewith 5 N as both bases and is directed 45° south ofeast. So the resultant is "5 2 1 5 2 5 "50 8 7.1 N45° south of east.13.equlibranta. Using the Pythagorean theorem,>@ f 1 @ 2 >1 @ f 2 @ 2 5 25 25 49>@ f 2 @ 5 7>b. The angle, u, between f 1 and the resultant isgiven bysin u5 @ f 225sin u5 725@ f 2>@ 2 5 25 2 2 @ f 1>@2>@u5sin 21 7258 16.3°>So the angle between f 1 and the equilibrant is180° 2 16.3° 5 163.7°.14. a.1 N5 25 2 2 24 260°= 24 N f 1f 21 Nresultant = 25 N60°60°1 NFor these three equal forces to be in equilibrium, theymust form an equilateral triangle. Since the resultantwill lie along one of these lines, and since all anglesof an equilateral triangle are 60°, the resultant will beat a 60° angle with the other two vectors.b. Since the equilibrant is directed opposite theresultant, the angle between the equilibrant and theother two vectors> >is 180° 2 60° 5 120°.15. Since f 1 and f 2 act opposite one another,>they>net a 10 N force directed west. Since f 3 and f 4 actopposite one another, they net a 10 N force directed45° north of east.>So using the cosine law to findthe resultant, f r ,>@ f r @ 2 5 10 2 1 10 2 2 2(10)(10) cos 45°5 200 2 200 cos 45°@ f r>@ 5Å200 2 200a "22 b8 7.65 NSince our net forces are equal at 10 N, the angle ofthe resultant is directed halfway>between>the two, or1at 2 (135°) 5 67.5° from f2 toward f 3 .16.T 2 T 1Let T 1 be the tension in the 30° rope and T 2 be thetension in the 45° rope.Since this system is in equilibrium, we know thatthe horizontal components of T 1 and T 2 are equaland opposite and the vertical components add to beopposite the action of the mass. Also, the forceproduced by the mass is (20 kg)(9.8 N>kg) 5 196 N.So we have a system of two equations: the first,(T 1 ) cos 30° 5 (T 2 ) cos 45° represents the balanceof the horizontal components, and the second,(T 1 ) sin 30° 1 (T 2 ) sin 45° 5 196 represents thebalance of the vertical components with the mass.So solving this system of two equations with twovariable gives the desired tensions.T 1 cos 30° 5 T 2 cos 45°cos 45°T 1 5 T 2cos 30°T 1 sin 30° 1 T 2 sin 45° 5 196aT 25 200 2 200a "22 b20 kgcos 45°cos 30° b sin 30° 1 T 2 sin 45° 5 196cos 45°T 2 aa b sin 30° 1 sin 45°b 5 196cos 30°7-4 <strong>Chapter</strong> 7: Applications of Vectors
T 2 (1.12) 5 196T 2 8 175.73 Ncos 45°T 1 5 (175.73)cos 30°8 143.48 NThus the tension in the 45° rope is 175.73 N and thetension in the 30° rope is 143.48 N.17.40 cmThus the tension in the 24 cm string is 39.2 N andthe tension in the 32 cm string is 29.4 N.18.xresultant35°u2x24 cm 32 cm5 kgFirst, use the Cosine Law to find the angles thestrings make at the point of suspension. Let u 1 be theangle made by the 32 cm string and u 2 be the anglemade by the 24 cm string.24 2 5 32 2 1 40 2 2 2(32)(40) cos u 122048 522560 cos u 1212048u 1 5 cos25608 36.9°32 2 5 24 2 1 40 2 2 2(24)(40) cos u 221152 521920 cos u 2211152u 2 5 cos19208 53.1°A keen eye could have recognized this triangle as a3-4-5 right triangle and simply used the Pythagoreantheorem as well. Now we set up the same system ofequations as in problem 16, with T 1 being the tensionin the 32 cm string and T 2 being the tension in the24 cm string, and the force of the mass being(5 kg)(9.8 N>kg) 5 49 N.cos 53.1°T 1 5 T 2cos 36.9°T 1 sin 36.9° 1 T 2 sin 53.1° 5 49cos 53.1°aT 2cos 36.9° b sin 36.9° 1 T 2 sin 53.1° 5 49cos 53.1°T 2 aa b sin 36.9° 1 sin 53.1°b 5 49cos 36.9°T 2 (1.25) 5 49T 2 8 39.2 Ncos 53.1°T 1 5 (39.2)cos 36.9°8 29.4 NCalculus and Vectors <strong>Solution</strong>s ManualT 1 cos 36.9° 5 T 2 cos 53.1°WNES(Port means left and starboard means right.) We arelooking for the resultant of these two force vectorsthat are 35° apart. We don’t know the exact valueof the force, so we will call it x. So the small tugis pulling with a force of x and the large tug ispulling with a force of 2x. To find the magnitudeof the resultant, r, in terms of x, we use the cosinelaw.r 2 5 x 2 1 (2x) 2 2 2(x)(2x) cos 145°5 x 2 1 4x 2 2 4x 2 cos 145°8 5x 2 2 4x 2 (20.8192)8 5x 2 1 3.2768x 28 8.2768x 2r 8 "8.2768x 28 2.8769xNow we use the cosine law again to find the angle,u, made by the resultant.x 2 5 r 2 1 (2x) 2 2 2(2.8769x)(2x) cos ux 2 5 8.2768x 2 1 4x 2 2 11.5076x 2 cos ux 2 5 12.2768x 2 2 11.5076x 2 cos u211.2768x 2 5211.5076x 2 cos ucos u5 11.276811.5076u5cos 21 a 11.276811.5076 b8 11.5° from the large tug toward thesmall tug, for a net of 8.5° to the starboard side.7-5
19.8N10N5Na. First we will find the resultant of the 5 N and8 N forces. Use the Pythagorean theorem to find themagnitude, m.m 2 5 5 2 1 8 25 25 1 645 89m 5 "89 8 9.4Next we use the Pythagorean theorem again to findthe magnitude, M, of the resultant of this net forceand the 10 N force.M 2 5 m 2 1 10 25 89 1 1005 189M 5 "189 8 13.75Since the equilibrant is equal in magnitude to theresultant, we have the magnitude of the equilibrantequal to approximately 13.75 N.b. To find each angle, use the definition of cosinewith respect each force as a leg and the resultant asthe hypotenuse. Let u 5N be the angle from the5 N force to the resultant, u 8N be the angle from the8 N force to the resultant, and u 10N be the anglefrom the 10 N force to the resultant.Let the sign of the resultant be negative, since it isin a direction away from the head of each of thegiven forces.5cos u 5N 5213.75u 5N 5 cos 21 5a213.75 b8 111.3°8cos u 8N 5213.75u 8N 5 cos 21 8a213.75 b8 125.6°cos u 10N 5 10213.75u 10N 5 cos 21 10a213.75 b8 136.7°20. We know that the resultant of these two forces isequal in magnitude and angle to the>diagonal>lineof the parallelogram formed with f and as legs> > 1 f 2and has diagonal length @ f 1 1 f2 @ . We also knowfrom the cosine law that> >@ f 1 1 f2 @ 2 >5 @ f 1 @ 2 >1 @ f 2 @ 2 > >2 2 @ f 1 @@f2@ cos fwhere f is the supplement to u in our parallelogram.Since we know f5180 2u, thencos f5cos (180 2u) 52cos u.Thus>we>have@ f 1 1 f2 @ 2 >5 @ f 1 @ 2 >1 @ f 2 @ 2 > >2 2 @ f 1 @@f2@ cos f5 @ f 1>@ 2 1 @ f 2>@ 2 1 2 @ f 1>@@f2>@ cos u@ f 1>1 f2>@ 5 " @ f1>@ 2 1 @ f 2>@ 2 1 2 @ f 1>@@f2>@ cos u7.2 Velocity, pp. 367–3701. a. Both the woman and the train’s velocities arein the same direction, so we add them.80 km>h 1 4 km>h 5 84 km>hb. The woman’s velocity is directed opposite that oftrain, so we subtract her velocity from the train’s.80 km>h 2 4 km>h 5 76 km>h. The resultant is inthe same direction as the train’s movement.2. a. The velocity of the wind is directed opposite thatof the airplane, so we subtract the wind’s velocityfrom the airplane’s.600 km>h 2 100 km>h 5 500 km>h north.b. Both the wind and the airplane’s velocities are inthe same direction, so we add them.600 km>h 1 100 km>h 5 700 km>h north.3. We use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, u, made.m 2 5 300 2 1 50 25 90 000 1 25005 92 500m 5 "92 5008 304.14 km>htan u5 503002150u5tan3008 9.5°. The resultant is 304.14 km>h, W 9.5° S.7-6 <strong>Chapter</strong> 7: Applications of Vectors
19.8N10N5Na. First we will find the resultant of the 5 N and8 N forces. Use the Pythagorean theorem to find themagnitude, m.m 2 5 5 2 1 8 25 25 1 645 89m 5 "89 8 9.4Next we use the Pythagorean theorem again to findthe magnitude, M, of the resultant of this net forceand the 10 N force.M 2 5 m 2 1 10 25 89 1 1005 189M 5 "189 8 13.75Since the equilibrant is equal in magnitude to theresultant, we have the magnitude of the equilibrantequal to approximately 13.75 N.b. To find each angle, use the definition of cosinewith respect each force as a leg and the resultant asthe hypotenuse. Let u 5N be the angle from the5 N force to the resultant, u 8N be the angle from the8 N force to the resultant, and u 10N be the anglefrom the 10 N force to the resultant.Let the sign of the resultant be negative, since it isin a direction away from the head of each of thegiven forces.5cos u 5N 5213.75u 5N 5 cos 21 5a213.75 b8 111.3°8cos u 8N 5213.75u 8N 5 cos 21 8a213.75 b8 125.6°cos u 10N 5 10213.75u 10N 5 cos 21 10a213.75 b8 136.7°20. We know that the resultant of these two forces isequal in magnitude and angle to the>diagonal>lineof the parallelogram formed with f and as legs> > 1 f 2and has diagonal length @ f 1 1 f2 @ . We also knowfrom the cosine law that> >@ f 1 1 f2 @ 2 >5 @ f 1 @ 2 >1 @ f 2 @ 2 > >2 2 @ f 1 @@f2@ cos fwhere f is the supplement to u in our parallelogram.Since we know f5180 2u, thencos f5cos (180 2u) 52cos u.Thus>we>have@ f 1 1 f2 @ 2 >5 @ f 1 @ 2 >1 @ f 2 @ 2 > >2 2 @ f 1 @@f2@ cos f5 @ f 1>@ 2 1 @ f 2>@ 2 1 2 @ f 1>@@f2>@ cos u@ f 1>1 f2>@ 5 " @ f1>@ 2 1 @ f 2>@ 2 1 2 @ f 1>@@f2>@ cos u7.2 Velocity, pp. 367–3701. a. Both the woman and the train’s velocities arein the same direction, so we add them.80 km>h 1 4 km>h 5 84 km>hb. The woman’s velocity is directed opposite that oftrain, so we subtract her velocity from the train’s.80 km>h 2 4 km>h 5 76 km>h. The resultant is inthe same direction as the train’s movement.2. a. The velocity of the wind is directed opposite thatof the airplane, so we subtract the wind’s velocityfrom the airplane’s.600 km>h 2 100 km>h 5 500 km>h north.b. Both the wind and the airplane’s velocities are inthe same direction, so we add them.600 km>h 1 100 km>h 5 700 km>h north.3. We use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, u, made.m 2 5 300 2 1 50 25 90 000 1 25005 92 500m 5 "92 5008 304.14 km>htan u5 503002150u5tan3008 9.5°. The resultant is 304.14 km>h, W 9.5° S.7-6 <strong>Chapter</strong> 7: Applications of Vectors
4. Adam must swim at an angle, u, upstream so asto counter the 1 km>h velocity of the stream. This isequivalent to Adam swimming along the hypotenuseof a right traingle with 1 km>h leg and a 2 km>hhypotenuse. So the angle is found using the definitionof cosine.cos u5 1 2u5cos 21 1 25 60° upstream5. a. 2 m>s forwardb. 20 m>s 1 2 m>s 5 22 m>s in the direction of the car6. Since the two velocities are at right angles wecan use the Pythagorean theorem to find themagnitude, m, of the resultant velocity and we usethe definition of sine to find the angle, u, made.m 2 5 12 2 1 5 25 144 1 255 169m 5 "1695 13 m>ssin u5 513u5sin 21 5138 22.6° from the direction of the boat towardthe direction of the current. This results in a net of22.6° 1 15° 5 37.6° , or N 37.6° W.7. a. First we find the components of the resultantdirected north and directed west. The componentdirected north is the velocity of the airplane, 800,minus 100 sin 45°, since the wind forms a 45°angle south of west. The western component ofthe resultant is simply 100 cos 45°. So we use thePythagorean theorem to find the magnitude, m, ofthe resultant and the definition of sine to find theangle, u, of the resultant.m 2 5 (800 2 100 sin 45°) 2 1 (100 cos 45°) 28 (729.29) 2 1 (71.71) 28 536 863.8082m 8 732.71 km>hUse the sine law to determine the direction.sin u sin 45°5100 732.71u 8 5.5°The direction is N 5.5° W.Calculus and Vectors <strong>Solution</strong>s Manualb. The airplane is travelling at approximately732.71 km>h, so in 1 hour the airplane will travelabout 732.71 km.8. a. First we find the velocity of the airplane. Weuse the Pythagorean theorem to find the magnitude,m, of the resultant.m 2 5 450 2 1 100 25 202 500 1 10 0005 212 500m 5 "212 5008 461 km>hSo in 3 hours, the airplane will travel about(461 km>h)(3 h) 5 1383 km.b. To find the angle, u, the airplane travels, we usethe definition of sine.sin u5 10046121100u5sin4618 12.5° east of north.9. a. To find the angle, u, at which to fly is theequivalent of the angle of a right triangle with 44 asthe opposite leg and 244 as the hypotenuse. So weuse the definition of sine to find this angle.sin u5 442442144u5sin2448 10.4° south of west.b. By the Pythagorean Theorem, the resultant groundspeed of the airplane is "(244 2 2 44 2 ) 5 240 km>h.Since time 5 distance>rate, the duration of theflight is simply (480 km)>(240 km>h) 5 2 h.10. a. Since Judy is swimming perpendicular tothe flow of the river, her resultant velocity is simplythe hypotenuse of a right triangle with 3 and 4 asbases, which is a 3-4-5 right triangle. Thus, Judy’sresultant velocity is 5 km>h. The direction isdetermined by tan u5 4 3. u 8 53.1° downstreamb. Judy’s distance traveled down the river would be the“4” leg of the 3-4-5 triangle formed by the vectors, butscaled down so that 1m (the width of the river) isequivalent to the “3” leg. So her distance traveled is43 8 1.33 km. This makes her about 0.67 km fromHelen’s cottage.c. While in the river, Judy is swimming at55 km>h for a distance of 3 km. Sincetime 5 distance>rate, her time taken is53 km5 km>h 5 1 3 hours 5 20 minutes.7-7
11.vu205 km/ha. and b. Here, 205 km>h directed 30° north of eastis the resultant of 212 km>h directed east, and thewind speed, v, directed at some angle. This problemis more easily approached finding the wind speed,v, first. So we will do that using the cosine law.v 2 5 205 2 1 212 2 2 2(205)(212) cos 30°5 42 025 1 44 944 2 86 920 cos 30°5 86 969 2 75 2755 11 694v 5 "11 6948 108 km>hNow to find the wind’s direction, we simply find theangle supplementary to the lesser angle, u, formedby the parallelogram of these three velocities. Wecan use the sine law for this.sin u sin 30°5205 108sin 30°sin u5205a108 bu5sin 21 sin 30°a205a108 bb8 71.6°Thus, the direction of v is the angle supplementaryto u in the parallelogram:180° 2 71.6° 5 108.4° 5 18.4° west of north.12.430°212 km/h5Since her swimming speed is a maximum of 4 km>h,this is her maximum resultant magnitude, which isalso the hypotenuse of the triangle formed by her andthe river’s velocity vector. Since one of these legs is5 km>h, we have a triangle with a leg larger than itshypotenuse, which is impossible.13. a. First we need to find Mary’s resultantvelocity, v. Since this resultant is the diagonal of theparallelogram formed by hers and the river’svelocity, we can use the cosine law with the angle, u,of the parallelogram adjacent 30°.v 2 5 3 2 1 4 2 2 2(3)(4) cos 150°5 9 1 16 2 24 cos 150°5 25 1 20.85 45.8v 5 "45.88 6.8 m>sSo in 10 seconds, Mary travels about(6.8 m>s)(10 s) 5 68 m.b. Since Mary is travelling at 3 m>s at an angle of 30°,to find the component of her velocity, v, perpendicularto the current, we use the definition of sine.v 5 3 sin 305 3a 1 2 b5 1.5 m>s perpendicular to the current.So since time 5 distance>rate, the time taken is(150 m)>(1.5 m>s) 5 100 s.14. a. So we have a 5.5 m>s vector and a 4 m>svector with a resultant vector that is directed 45°south of west. Letting u be the angle between the4 km>h vector and the resultant, we can constructa parallelogram using these three vectors and asubsequent triangle with u opposite the 5.5 m>svector and 45° opposite the 4 m>s vector. We nowuse the sine law to find u.sin u sin 45°55.5 4sin 45°sin u55.5a b4u5sin 21 sin 45°a5.5a bb48 76.5° from the resultant.Since the resultant is 45° west of south, Dave’sdirection is 76.5° 1 45° 5 121.5° west of south,which is equivalent to about 180° 2 121.5° 5 58.5°upstream.b. First, we find the magnitude, m, of Dave’s 4 m>svelocity in the direction perpendicular to the river.This is done using the definition of sine.m 5 4 sin 58.5°8 3.41 m>s perpendicular to the river.Since time is distance>rate, we have(200 m)>(3.41 m>s) 8 58.6 s.15. Let b represent the speed of the steamboat and crepresent the speed of the current. On the waydownstream, the effective speed is b 1 c, andupstream is b 2 c. The distance upstream anddownstream is the same, so 5(b 1 c) 5 7(b 2 c).So, b 5 6c. This means that the speed of the boatis 6 times the speed of the current. So, (6c 1 c) ? 57-8 <strong>Chapter</strong> 7: Applications of Vectors
or 35c is the distance. This means that it would takea raft 35 hours moving with the speed of the currentto get from A to B.7.3 The Dot Product of Two GeometricVectors, pp. 377–3781. a > This meansor @b > ? b > 5 0 a > 0 @b > @ cos u50.0 a > 0 5 0,@ 5 0, or cos u50.To be guaranteed that thetwo vectors are perpendicular, the vectors must benonzero.2. a > ? b > is a scalar, and a dot product is only definedfor vectors, so (a > ? b > ) ? c > is meaningless.3. Answers may vary. Leta > ? b > but4. a > ? b > b >5 because5. Since a > b > ? c >? a >and b > 5 b > ? c > a > a > 5 i > ,52c > b > 5 j > , c > 52i > .5 0, 5 0,.c > 5are unit vectors, 0 a > a >0 5 @b > @ 5 1 andsince they are pointing in opposite directions thenu5180° soTherefore6. a. p > ? q > cos5 0 p > u00q > 521. a > ? b > 521.0cos u5 (4)(8) cos (60°)5 (32)(.5)b. x > ? y > 55 0 x > 1600y > 0cos u5 (2)(4) cos (150°)c.d.e.f.a > ? b > 8 26.935 0 a > 0 @b > @ cos u5 (0)(8) cos (100°)p > ? q > 5 05 0 p > 00q > 0cos u5 (1)(1) cos (180°)5 (1)(21)m > ? n > 5215 0 m > 00n > 0cos u5 (2)(5) cos (90°)5 (10)(0)u > ? v > 5 05 0 u > 00v > 0cos u5 (4)(8) cos 145°x > 8? y > 226.25 0 x > 00y > 0cos u7. a.5 (8)a2 "32 b12"3 5 (8)(3) cos u"32 5 cos uu530°b. m > ? n > 5 0 m > 00n > 0cos u(6) 5 (6)(6) cos u16 5 cos uc. p > ? q > u 8 80°5 0 p > 00q > 0cos u3 5 (5)(1) cos u35 5 cos ud. p > u? q > 8 53°5 0 p > 00q > 0cos u23 5 (5)(1) cos ue.f.8.2 3 5 5 cos ua > u? b > 8 127°5 0 a > 0 @b > @ cos u10.5 5 (7)(3) cos u12 5 cos uu > u560°? v > 5 0 u > 00v > 0cos u250 5 (10)(10) cos u2 1 2 5 cos ua > u5120°? b > 5 0 a > 0 @b > @ cos u5 (7.5)(6) cos (180° 2 120°)5 (45)a 1 2 b5 22.5Note: u is the angle between the two vectors whenthey are tail to tail, so9. a. (a > 1 5b > ) ? (2a > u 22 3b > 120°.) 5 a > ? 2a > 21 5b >5 20 a > ? 2a > a > ? 3b >202 2 15 @b > 5b > ? 3b >2@2 3a >5 20 a > ? b > 1 10a >02 2 15 @b > ? b >2@3x > ? (x > 2 3y > ) 2 (x > 2 3y > 1 7a > ?) ? (23x > b >1 y > )b.5 30 x > 021 305 60 x > y > 2 3x > ? 3y > 1 30 x > 0 2 2 x > ? y > 2 (23y > ? 23x > )205 60 x > 0 2 2 9x > ?0 2 2 19x > y > 2? y > x > ? y >1 30 y > 2 9x > ? y > 1 30 y > 202010. so the dot product of any vector with0 > @0 > @ 5 0is 0.Calculus and Vectors <strong>Solution</strong>s Manual7-9
or 35c is the distance. This means that it would takea raft 35 hours moving with the speed of the currentto get from A to B.7.3 The Dot Product of Two GeometricVectors, pp. 377–3781. a > This meansor @b > ? b > 5 0 a > 0 @b > @ cos u50.0 a > 0 5 0,@ 5 0, or cos u50.To be guaranteed that thetwo vectors are perpendicular, the vectors must benonzero.2. a > ? b > is a scalar, and a dot product is only definedfor vectors, so (a > ? b > ) ? c > is meaningless.3. Answers may vary. Leta > ? b > but4. a > ? b > b >5 because5. Since a > b > ? c >? a >and b > 5 b > ? c > a > a > 5 i > ,52c > b > 5 j > , c > 52i > .5 0, 5 0,.c > 5are unit vectors, 0 a > a >0 5 @b > @ 5 1 andsince they are pointing in opposite directions thenu5180° soTherefore6. a. p > ? q > cos5 0 p > u00q > 521. a > ? b > 521.0cos u5 (4)(8) cos (60°)5 (32)(.5)b. x > ? y > 55 0 x > 1600y > 0cos u5 (2)(4) cos (150°)c.d.e.f.a > ? b > 8 26.935 0 a > 0 @b > @ cos u5 (0)(8) cos (100°)p > ? q > 5 05 0 p > 00q > 0cos u5 (1)(1) cos (180°)5 (1)(21)m > ? n > 5215 0 m > 00n > 0cos u5 (2)(5) cos (90°)5 (10)(0)u > ? v > 5 05 0 u > 00v > 0cos u5 (4)(8) cos 145°x > 8? y > 226.25 0 x > 00y > 0cos u7. a.5 (8)a2 "32 b12"3 5 (8)(3) cos u"32 5 cos uu530°b. m > ? n > 5 0 m > 00n > 0cos u(6) 5 (6)(6) cos u16 5 cos uc. p > ? q > u 8 80°5 0 p > 00q > 0cos u3 5 (5)(1) cos u35 5 cos ud. p > u? q > 8 53°5 0 p > 00q > 0cos u23 5 (5)(1) cos ue.f.8.2 3 5 5 cos ua > u? b > 8 127°5 0 a > 0 @b > @ cos u10.5 5 (7)(3) cos u12 5 cos uu > u560°? v > 5 0 u > 00v > 0cos u250 5 (10)(10) cos u2 1 2 5 cos ua > u5120°? b > 5 0 a > 0 @b > @ cos u5 (7.5)(6) cos (180° 2 120°)5 (45)a 1 2 b5 22.5Note: u is the angle between the two vectors whenthey are tail to tail, so9. a. (a > 1 5b > ) ? (2a > u 22 3b > 120°.) 5 a > ? 2a > 21 5b >5 20 a > ? 2a > a > ? 3b >202 2 15 @b > 5b > ? 3b >2@2 3a >5 20 a > ? b > 1 10a >02 2 15 @b > ? b >2@3x > ? (x > 2 3y > ) 2 (x > 2 3y > 1 7a > ?) ? (23x > b >1 y > )b.5 30 x > 021 305 60 x > y > 2 3x > ? 3y > 1 30 x > 0 2 2 x > ? y > 2 (23y > ? 23x > )205 60 x > 0 2 2 9x > ?0 2 2 19x > y > 2? y > x > ? y >1 30 y > 2 9x > ? y > 1 30 y > 202010. so the dot product of any vector with0 > @0 > @ 5 0is 0.Calculus and Vectors <strong>Solution</strong>s Manual7-9
11.0 a > (a > 20 2 2 a > 5b >? b > ) ? (a >2 5b > 2 b >?0 a > a > ) 5 @a >1 5 @b > 2 5b > @@a > 2 b > @ cos (90°)@ 2 5 002 1 5 @b > @ 2 5 6a > ? b >12. a.b.13. a.b. b > ? c > 5 @b > @Therefore 0 a > 0 c > 0cos (90°) 502 5 @b > @ 2 1 0 c > 002 .This is just what the Pythagorean theorem says,where and are the legs of the right triangle.14. (u > b >5 u > 1?15 0 u > v > u > v > c >11? w > u > w >? v > ) ? (u >10 2 1 0 v > w > 1? u > u > 1? w > v > 10 2 1 0 w > 1 w > 1? v > v > w > )? u >0 2 1 20 u > 100v > w > 1 v >? w > ? v >0cos (90°)1 20 u > 00w > 0cos (90°) 1 20 v > 00w > 0cos (90°)5 (1) 2 1 (2) 2 1 (3) 25 1415. 0 u > 1 v > 0 2 1 0 u > 2 v > 205 (u > 1 v > ) ? (u > 1 v > ) 1 (u > 2 v > ) ? (u > 2 v > )5 0 u > 0 2 1 2u > ? v > 1 0 v > 0 2 1 0 u > 0 2 2 2u > ? v > 1 0 v > 205 20 u > 0 2 1 20 v > 2016. (a >5 0 a > 1 b > )02 1 a > ? (a > 1? b > 1 a > b > 1 c > )? c > 1 b > ? a > 1 @b > @ 2 1 b > ? c >5 1 1 20 a > 0 @b > @ cos (60°) 1 0 a > 00c > 0cos (60°) 1 11 @b > @ 0 c > 0cos (120°)5 2 1 2a 1 2 b 1 1 2 2 1 25 317.a > a >? (a > 1 b >1 b > 1 c >1 c > 5 0 >) 1 b > ? (a > 1 b > 1 c > )10 a > c > ? (a >02 1 a > 1 b > 1? b > 1 a > c > )? c > 5 01 b > ? a > 1 @b > @ 2 1 b > ? c >1 c > ? a > 1 c > ? b > 1 0 c > 0 2 5 0a > ? b > 5 1 6 ( 0 a> 0 2 1 5 @b > @ 2 )(a > 1 b > ) ? (a > 1 b > ) 5 a > 5? a > 11 a > ? b >15 0 a > b > ? a >02 1 a > 1 b >? b > ? b >1 a > ? b >1(a > 1 b > ) ? (a > 2 b > ) 5 a > 5 0? a > a > @b > 2@02 12 a > 2a >? b > ? b > 11 b > @b > 2? a > @25 0 a > b > ? b >0 a > 02 5 a > ?5 (b > a > 5 0 a > 0 2 2 a >02 2 @b > ? b > 1 a > ? b > 2 @b > 2@2@5 @b > 1 c > ) ?@ 2 1 2b > (b >? c > 1 c > )1 0 c > 201 1 4 1 9 1 2(a > ? b > 1 a > ? c > 1 b > ? c > ) 5 018. d >c > b > 5 b >? a > 5 d > 2 c > a > ? b > 1 a > ? c > 1 b > ? c > 52715 ((b > c >? a > ) a > ) ? a >c > ? a > 5 (b > ? a > )(a > ? a > ) because b > ? a > is a scalarc > c > ? a > becausec > ? a > 5 (b > ? a > )0d > ? a > 5 (d > 1? a > 5 d > ? a > c > a > 20) ?1 c > a >? a > 0 a > 0 5 15 07.4 The Dot Product for AlgebraicVectors, pp. 385–3871. a > ? b > 5 0(21)b 1 1 b 2 5 0b 2 5 b 1Any vector of the form (c, c) is perpendicularto a > . Therefore there are infinitely many vectorsperpendicular to a > . Answers may vary. For example:(1, 1), (2, 2), (3, 3).2. a. a > ? b > 5 (22)(1) 1 (1)(2)5 0b. a > ? b > u590°5 (2)(4) 1 (3)(3) 1 (21)(217)5 8 1 9 1 175 34 . 0cos u .02(a > ? b > 1 a > ? c > 1 b > ? c > ) 5214u is acutec. a > ? b > 5 (1)(3) 1 (22)(22) 1 (5)(22)5 3 1 4 2 10523 , 0cos u ,0u is obtuse3. Any vector in the xy-plane is of the forma > 5 (a Let b > 1 , a 2 , 0). 5 (0, 0, 1).a > ? b > 5 (0)(a 1 ) 1 (0)(a 2 ) 1 (0)(1)5 0Therefore (0, 0, 1) is perpendicular to every vectorin the xy-plane.Any vector in the xz-plane is of the formc > 5 (c Let d > 1 , 0, c 3 ). 5 (0, 1, 0).c > ? d > 5 (0)(c 1 ) 1 (0)(1) 1 (0)(c 3 )5 0Therefore (0, 1, 0) is perpendicular to every vectorin the xz-plane.7-10 <strong>Chapter</strong> 7: Applications of Vectors
11.0 a > (a > 20 2 2 a > 5b >? b > ) ? (a >2 5b > 2 b >?0 a > a > ) 5 @a >1 5 @b > 2 5b > @@a > 2 b > @ cos (90°)@ 2 5 002 1 5 @b > @ 2 5 6a > ? b >12. a.b.13. a.b. b > ? c > 5 @b > @Therefore 0 a > 0 c > 0cos (90°) 502 5 @b > @ 2 1 0 c > 002 .This is just what the Pythagorean theorem says,where and are the legs of the right triangle.14. (u > b >5 u > 1?15 0 u > v > u > v > c >11? w > u > w >? v > ) ? (u >10 2 1 0 v > w > 1? u > u > 1? w > v > 10 2 1 0 w > 1 w > 1? v > v > w > )? u >0 2 1 20 u > 100v > w > 1 v >? w > ? v >0cos (90°)1 20 u > 00w > 0cos (90°) 1 20 v > 00w > 0cos (90°)5 (1) 2 1 (2) 2 1 (3) 25 1415. 0 u > 1 v > 0 2 1 0 u > 2 v > 205 (u > 1 v > ) ? (u > 1 v > ) 1 (u > 2 v > ) ? (u > 2 v > )5 0 u > 0 2 1 2u > ? v > 1 0 v > 0 2 1 0 u > 0 2 2 2u > ? v > 1 0 v > 205 20 u > 0 2 1 20 v > 2016. (a >5 0 a > 1 b > )02 1 a > ? (a > 1? b > 1 a > b > 1 c > )? c > 1 b > ? a > 1 @b > @ 2 1 b > ? c >5 1 1 20 a > 0 @b > @ cos (60°) 1 0 a > 00c > 0cos (60°) 1 11 @b > @ 0 c > 0cos (120°)5 2 1 2a 1 2 b 1 1 2 2 1 25 317.a > a >? (a > 1 b >1 b > 1 c >1 c > 5 0 >) 1 b > ? (a > 1 b > 1 c > )10 a > c > ? (a >02 1 a > 1 b > 1? b > 1 a > c > )? c > 5 01 b > ? a > 1 @b > @ 2 1 b > ? c >1 c > ? a > 1 c > ? b > 1 0 c > 0 2 5 0a > ? b > 5 1 6 ( 0 a> 0 2 1 5 @b > @ 2 )(a > 1 b > ) ? (a > 1 b > ) 5 a > 5? a > 11 a > ? b >15 0 a > b > ? a >02 1 a > 1 b >? b > ? b >1 a > ? b >1(a > 1 b > ) ? (a > 2 b > ) 5 a > 5 0? a > a > @b > 2@02 12 a > 2a >? b > ? b > 11 b > @b > 2? a > @25 0 a > b > ? b >0 a > 02 5 a > ?5 (b > a > 5 0 a > 0 2 2 a >02 2 @b > ? b > 1 a > ? b > 2 @b > 2@2@5 @b > 1 c > ) ?@ 2 1 2b > (b >? c > 1 c > )1 0 c > 201 1 4 1 9 1 2(a > ? b > 1 a > ? c > 1 b > ? c > ) 5 018. d >c > b > 5 b >? a > 5 d > 2 c > a > ? b > 1 a > ? c > 1 b > ? c > 52715 ((b > c >? a > ) a > ) ? a >c > ? a > 5 (b > ? a > )(a > ? a > ) because b > ? a > is a scalarc > c > ? a > becausec > ? a > 5 (b > ? a > )0d > ? a > 5 (d > 1? a > 5 d > ? a > c > a > 20) ?1 c > a >? a > 0 a > 0 5 15 07.4 The Dot Product for AlgebraicVectors, pp. 385–3871. a > ? b > 5 0(21)b 1 1 b 2 5 0b 2 5 b 1Any vector of the form (c, c) is perpendicularto a > . Therefore there are infinitely many vectorsperpendicular to a > . Answers may vary. For example:(1, 1), (2, 2), (3, 3).2. a. a > ? b > 5 (22)(1) 1 (1)(2)5 0b. a > ? b > u590°5 (2)(4) 1 (3)(3) 1 (21)(217)5 8 1 9 1 175 34 . 0cos u .02(a > ? b > 1 a > ? c > 1 b > ? c > ) 5214u is acutec. a > ? b > 5 (1)(3) 1 (22)(22) 1 (5)(22)5 3 1 4 2 10523 , 0cos u ,0u is obtuse3. Any vector in the xy-plane is of the forma > 5 (a Let b > 1 , a 2 , 0). 5 (0, 0, 1).a > ? b > 5 (0)(a 1 ) 1 (0)(a 2 ) 1 (0)(1)5 0Therefore (0, 0, 1) is perpendicular to every vectorin the xy-plane.Any vector in the xz-plane is of the formc > 5 (c Let d > 1 , 0, c 3 ). 5 (0, 1, 0).c > ? d > 5 (0)(c 1 ) 1 (0)(1) 1 (0)(c 3 )5 0Therefore (0, 1, 0) is perpendicular to every vectorin the xz-plane.7-10 <strong>Chapter</strong> 7: Applications of Vectors
Any vector in the yz-plane is of the forme > 5 (0, e Let f > 2 , e 3 ). 5 (1, 0, 0) .e > ? f > 5 (1)(0) 1 (0)(e 2 ) 1 (0)(e 3 )5 0Therefore (1, 0, 0) is perpendicular to every vectorin the yz-plane.4. a. (1, 2, 21) ? (4, 3, 10) 5 4 1 6 2 105 0( 24, 25, 26) ? a5, 23, 2 5 b 5220 1 15 1 565 0b. If any of the vectors were collinear then onewould be a scalar multiple of the other. Comparingthe signs of the individual components of eachvector eliminates (1, 2, 21) and (5, 23, 2 5 6). All ofthe components of (24, 25, 26) have the samesign and the same is true for (4, 3, 10), but (4, 3, 10)is not a scalar multiple of (24, 25, 26). Thereforenone of the vectors are collinear.5. a. Using the strategy of Example 5 yields(x, y) ? (1, 22) 5 0 and (x, y) ? (1, 1) 5 0x 2 2y 5 0 and x 1 y 5 03y 5 0Therefore the only result is x 5 y 5 0, or (0, 0).This is because (1, 22) and (1, 1) both lie on thexy-plane and are not collinear, so any vector that isperpendicular to both vectors must be in R 3 whichdoes not exist in R 2 .b. If we select any two vectors that are not collinearin R 2 , then any vector that is perpendicular to bothcannot be in R 2 and must be in R 3 This is notpossible since R 3 does not exist in.R6. a. cos u5 a> ? b >2 .0 a > 0 @b > @(5)(21) 1 (3)(22)5"25 1 9"1 1 45 211"(34)(5)5 211"170u 8 148°b. cos u5 a> ? b >0 a > 0 @b > @(21)(6) 1 (4)(22)5"1 1 16"36 1 45 214"680u 8 123°Calculus and Vectors <strong>Solution</strong>s Manualc. cos u5 a> ? b >0 a > 0 @b > @(2)(2) 1 (2)(1) 1 (1)(22)5"4 1 4 1 1"4 1 1 1 45 4(3)(3)d. cos u5 a> ? b >0 a > 0 @b > @(2)(25) 1 (3)(0) 1 (26)(12)5"4 1 9 1 36"25 1 1445 282(7)(13)5 28291u 8 154°7. a.a > ? b > 5 0 a >(21)(26k) 1 (2)(21) 1 (23)(k) 5 0 a > 0 @b > @0 @b > cos u@b.(1)(0) 1 (1)(k) 5 "1 1 1"k 2 cos (45°)1k 5 "2 0 k 0"28. a.b.5 4 9u 8 64°–2 –1a > ? b > 5 0 a > 0 @b > @ cos u2(0, 1)1(1, 0) x0 1 2–1–2k 5 0 k 0k $ 0yy2(0, 1)1(1, 0) x–2 –1 0–1–21 26 k 2 2 2 3k 5 03 k 5 2k 5 2 3cos (90°)7-11
The diagonals are (1, 0) 1 (0, 1) 5 (1, 1) and(1, 0) 2 (0, 1) 5 (1, 21) or(1, 0) 1 (0, 1) 5 (1, 1) and(0, 1) 2 (1, 0) 5 (21, 0) .c. (1, 1) ? (1, 21)5 1 2 15 0or (1, 1) ? (21, 1)521 1 15 09. a. cos u5 a> ? b >0 a > 0 @b > @5 (1 2 "2)(1) 1 ("2 2 1)(1)0 a > 0 @b > @5 0u590°b. cos u5 a> ? b >0 a > 0 @b > @"2 2 1 1 "2 1 1 1 "25"(2 2 2"2 1 1) 1 (2 1 2"2 1 1) 1 2"1 1 1 1 15 3"2"8"35 "32u530°10. a. i. a > 5 kb >8 5 12kk 5 2 3p 5 4a 2 3 bp 5 8 32 5 2 3 qq 5 3ii. Answers may vary. For example:a > ? b > 5 02q 1 4p 1 96 5 0q 522p 2 48Let p 5 1q 5250b. In part a., the values are unique because bothvectors have their third component specified, andthe ratios must be the same for each component b > .In part b. the values are not unique; any value ofp could have been chosen, each resulting in adifferent value of q.11. AB > 5 (2, 6), BC > 5 (25, 25), CA > 5 (3, 21)cos (180° 2u A ) 5 AB> ? CA >@AB > @@CA > @5 6 2 6@AB > @@CA > @5 0180° 2u A 5 90°u A 5 90°cos (180° 2u B ) 5 AB> ? BC >@AB > @@BC > @210 2 305"4 1 36"25 1 252405"(40)(50)452 Å 5180° 2u B 8 153.4°u B 8 26.6°u C 5 180° 2u A 2u Bu C 8 63.4°12. a. O 5 (0, 0, 0), A 5 (7, 0, 0), B 5 (7, 4, 0),C 5 (0, 4, 0), D 5 (7, 0, 5), E 5 (0, 4, 5),F 5 (0, 0, 5)b.AE > ? BF > 5 @AE > @@BF > @ cos u(27, 4, 5) ? (27, 24, 5) 5 "49 1 16 1 253 "49 1 16 1 25 cos u49 2 16 1 25 5 90 cos u5890 5 cos uu 8 50°13. a. Answers may vary. For example:( x, y, z) ? (21, 3, 0) 5 02x 1 3y 5 0x 5 3y( x, y, z) ? (1, 25, 2) 5 0x 2 5y 1 2z 5 022y 1 2z 5 0y 5 zLet y 5 1.(3, 1, 1) is perpendicular to (21, 3, 0) and(1, 25, 2).b. Answers may vary. For example:( x, y, z) ? (1, 3, 24) 5 0x 1 3y 2 4z 5 0x 5 4z 2 3y( x, y, z) ? (21, 22, 3) 5 07-12 <strong>Chapter</strong> 7: Applications of Vectors
2x 2 2y 1 3z 5 03y 2 4z 2 2y 1 3z 5 0y 5 zLet y 5 1.(1, 1, 1) is perpendicular to (1, 3, 24) and(21, 22, 3).14. (p, p, 1) ? (p, 22, 23) 5 0p 2 2 2p 2 3 5 0p 5 2 6 "22 2 4(23)2p 5 1 6 2p 5 3 or 2115. a. (23, p, 21) ? (1, 24, q) 5 023 2 4p 2 q 5 03 1 4p 1 q 5 0b. 3 1 4p 2 3 5 0p 5 016. Answers may vary. For example: Note thats > 522r > , so they are collinear. Therefore anyvector that is perpendicular to is alsoperpendicular to r > s >.( x, y, z) ? (1, 2, 21) 5 0x 1 2y 2 z 5 0Let x 5 z 5 1.(1, 0, 1) is perpendicular to (1, 2, 21) and(22, 24, 2).Let x 5 y 5 1.(1, 1, 3) is perpendicular to (1, 2 2 1) and(22,17. x > 24,? y > 2).5 0 x > 00y > 0cos u(24, p, 22) ? (22, 3, 6)5 "16 1 p 2 1 4"4 1 9 1 36 cos u9p 2 2 24p 1 16 5 49(20 1 p 2 )a 4 21 b 29p 2 2 24p 1 16 5 3209 1 1665p 2 2 216p 2 176 5 0p 5 216 6 "(2216)2 2 4(65)(2176)2(65)p 5 4 or8 1 3p 2 12 5 "20 1 p 2 (7) cos u2(3p 2 4) 2 5 a7"20 1 p 2 cos ub2 44659 p218. a. a > ? b > 523 1 35 0Therefore, since the two diagonals are perpendicular,all the sides must be the same length.Calculus and Vectors <strong>Solution</strong>s Manualb. AB > 5 1 2 (a> 1 b > )5 (1, 2, 21)BC > 5 1 2 (a> 2 b > )5 (2, 1, 1)@AB > @ 5 @BC > @ 5 "6c. AB > ? BC > 5 @AB > @@BC > @ cos u 12 1 2 2 1 5 6 cos u 112 5 cos u 1u 1 5 60°2u 1 1 2u 2 5 360°u 2 5 120°19. a. AB > 5 (3, 4, 212), DA > 5 (24, 2 2 q, 25)AB > ? DA > 5 0212 1 8 2 4q 1 60 5 021 2 q 1 15 5 0q 5 14DA > 5 CB >( 24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)x 5 6, y 5 18, z 524The coordinates of vertex C are (6, 18, 24).b. AC > ? BD > 5 @AC > @@BD > @ cos u(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 493 "1 1 64 1 289 cos u7 1 128 2 119 5 354 cos u16354 5 cos uu 8 87.4°20. The two vectors representing the body diagonalsare (0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1) and(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)( 21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u1 2 1 1 1 5 3 cos u13 5 cos uu 8 70.5°a5180° 2ua 8 109.5°Mid-<strong>Chapter</strong> Review, pp. 388–3891. a. a > ? b > 5 (3)(2) cos (60°)5 (6) 1 25 37-13
2x 2 2y 1 3z 5 03y 2 4z 2 2y 1 3z 5 0y 5 zLet y 5 1.(1, 1, 1) is perpendicular to (1, 3, 24) and(21, 22, 3).14. (p, p, 1) ? (p, 22, 23) 5 0p 2 2 2p 2 3 5 0p 5 2 6 "22 2 4(23)2p 5 1 6 2p 5 3 or 2115. a. (23, p, 21) ? (1, 24, q) 5 023 2 4p 2 q 5 03 1 4p 1 q 5 0b. 3 1 4p 2 3 5 0p 5 016. Answers may vary. For example: Note thats > 522r > , so they are collinear. Therefore anyvector that is perpendicular to is alsoperpendicular to r > s >.( x, y, z) ? (1, 2, 21) 5 0x 1 2y 2 z 5 0Let x 5 z 5 1.(1, 0, 1) is perpendicular to (1, 2, 21) and(22, 24, 2).Let x 5 y 5 1.(1, 1, 3) is perpendicular to (1, 2 2 1) and(22,17. x > 24,? y > 2).5 0 x > 00y > 0cos u(24, p, 22) ? (22, 3, 6)5 "16 1 p 2 1 4"4 1 9 1 36 cos u9p 2 2 24p 1 16 5 49(20 1 p 2 )a 4 21 b 29p 2 2 24p 1 16 5 3209 1 1665p 2 2 216p 2 176 5 0p 5 216 6 "(2216)2 2 4(65)(2176)2(65)p 5 4 or8 1 3p 2 12 5 "20 1 p 2 (7) cos u2(3p 2 4) 2 5 a7"20 1 p 2 cos ub2 44659 p218. a. a > ? b > 523 1 35 0Therefore, since the two diagonals are perpendicular,all the sides must be the same length.Calculus and Vectors <strong>Solution</strong>s Manualb. AB > 5 1 2 (a> 1 b > )5 (1, 2, 21)BC > 5 1 2 (a> 2 b > )5 (2, 1, 1)@AB > @ 5 @BC > @ 5 "6c. AB > ? BC > 5 @AB > @@BC > @ cos u 12 1 2 2 1 5 6 cos u 112 5 cos u 1u 1 5 60°2u 1 1 2u 2 5 360°u 2 5 120°19. a. AB > 5 (3, 4, 212), DA > 5 (24, 2 2 q, 25)AB > ? DA > 5 0212 1 8 2 4q 1 60 5 021 2 q 1 15 5 0q 5 14DA > 5 CB >( 24, 212, 25) 5 (2 2 x, 6 2 y, 29 2 z)x 5 6, y 5 18, z 524The coordinates of vertex C are (6, 18, 24).b. AC > ? BD > 5 @AC > @@BD > @ cos u(7, 16, 27) ? (1, 8, 17) 5 "49 1 256 1 493 "1 1 64 1 289 cos u7 1 128 2 119 5 354 cos u16354 5 cos uu 8 87.4°20. The two vectors representing the body diagonalsare (0 2 1, 1 2 0, 1 2 0) 5 (21, 1, 1) and(0 2 1, 0 2 1, 1 2 0) 5 (21, 21, 1)( 21, 1, 1) ? (21, 21, 1) 5 "3"3 cos u1 2 1 1 1 5 3 cos u13 5 cos uu 8 70.5°a5180° 2ua 8 109.5°Mid-<strong>Chapter</strong> Review, pp. 388–3891. a. a > ? b > 5 (3)(2) cos (60°)5 (6) 1 25 37-13
. (3a > 1 2b > ) ? (4a > 2 3b > ) 5 120 a > 021 8b > 2? a > 9a > ? b >2 6 @b > 2@2.5 8120 cmu 225 cmLet T 1 be the tension in the 15 cm cord and T 2 bethe tension in the 20 cm cord. Let u 1 be the anglethe 15 cm cord makes with the ceiling and u 2 be theangle the 20 cm cord makes with the ceiling. By thecosine law:(15) 2 5 (20) 2 1 (25) 2 2 2(20)(25) cos (u 2 )cos (u 2 ) 5 0.8Therefore the tension in the 15 cm cord is 117.60 Nand the tension in the 20 cm cord is 88.20 N.3. The diagonals of a square are perpendicular, sothe dot product is 0.4. a.v135°a0 v > w0 5 500, 0 w > 0 5 100By the cosine law:0 v > 1 w > 0 2 5 (500) 2 1 (100) 20 v > 1 w > 2 2(500)(100) cos (135°)0 8 575.1u 1sin (u 2 ) 5 "1 2 cos 2 (u 2 )sin (u 2 ) 5 0.6(20) 2 5 (15) 2 1 (25) 2 2 (2)(15)(25) cos (u 1 )cos (u 1 ) 5 0.6sin (u 1 ) 5 0.8Horizontal Components:2T 1 cos (u 1 ) 1 T 2 cos (u 2 ) 5 0(0.8)T 2 5 (0.6)T 1T 2 5 (0.75)T 1Vertical Components:T 1 sin (u 1 ) 1 T 2 sin (u 2 ) 2 (15)(9.8) 5 0(0.8)T 1 1 (0.6)(0.75)T 1 5 147(1.25)T 1 5 147T 1 5 117.6 NT 2 5 (0.75)T 1T 2 5 88.2 Nv+ w5 12(3) 2 2 3 2 6(2) 215 cmBy the cosine law:sin (a) sin (135°)5100 575.1sin (a) 8 0.123a 8 7.06°The resultant velocity of the airplane is 575.1 km>hat S7.06°Eb. (distance) 5 (rate)(time)t 8 1000575.1 ? km(km/h)t 8 1.74 hours5. a.@E '>@ 5 @E>@ cos (40°)@E '>@ 5 (9.8)(15)cos (40°)@E '>@ 8 112.61 Nb.EF@F > @ 5 @E > @ sin (40°)@F > @ 8 94.49 N6. 6u 5360°a > u560°? b > 5 0 a > 0 @b > @ cos (60°)5 (3)(3)(0.5)57. a. a > 4.5? b > 5 (4)(1) 1 (25)(2) 1 (20)(2)b. a > ? b > 55 0 a > 340 @b > @ cos (u)34 5 "16 1 25 1 400 "1 1 4 1 4 cos (u)cos (u) 5 348. a. a > ? b > 635 (i > 1 2j > 1 k > ) ? (2i > 2 3j > 1 4k > )5 2 2 6 1 4b. b > ? c > 5 05 (2i > 2 3j > 1 4k > ) ? (3i > 2 j > 2 k > )5 6 1 3 2 4c. b > 1 c > 5 55 (2i >d. a > ? (b > 5 5i > 2 3j >1 c > 2 4j > 1 4k >) 5 (i > 1 3k > ) 1 (3i > 2 j > 2 k > )1 2j > 1 k > ) ? (5i > 2 4j > 1 3k > )5 5 2 8 1 3e. (a > 1 b > 5) ? (b > 01 c > ) 5 (3i > 2? (5j > j >12 4j > 5k > )1 3k > )5 15 1 4 1 155 347-14 <strong>Chapter</strong> 7: Applications of Vectors
f. (2a > 2 3b > ) ? (2a > 1 c > ) 5 (( 2i > 12 (6i > 4j > 1? (( 2i > 2 9j > 2k > )1 (3i > 1 4j > 1 12k >5 (24i > 2 j > 1 2k > ))1? (5i > 1 13j > k > )))1 3j > 21 k > 10k > ))5220 1 39 2 109. a.(xi > p >1 j > ? q > 5 951 3k > 0) ? (3xi > 1 10xj > 1 k > ) 5 03x 2 1 10x 1 3 5 0x 5 210 6 "(10)2 2 4(3)(3)2(3)x 5 210 6 86x 523 or x 52 1b. If p > and q > 3are parallel then one is a scalarmultiple of the other.p > where n is a constantxi > 5 nq >1 j > 1 3k > 5by the k > n(3xi > 1 10xj > 1 k > )n 5 3 componentx 5 9x by the i >componentx 5 01 5 30(0) by the j >component1 2 0Therefore there is no value of x that will make thesetwo vectors parallel.10. a. 3x > 2 2y > 5 (3i >b. 3x > ? 2y > 55 (3i > i > 222 6j > 4j > 6j > 222 3k > k > 3k > ) 2 (2i > 2 2j > 2 2k > )) ? (2i > 2 2j > 2 2k > )5 6 1 12 1 6c. 0 x > 2 2y > 5 240 5 @ (i >5 @2i > 2 2j >1 k > 2 k > ) 2 (2i > 2 2j > 2 2k > )@@5 "(2i > 1 k > ) ? (2i > 1 k > )or 1.41d. (2x > 52 3y > "2) ? (x > 1 4y > ) 5 (( 2i > 22 (3i > 4j > 21 (( i > 2 3j > 2k > )21 (4i > 2j > 2 3k >25 (2i > 2? (5i > 2 j > 4j > k > )) ?)212 6j > k > 4k > ))2 5k > )525 1 6 2 5524e. 2x > ? y > 2 5y > ? x > 5 2x > ?523x > y > 2523(i > ? y > 5x > ? y >2 2j > 2 k) ? (i > 2 j > 2 k > )523(1 1 2 1 1)521211.4 N(4) 2 5 (5) 2 1 (3) 2 2 2(3)(5) cos (180° 2u)0.6 5 cos (180° 2u)180° 2u8 53.1u 8 126.9°12. (F) 2 5 (3) 2 1 (4) 2 2 2(3)(4) cos (180° 2 60°)( F) 2 5 25 2 24 cos (120°)( F) 2 5 37F 8 6.08 N(3) 2 5 (4) 2 1 ("37) 2 2 2(4)("37) cos ucos u 5 448"375 N180° - u3 NF > u 8 25.3°8 6.08 N, 25.3° from the 4 N force towards the3 N force.E > 8 6.08 N, 180° 2 25.3° 5 154.7° from the4 N force away from the 3 N force.13. a. The diagonals are andm >(m > 1 n > m > 1 n > m > 1 n > m > 2 n >) ? (m > 22 n > n > 5 (1, 4, 10)5 (3,) 5 0 m > 210,1 n > 0)00m > 2 n > 0cos u3 2 40 5 "1 1 16 1 100 "9 1 100 cos ucos u 8 20.3276b. 0 m > u 82 n > 109.1°0 2 5 0 m > 0 2 1 0 n > 0 2 2 20 m > 00n > 0cos u(9 1 100) 5 (4 1 9 1 25) 1 (1 1 49 1 25)2 2"38 "75 cos ucos u 8 0.0374u 8 87.9°Calculus and Vectors <strong>Solution</strong>s Manual7-15
14. a. 45 sin (150°) 5 500 sin uu 8 N 2.6° Eb. v 5 500 cos (2.6°) 2 45 cos (30°)8 460.5 km>ht 8 1000460.5t 8 2.17 hours15.a > ? x > 5 02x 1 1 2x 2 1 5x 3 5 0b > x 1? x > 5 2x 2 1 5x 35 0x 1 1 3x 2 1 5x 3 5 02x 2 1 5x 3 1 3x 2 1 5x 3 5 0x 2 1 2x 3 5 0choose x 3 5 1x 2 522x 1 5 1x > 5 1 (1, 22, 1)!6x> or a2 1 !6 , 25 a 1 !6 , 2 1 !6 ,2 2 !6 , 1!6 b!6 b16. a. v 5 4 1 3 cos (45°)8 6.12 ms >d 8 (6.12)(10)8 61.2 mb. w 5 3 sin (45°)8 2.12 ms >t 8 1802.12t 8 84.9 seconds17. a.0 x > 0 2 2 x > (x > 1? y > y > )1 y > ? (x >? x > 2 y >2 0 y > ) 5 0( x > 1 y > ) ? (x > 0 x > 0 2 5 02 y > 0 2 5 0 y > 20) 5 0 when x > and y > havethe same length.b. Vectors and determine a parallelogram. Theirsum a > 1 b > a > b >is one diagonal of the parallelogramformed, with its tail in the same location as the tailsof a > and b > . Their difference a > 2 b >is the otherdiagonal of the parallelogram.18. @F > @ 5 350 cos (40°)8 268.12 N7.5 Scalar and Vector Projections,pp. 398–4001. a. Scalar projection of a > on b >is wherea > 5 (2, 3) and is the positive x-axis (X, 0).a > ? b > 5 (2X) 1 (3 3 0)5 2X 1 05 2X@b > @ 5 "X 2 1 0 2a > ? b > 5 X@b > 5 2X @ X5 2;The vector projection is the scalar projectionmultiplied by where is the x-axis divided bythe magnitude of the x-axis which is equal toThe scalar projection of 2 multiplied by equalsa > i > i > .2i > .>?bb. Scalar projection of a > on b >is wherea > 5 (2, 3) and>b@ b > @a > ? b > 5 (2 3 0) 1 (3Y)5 0 1 3Y@b > 5 3Y@ 5 "0 2 1 Y 25 Y>bis now the positive y-axis (0, Y).a > @b > @@b > 5 3Y @ Y5 3;The vector projection is the scalar projectionmultiplied by where is the y-axis divided@ b > @b >b >>b@ b > @>b@ b > @by the magnitude of the y-axis which is equal toThe scalar projection of 3 multiplied by j > j > .equals 3j > .2. Using the formula would cause a division by 0.Generally the 0 > has any direction and 0 magnitude.You can not project onto nothing.3. You are projecting a > onto the tail of b >whichis a point with magnitude 0. Therefore it is theprojections of onto the tail of are also 0and 0 > b > a > 0 > ;.@b > @a > >?b@b > @7-16 <strong>Chapter</strong> 7: Applications of Vectors
14. a. 45 sin (150°) 5 500 sin uu 8 N 2.6° Eb. v 5 500 cos (2.6°) 2 45 cos (30°)8 460.5 km>ht 8 1000460.5t 8 2.17 hours15.a > ? x > 5 02x 1 1 2x 2 1 5x 3 5 0b > x 1? x > 5 2x 2 1 5x 35 0x 1 1 3x 2 1 5x 3 5 02x 2 1 5x 3 1 3x 2 1 5x 3 5 0x 2 1 2x 3 5 0choose x 3 5 1x 2 522x 1 5 1x > 5 1 (1, 22, 1)!6x> or a2 1 !6 , 25 a 1 !6 , 2 1 !6 ,2 2 !6 , 1!6 b!6 b16. a. v 5 4 1 3 cos (45°)8 6.12 ms >d 8 (6.12)(10)8 61.2 mb. w 5 3 sin (45°)8 2.12 ms >t 8 1802.12t 8 84.9 seconds17. a.0 x > 0 2 2 x > (x > 1? y > y > )1 y > ? (x >? x > 2 y >2 0 y > ) 5 0( x > 1 y > ) ? (x > 0 x > 0 2 5 02 y > 0 2 5 0 y > 20) 5 0 when x > and y > havethe same length.b. Vectors and determine a parallelogram. Theirsum a > 1 b > a > b >is one diagonal of the parallelogramformed, with its tail in the same location as the tailsof a > and b > . Their difference a > 2 b >is the otherdiagonal of the parallelogram.18. @F > @ 5 350 cos (40°)8 268.12 N7.5 Scalar and Vector Projections,pp. 398–4001. a. Scalar projection of a > on b >is wherea > 5 (2, 3) and is the positive x-axis (X, 0).a > ? b > 5 (2X) 1 (3 3 0)5 2X 1 05 2X@b > @ 5 "X 2 1 0 2a > ? b > 5 X@b > 5 2X @ X5 2;The vector projection is the scalar projectionmultiplied by where is the x-axis divided bythe magnitude of the x-axis which is equal toThe scalar projection of 2 multiplied by equalsa > i > i > .2i > .>?bb. Scalar projection of a > on b >is wherea > 5 (2, 3) and>b@ b > @a > ? b > 5 (2 3 0) 1 (3Y)5 0 1 3Y@b > 5 3Y@ 5 "0 2 1 Y 25 Y>bis now the positive y-axis (0, Y).a > @b > @@b > 5 3Y @ Y5 3;The vector projection is the scalar projectionmultiplied by where is the y-axis divided@ b > @b >b >>b@ b > @>b@ b > @by the magnitude of the y-axis which is equal toThe scalar projection of 3 multiplied by j > j > .equals 3j > .2. Using the formula would cause a division by 0.Generally the 0 > has any direction and 0 magnitude.You can not project onto nothing.3. You are projecting a > onto the tail of b >whichis a point with magnitude 0. Therefore it is theprojections of onto the tail of are also 0and 0 > b > a > 0 > ;.@b > @a > >?b@b > @7-16 <strong>Chapter</strong> 7: Applications of Vectors
4. Answers may vary. For example:q > 5 AB > p > 5 AE > ,pEDA C q BScalar projection p > onVector projection p > q >onScalar projection q > q > 5 @AC > @onVector projection on5. When a > q > p > 5 AC > ;;p > 5 @AD > @5 AD > ;5 (21, 2, 5) and b > 5 (1, 0, 0) thena > ? b > 5 (21 3 1 1 2 3 0 1 5 3 0)@b > 521@ 5 "1 2 1 0 2 1 0 25 1a > ? b >Therefore the scalar projection is@b > 5 21@ 1521;The vector equation isUnder the same approach, whenand b > 5 (0, 1, 0), thena > ? b > 5 (21 3 0 1 2 3 1 1 5 3 0)@b > 5 2@ 5 "0 2 1 1 1 0 25 1a > ? b >Therefore the scalar projection is@b > 5 2 @ 1The vector equation is21 3 b>@b > @2 3 b>@b > @521 3521;a > 5 (21, 2, 5)5 2 35 2,(0, 1, 0)1(1, 0, 0)1The same is also true whenandb > a > 5 2;5 (21, 2, 5)thena > 5? b > (0, 0, 1)5 (21 3 0 1 2 3 0 1 5 3 1)@b > 5 5@ 5 "0 2 1 0 2 1 1 25 1a > ? b >Therefore the scalar projection is@b > 5 5 @ 15 5,The vector equation is 5 3 b> (0, 0, 1)@b > 5 5 3@15 5;Calculus and Vectors <strong>Solution</strong>s ManualWithout having to use formulae, a projection of(21, 2, 5) on i > , j > , or is the same as a projectionof on i > k >(21, 0, 0) , (0, 2, 0) on j > , and (0, 0, 5) onwhich intuitively yields the same result.6. a. p > ? q > 5 (3 324) 1 (6 3 5)1 (222 3220)5212 1 30 1 4400 q > 5 4580 5 "(24) 2 1 5 2 1 (220) 25 "16 1 25 1 4005 "4415 21p > ? q >Therefore the scalar projection is0 q > 0The vector equation 5 45821 3 q>0 q > 05 45821b. Direction angles for p > where p > 5 (a, b, c)ainclude a, b, and g. cos a5"a 2 1 b 2 1 c 235"3 2 1 6 2 1 (222) 235"9 1 36 1 4845 3"529Therefore a5cos 21 a 38 82.5°;bcos b5"a 2 1 b 2 1 c 265"3 2 1 6 2 1 (222) 265"9 1 36 1 4845 6"5295 623 ,Therefore b5cos 21 a 623 b8 74.9°;5 458 (24, 5, 20).44123 b 5 323 ,5 45821 ,(24, 5, 220).21k >7-17
ccos g5"a 2 1 b 2 1 c 22225"3 2 1 6 2 1 (222) 22225"9 1 36 1 4845 222"5295 22223 ,Therefore g5cos 21 a 22223 b7. a. x > ? y > 8 163.0°5 (1 3 1) 1 (1 321)5 1 1 (21)0 y > 5 00 5 "1 2 1 (21) 25 "2x > ? y >The scalar projection is0 y > 5 0 0 "25 0;The vector projection is 0 3 y>b. x > ? y > 0 y > 5 0 >05 (2 3 1) 1 (2"3 3 0)5 20 y > 0 5 "1 2 1 0 25 1x > ? y >The scalar projection is0 y > 5 2 0 15 2;The vector projection is 2 3 y>0 y > 5 2 30c. x > ? y > 5 2i >5 (2 325) 1 (5 3 12)5210 1 600 y > 5 500 5 "(25) 2 1 12 25 "25 1 1445 "1695 13x > ? y >The scalar projection is0 y > 0The vector projection is5 5013 .5013 3 y>0 y > 0(1, 0)15 50 (25, 12)313 135 50 (25, 12)1698. a. The scalar projection of a > on the x-axis(X, 0, 0) isa > ? (X, 0, 0)0 (X, 0, 0) 0a > ? (X, 0, 0)0 (X, 0, 0) 0(21 3 X) 1 (2 3 0) 1 (4 3 0)5"X 2 1 0 2 1 0 25 2XX521;The vector projection of a > on the x-axis is(X, 0, 0)(X, 0, 0)21 3521 3"X 2 1 0 2 21 052i > XThe scalar projection of on the y-axis isa > a > ;(0, Y, 0)? (0, Y, 0) (21 3 0) 1 (2 3 Y) 1 (4 3 0)50 (0, Y, 0) 0"0 2 1 Y 2 1 0 25 2Y Y5 2The vector projection of a > on the y-axis is(0, Y, 0)(0, Y, 0)2 35 2 3"0 2 1 Y 2 21 05 2j > YThe scalar projection of on the z-axis isa > a > ;(0, 0, Z)? (0, 0, Z) (21 3 0) 1 (2 3 0) 1 (4 3 Z)50 (0, 0, Z) 0"0 2 1 0 2 1 Z 25 4Z Z5 4;The vector projection of a > on the z-axis is(0, 0, Z)(0, 0, Z)4 35 4 3"0 2 1 0 2 21 Z5 4k > Z.b. The scalar projection of m a > on the x-axis(X, isma > 0, 0)? (X, 0, 0) (2m 3 X) 1 (2m 3 0)50 (X, 0, 0) 0 "X 2 1 0 2 1 0 2(4m 3 0)1"X 2 1 0 2 1 0 25 2mXX52mThe vector projection of ma > on the x-axis is(X, 0, 0)(X, 0, 0)2m 352m 3"X 2 1 0 2 21 052mi > X;7-18 <strong>Chapter</strong> 7: Applications of Vectors
The scalar projection of on the y-axis isma > (0, Y, 0)? (0, Y, 0) (2m 3 0) 1 (2m 3 Y)50 (0, Y, 0) 0 "0 2 1 Y 2 1 0 2(4m 3 0)1"0 2 1 y 2 1 0 25 2mYY5 2m;The vector projection of ma > on the y-axis is(0, Y, 0)(0, Y, 0)2m 35 2m 3"0 2 1 Y 2 21 05 2mj > YThe scalar projection of on the z-axis isma > ma > ;(0, 0, Z)? (0, 0, Z) (2m 3 0) 1 (2m 3 0)50 (0, 0, Z) 0 "0 2 1 0 2 1 Z 2(4m 3 Z)1"0 2 1 0 2 1 Z 25 4mZZ5 4m;The vector projection of ma > on the z-axis is(0, 0, Z)(0, 0, Z)4m 35 4m 3"0 2 1 0 2 21 Z5 4mk > Z.9. a.a11. a. AB > 5 Point B 2 Point A5 (21, 3, 4) 2 (1, 2, 2)5 (22, 1, 2)The scalar projection of on the x-axis isa > AB >(X, 0, 0)? (X, 0, 0) (22 3 X) 1 (1 3 0) 1 (2 3 0)50 (X, 0, 0) 0"X 2 1 0 2 1 0 25 22XX522;The vector projection of AB >on the x-axis is(X, 0, 0)(X, 0, 0)22 3522 3"X 2 1 0 2 21 0522i > XThe scalar projection of on the y-axis isa > AB > ;(0, Y, 0)? (0, Y, 0) (22 3 0) 1 (1 3 Y) 1 (2 3 0)50 (0, Y, 0) 0"0 2 1 Y 2 1 0 25 Y Y5 1;The vector projection of AB >on the y-axis is(0, Y, 0)(0, Y, 0)1 35 1 3"0 2 1 Y 2 21 05 j > Y;The scalar projection of on the z-axis isa > AB >(0, 0, Z)? (0, 0, Z) (22 3 0) 1 (1 3 0) 1 (2 3 Z)50 (0, 0, Z) 0"0 2 1 0 2 1 Z 2ma > 7-19a > projected onto itself will yield itself. The scalarprojection will be the magnitude of itself.b. Using the formula for the scalar projection0 a > 0cos u50 a >5 0 a > 0cos 05 0 a > 0 (1)0.The vector projection is the scalar projectionmultiplied by10. a.a >0 a > ,0B –a O a A(2a > ) ? a >b.0 a > 2 0 a > 2050 0 a >520 a > 000 a > 0 3 a>0 a > 05 a > .So the vector projection is 2 0 a > 0a 0 a> 0b 5 2a > .0 a > 05 2Z Z5 2;The vector projection of AB >on the z-axis is(0, 0, Z)(0, 0, Z)2 35 2 3"0 2 1 0 2 21 Z5 2k > Zb. The angle made with the y-axis is bbcos b5"a 2 1 b 2 1 c 215"(22) 2 1 1 2 1 2 215"4 1 1 1 45 1 "95 1 3 ,Calculus and Vectors <strong>Solution</strong>s Manual
Therefore b5cos 21 a 1 3 b12. a.b.@BD > @@BD > @B8 70.5°Bbuc. In an isosceles triangle, CD is a median and aright bisector of BA. Therefore and have thesame magnitude projected on c > a > b >.d. Yes, not only do they have the same magnitude,but they are in the same direction as well whichmakes them have equivalent vector projections.13. a. Use the formula for the scalar projection of onb > 5 0 a > a >0cos u5 10 cos 135°527.07And the formula for the scalar projection of ona > 5 @b > b >@ cos u5 12 cos 135°528.49b. b12aCD135°10aQ OPOQ > is the vector projection of onOP > b >is the vector projection of a > a >on b >14. a. AB > 5 Point B 2 Point A5 (1, 3, 3) 2 (22, 1, 4)5 (3, 2, 21)The scalar projection of on isAB > ? OD >AB >OD >(3 321) 1 (2 3 2) 1 (21 3 2)@ OD > 5@"(21) 2 1 2 2 1 2 2(23) 1 4 1 (22)5"1 1 4 1 4Cabu u AD ccbuAb.52 1BC > 35 Point C 2 Point B5 (26, 7, 5) 2 (1, 3, 3)5 (27, 4, 2)The scalar projection of on isBC > ? OD >BC >OD >(27 321) 1 (4 3 2) 1 (2 3 2)@ OD > 5@"(21) 2 1 2 2 1 2 25 7 1 8 1 4"1 1 4 1 4AB > ? OD >@OD > @5 21"95 19"95 1931 BC> ? OD >@OD > @AC > 5 65 Point C 2 Point A5 (26, 7, 5) 2 (22, 1, 4)5 (24, 6, 1)The scalar projection of on isAC > ? OD >AC >OD >(24 321) 1 (6 3 2) 1 (1 3 2)@OD > 5@"(21) 2 1 2 2 1 2 25 4 1 12 1 2"1 1 4 1 45 18"952 1 3 1 1935 1835 1835 6c. Same lengths and both are in the direction of OD > .Add to get one vector.15. a. 1 5 cos 2 a1cos 2 b1cos 2 g22ab5 a"a 2 1 b 2 1 c 2b 1 a"a 2 1 b 2 1 c 2b2c1 a"a 2 1 b 2 1 c 2ba 25a 2 1 b 2 1 c 2 1 b 2a 2 1 b 2 1 c 2c 21a 2 1 b 2 1 c 27-20 <strong>Chapter</strong> 7: Applications of Vectors
5 a2 1 b 2 1 c 25 1b. a590°, b530°, g560°cos a5cos 90°5 0,x 5 0cos b5cos 30°5 "32 ,"3y is a multiple of 2 .cos g5cos 60°5 1 2 ,18. Answers may vary. For example:zB (0, c, d)yxA (a, b, 0)a 2 1 b 2 1 c 2 7-211z is a multiple of2 .Answers include Q0, "32 , 1 2R, Q0, "3, 1R, etc.c. If two angles add to 90° , then all three will add to180° .16. a. a5b5gcos a5cos b5cos gcos 2 a5cos 2 b5cos 2 g1 5 cos 2 a1cos 2 b1cos 2 g1 5 3 cos 2 x13 5 cos2 x1Å 3 5 cos x1x 5 cos 21 Å 3x 8 54.7°1b. For obtuse, use cos x 52 Å 3 .x 5 cos 21 1a2 Å 3 bx 8 125.3°17. cos 2 x 1 sin 2 x 5 1cos 2 x 5 1 2 sin 2 x1 5 cos 2 a1cos 2 b1cos 2 g1 5 (1 2 sin 2 a) 1 (1 2 sin 2 b) 1 (1 2 sin 2 g)1 5 3 2 (sin 2 a1sin 2 b1sin 2 g)sin 2 a1sin 2 b1sin 2 g527.6 The Cross Product of TwoVectors, pp. 407–4081. a.xa x bzaba > 3 b >is perpendicular to a > . Thus, their dot productmust equal 0. The same applies to the second case.za3bba1byab. is still in the same plane formed by andb > a > x1 b >thus a > 1 b >is perpendicular to a > 3 b > a >,making thedot product 0 again.c. Once again, is still in the same planeformed by and thus is perpendicular toa > 3 making the dot product 0 again.2. a > b > a > a > 2b > b >, a > 2 b >3 b >produces a vector, not a scalar. Thus, theequality is meaningless.3. a. It’s possible because there is a vector crossedwith a vector, then dotted with another vector,producing a scalar.b. This is meaningless because a > ? b >produces ascalar. This results in a scalar crossed with a vector,which is meaningless.yCalculus and Vectors <strong>Solution</strong>s Manual
5 a2 1 b 2 1 c 25 1b. a590°, b530°, g560°cos a5cos 90°5 0,x 5 0cos b5cos 30°5 "32 ,"3y is a multiple of 2 .cos g5cos 60°5 1 2 ,18. Answers may vary. For example:zB (0, c, d)yxA (a, b, 0)a 2 1 b 2 1 c 2 7-211z is a multiple of2 .Answers include Q0, "32 , 1 2R, Q0, "3, 1R, etc.c. If two angles add to 90° , then all three will add to180° .16. a. a5b5gcos a5cos b5cos gcos 2 a5cos 2 b5cos 2 g1 5 cos 2 a1cos 2 b1cos 2 g1 5 3 cos 2 x13 5 cos2 x1Å 3 5 cos x1x 5 cos 21 Å 3x 8 54.7°1b. For obtuse, use cos x 52 Å 3 .x 5 cos 21 1a2 Å 3 bx 8 125.3°17. cos 2 x 1 sin 2 x 5 1cos 2 x 5 1 2 sin 2 x1 5 cos 2 a1cos 2 b1cos 2 g1 5 (1 2 sin 2 a) 1 (1 2 sin 2 b) 1 (1 2 sin 2 g)1 5 3 2 (sin 2 a1sin 2 b1sin 2 g)sin 2 a1sin 2 b1sin 2 g527.6 The Cross Product of TwoVectors, pp. 407–4081. a.xa x bzaba > 3 b >is perpendicular to a > . Thus, their dot productmust equal 0. The same applies to the second case.za3bba1byab. is still in the same plane formed by andb > a > x1 b >thus a > 1 b >is perpendicular to a > 3 b > a >,making thedot product 0 again.c. Once again, is still in the same planeformed by and thus is perpendicular toa > 3 making the dot product 0 again.2. a > b > a > a > 2b > b >, a > 2 b >3 b >produces a vector, not a scalar. Thus, theequality is meaningless.3. a. It’s possible because there is a vector crossedwith a vector, then dotted with another vector,producing a scalar.b. This is meaningless because a > ? b >produces ascalar. This results in a scalar crossed with a vector,which is meaningless.yCalculus and Vectors <strong>Solution</strong>s Manual
c. This is possible. produces a vector, andc > 1 d > a > 3 b >also produces a vector. The result is a vectordotted with a vector producing a scalar.d. This is possible. produces a scalar, andc > 3 d > a > ? b >produces a vector. The product of a scalarand vector produces a vector.e. This is possible. produces a vector, andc > 3 d > a > 3 b >produces a vector. The cross product of avector and vector produces a vector.f. This is possible. a > 3 b >produces a vector. Whenadded to another vector, it produces another vector.4. a. (2, 23, 5) 3 (0, 21, 4)5 (23(4) 2 5(21), 5(0) 2 2(4),2(21) 2 (23)(0))5 (27, 28, 22)(2, 23, 5) ? (27, 28, 22) 5 0(0, 21, 4) ? (27, 28, 22) 5 0b. (2, 21, 3) 3 (3, 21, 2)5 (21(2) 2 3(21), 3(3) 2 2(2),2(21) 2 (21)(3))5 (1, 5, 1)(2, 21, 3) ? (1, 5, 1) 5 0(3, 21, 2) ? (1, 5, 1) 5 0c. (5, 21, 1) 3 (2, 4, 7)5 (21(7) 2 1(4), 1(2) 2 5(7),5(4) 2 (21)(2))5 (211, 233, 22)(5, 21, 1) ? (211, 233, 22) 5 0(2, 4, 7) ? (211, 233, 22) 5 0d. (1, 2, 9) 3 (22, 3, 4)5 (2(4) 2 9(3), 9(22) 2 1(4),1(3) 2 2(22))5 (219, 222, 7)(1, 2, 9) ? (219, 222, 7) 5 0(22, 3, 4) ? (219, 222, 7) 5 0e. (22, 3, 3) 3 (1, 21, 0)5 (3(0) 2 3(21), 3(1) 2 (22)(0),22(21) 2 3(1))5 (3, 3, 21)(22, 3, 3) ? (3, 3, 21) 5 0(1, 21, 0) ? (3, 3, 21) 5 0f. (5, 1, 6) 3 (21, 2, 4)5 (1(4) 2 6(2), 6(21) 2 5(4),5(2) 2 1(21))5 (28, 226, 11)(5, 1, 6) ? (28, 226, 11) 5 0(21, 2, 4) ? (28, 226, 11) 5 05. (21, 3, 5) 3 (0, a, 1)5 (3(1) 2 5(a), 5(0) 2 (21)(1),21(a) 2 3(0))If we look at the x component, we know that:3(1) 2 5(a) 52225(a) 5256. a. a > 3 b > a 5 15 (1(1) 2 1(5), 1(0) 2 0(1),0(5) 2 0(1))5 (24, 0, 0)b. Vectors of the form (0, b, c) are in theyz-plane. Thus, the only vectors perpendicular to theyz-plane are those of the form (a, 0, 0) because theyare parallel to the x-axis.7. a. (1, 2, 1) 3 (2, 4, 2)5 (2(2) 2 1(4), 1(2) 2 1(2), 1(4) 2 2(2))5 (0, 0, 0)b. (a, b, c) 3 (ka, kb, kc)5 (b(kc) 2 c(kb), c(ka) 2 a(kc),a(kb) 2 b(ka))Using the commutative law of multiplication wecan rearrange this:5 (bck 2 bck, ack 2 ack, abk 2 abk)5 (0,8. a. p > 0, 0)3 (q > 1 r > ) 5 (1, 22, 4) 3 3(1, 2, 7)1 (21, 1, 0)45 (1, 22, 4) 3 (1 2 1, 2 1 1, 7 1 0)5 (1, 22, 4) 3 (0, 3, 7)5 (22(7) 2 4(3), 4(0) 2 1(7),1(3) 1 2(0))p > 53 q > (226,1 p > 27,3 r > 3)5 (22(7) 2 4(2),b.p > 3 (q > 1 r > 5 (226, 27, 3)) 5 (4, 1, 2) 3 3(3, 1, 21)1 (0, 1, 2)45 (4, 1, 2) 3 (3, 1 1 1, 21 1 2)5 (4, 1, 2) 3 (3, 2, 1)5 (1(1) 2 2(2), 3(2) 2 4(1),4(2) 2 1(3))p > 3 q > 1 p > 3 r > 5 (23, 2, 5)5 (1(21) 2 2(1), 2(3) 2 4(21),4(1) 2 1(3)) 1 (1(2) 2 2(1),9. a.4(1) 2 1(7), 1(2) 1 2(1))1 (22(0) 2 4(1),4(21) 2 1(0), 1(1) 1 2(21))5 (222, 23, 4) 1 (24, 24, 21)2(0) 2 4(2), 4(1) 2 1(0))5 (23, 10, 1) 1 (0, 28, 4)i > 3 j > 5 (23, 2, 5)5 (1, 0, 0) 3 (0, 1, 0)5 (0 2 0, 0 2 0, 1 2 0)5 (0,5 k > 0, 1)7-22 <strong>Chapter</strong> 7: Applications of Vectors
.10.2j > 3 i > 5 (0, 21, 0) 3 (1, 0, 0)5 (0 2 0, 0 2 0, 0 2 (21))5 (0,j > 3 k > 5 k > 0, 1)5 (0, 1, 0) 3 (0, 0, 1)5 (1 2 0, 0 2 0, 0 2 0)5 (1,2k > 3 j > 5 i > 0, 0)5 (0, 0, 21) 3 (0, 1, 0)5 (0 2 (21), 0 2 0, 0 2 0)5 (1,c. k > 3 i > 5 i > 0, 0)5 (0, 0, 1) 3 (1, 0, 0)5 (0 2 0, 1 2 0, 0 2 0)5 (0,2i > 3 k > 5 j > 1, 0)5 (21, 0, 0) 3 (0, 0, 1)b.(by part a.)c. All the vectors are in the xy-plane. Thus, the crossproduct in part b. is between vectors parallel to thez-axis and so parallel to each other. The crossproduct of parallel vectors is12. Let x > 0 > .y > 5 (1, 0, 1)5 (1, 1, 1)Then5 (0 2 0, 0 2 (21), 0 2 0)5 (0,5 j > 1, 0)k(a 2 b 3 2 a 3 b 2 , a 3 b 1 2 a 1 b 3 , a 1 b 2 2 a 2 b 1 )? (a 1 , a 2 , a 3 )5 k(a 1 a 2 b 3 2 a 1 a 3 b 2 1 a 2 a 3 b 1 2 a 2 a 1 b 31 a 3 a 1 b 2 2 a 3 a 2 b 1 )5 k(0)a > 5 0is perpendicular to11. a. a > 3 b > k(a > 3 b > ).5 (2, 0, 0) 3 (0, 3, 0)5 (0 2 0, 0 2 0, 6 2 0)c > 3 d > 5 (0, 0, 6)5 (2, 3, 0) 3 (4, 3, 0)5 (0 2 0, 0 2 0, 6 2 12)(a > 3 b > 5 (0,) 3 (c > 0,3 d > 26)) 5 (0, 0, 6) 3 (0, 0, 26)5 (0 2 0, 0 2 0, 0 2 0)5 (0, 0, 0)z >x > 53 y > (1, 2, 3)5 (0 2 1, 1 2 1, 1 2 0)5 (21, 0, 1)(x > 3 y > ) 3 z > 5 (0 2 2, 1 2 (23), 23 2 0)5 (22, 4, 23)y > 3 z > 5 (3 2 2, 1 2 3, 2 2 1)x > 3 (y > 3 z > 5 (1, 22, 1)) 5 (0 1 2, 1 2 1, 22 2 0)Thus (x >13. (a > 32 b > y > 5 (2,) 3 z > 0, 22)) 3 (a > 2 x >1 b > 3 (y > 3 z > ).)By the distributive property of cross product:5 (a > 2 b > ) 3 a > 1 (a > 2 b > ) 3 b >By the distributive property again:5 a > 3 a > 2 b > 3 a > 1 a > 3 b > 2 b > 3A vector crossed with itself equals thus:52b >5 a > 35 a > 3 b > a > 135 2a > b > 2 b > a > 3323 b > (2a > a > b > 0 > b >,3 b > )7.7 Applications of the Dot Productand Cross Product, pp. 414–4151. By pushing as far away from the hinge aspossible, 0 r > 0 is increased making the cross productbigger. By pushing at right angles, sine is its largestvalue, 1, making the cross product larger.2. a. a > 3 b > 5 (1, 2, 1) 3 (2, 4, 2)5 (2(2) 2 1(4), 1(2)2 1(2), 1(4) 2 2(2))@a > 3 b > 5 (0, 0, 0)@ 5 0b. This makes sense because the vectors lie on thesame line. Thus, the parallelogram would just be aline making its area 0.3. a.b.f > ? s > 5 3 ? 150 5 450 Jy x40 m50° 392 NThe axes are tilted to illustrate the force of gravitycan be split up into components to find the part inthe direction of the motion. Let x be the componentof force going in the motion’s direction.cos (50°) 5x392x 5 (392) cos (50°)Now we have our force, so:(392) cos 50° N ? 40 m 8 10 078.91 JCalculus and Vectors <strong>Solution</strong>s Manual7-23
.10.2j > 3 i > 5 (0, 21, 0) 3 (1, 0, 0)5 (0 2 0, 0 2 0, 0 2 (21))5 (0,j > 3 k > 5 k > 0, 1)5 (0, 1, 0) 3 (0, 0, 1)5 (1 2 0, 0 2 0, 0 2 0)5 (1,2k > 3 j > 5 i > 0, 0)5 (0, 0, 21) 3 (0, 1, 0)5 (0 2 (21), 0 2 0, 0 2 0)5 (1,c. k > 3 i > 5 i > 0, 0)5 (0, 0, 1) 3 (1, 0, 0)5 (0 2 0, 1 2 0, 0 2 0)5 (0,2i > 3 k > 5 j > 1, 0)5 (21, 0, 0) 3 (0, 0, 1)b.(by part a.)c. All the vectors are in the xy-plane. Thus, the crossproduct in part b. is between vectors parallel to thez-axis and so parallel to each other. The crossproduct of parallel vectors is12. Let x > 0 > .y > 5 (1, 0, 1)5 (1, 1, 1)Then5 (0 2 0, 0 2 (21), 0 2 0)5 (0,5 j > 1, 0)k(a 2 b 3 2 a 3 b 2 , a 3 b 1 2 a 1 b 3 , a 1 b 2 2 a 2 b 1 )? (a 1 , a 2 , a 3 )5 k(a 1 a 2 b 3 2 a 1 a 3 b 2 1 a 2 a 3 b 1 2 a 2 a 1 b 31 a 3 a 1 b 2 2 a 3 a 2 b 1 )5 k(0)a > 5 0is perpendicular to11. a. a > 3 b > k(a > 3 b > ).5 (2, 0, 0) 3 (0, 3, 0)5 (0 2 0, 0 2 0, 6 2 0)c > 3 d > 5 (0, 0, 6)5 (2, 3, 0) 3 (4, 3, 0)5 (0 2 0, 0 2 0, 6 2 12)(a > 3 b > 5 (0,) 3 (c > 0,3 d > 26)) 5 (0, 0, 6) 3 (0, 0, 26)5 (0 2 0, 0 2 0, 0 2 0)5 (0, 0, 0)z >x > 53 y > (1, 2, 3)5 (0 2 1, 1 2 1, 1 2 0)5 (21, 0, 1)(x > 3 y > ) 3 z > 5 (0 2 2, 1 2 (23), 23 2 0)5 (22, 4, 23)y > 3 z > 5 (3 2 2, 1 2 3, 2 2 1)x > 3 (y > 3 z > 5 (1, 22, 1)) 5 (0 1 2, 1 2 1, 22 2 0)Thus (x >13. (a > 32 b > y > 5 (2,) 3 z > 0, 22)) 3 (a > 2 x >1 b > 3 (y > 3 z > ).)By the distributive property of cross product:5 (a > 2 b > ) 3 a > 1 (a > 2 b > ) 3 b >By the distributive property again:5 a > 3 a > 2 b > 3 a > 1 a > 3 b > 2 b > 3A vector crossed with itself equals thus:52b >5 a > 35 a > 3 b > a > 135 2a > b > 2 b > a > 3323 b > (2a > a > b > 0 > b >,3 b > )7.7 Applications of the Dot Productand Cross Product, pp. 414–4151. By pushing as far away from the hinge aspossible, 0 r > 0 is increased making the cross productbigger. By pushing at right angles, sine is its largestvalue, 1, making the cross product larger.2. a. a > 3 b > 5 (1, 2, 1) 3 (2, 4, 2)5 (2(2) 2 1(4), 1(2)2 1(2), 1(4) 2 2(2))@a > 3 b > 5 (0, 0, 0)@ 5 0b. This makes sense because the vectors lie on thesame line. Thus, the parallelogram would just be aline making its area 0.3. a.b.f > ? s > 5 3 ? 150 5 450 Jy x40 m50° 392 NThe axes are tilted to illustrate the force of gravitycan be split up into components to find the part inthe direction of the motion. Let x be the componentof force going in the motion’s direction.cos (50°) 5x392x 5 (392) cos (50°)Now we have our force, so:(392) cos 50° N ? 40 m 8 10 078.91 JCalculus and Vectors <strong>Solution</strong>s Manual7-23
c.140 N20°250 mFirst find the x component of the force:(140) cos (20°) 5 xCalculate work:140 cos 20° N ? 250 m 8 32 889.24 Jd.100 N45°500 mFirst calculate the x component of the force:x 5 (100) cos (45°)Calculate work:100 cos4. a. i > 45°3 j > ? 5005 k > m 5 35 355.34 JThe square formed by the 2 vectors has an area of 1.The 2 vectors are on the xy-plane, thus, the crossproduct must be by the right hand rule.b. 2i > 3 j > 52k > k>Once again, the area is 1, making the possible vectorhave a magnitude of 1. Also, the 2 vectors are on thexy-plane again so the cross product must lie on thez axis. However, because of the right hand rule, theproduct must be this time.c. i > 3 k > 52j > 2k >The square has an area of 1, so the magnitude of thevector produced must be 1. The 2 vectors are on thexz-plane. The new vector must be on the y axismaking it – because of the right hand rule.d. 2i > 3 k > j >52j >The square has an area of 1. The 2 vectors are onthe xz-plane. So the new vector must be j >becauseof the right hand rule.5. a. a > 3 b > 5 (1, 1, 0) 3 (1, 0, 1)5 (1 2 0, 0 2 1, 0 2 1)@a > 3 b > 5 (1, 21, 21)@ 5 "1 1 1 1 1 5 "3So the area of the parallelogram is square units.b. a > 3 b > "35 (1, 22, 3) 3 (1, 2, 4)5 (28 2 6, 3 2 4, 2 1 2)@a > 3 b > 5 (214, 21, 4)@ 5 "196 1 1 1 16 5 "213So the area of the parallelogram is "213 square units.6. p > 3 q > 5 (a, 1, 21) 3 (1, 1, 2)5 (2 1 1, 22a 2 1, a 2 1)0 p > 3 q > 5 (3, 2a 1 1, a 2 1)0 5 "9 1 (2a 1 1) 2 1 (a 2 1) 2 5 "359 1 (2a 1 1) 2 1 (a 2 1) 2 5 359 1 4a 2 1 4a 1 1 1 a 2 2 2a 1 1 5 355a 2 1 2a 2 24 5 0a 5 22 6 "22 2 4(5)(224)2(5)22 6 225102125 2,57. a.BABAAC CAs we see from the picture, the area of the triangleABC is just half the area of the parallelogramdetermined by vectors AB >and AC > . Thus, we use themagnitude of the cross product to calculate the area.AB >AC > 5 (1 1 2, 0 2 1, 1 2 3) 5 (3, 21, 22)AB > 5 (2 1@AB > 3 AC > 2, 3 2 1, 2 2 3) 5 (4, 2, 21)3 AC > 5 (1 1 4, 23 1 8, 6 1 4) 5 (5, 5, 10)@ 5 "25 1 25 1 100 5 5"6Since triangle ABC is half the area of the5"6parallelogram, its area is 2 square units.b. This is just a different way of describing the first5"6triangle, thus the area is 2 square units.c. Any two sides of a triangle can be used tocalculate its area.8. @r > 3 f > @ 5 ( 0 r > 0sin (u)) @ f > @5 (0.14) sin (45°) ? 108 0.99 J9.AOA BNOCOB BWe know that the area of a parallelogram is equal toits height multiplied with its base. Its height is BNand its base is AC > 5 OB >as can be seen from thepicture. We can calculate the area using the givenvectors, then use the area to find BN.OA > 3 OB > 5 (8 2 4, 12 2 16, 4 2 6)5 (4, 24, 22)@ OA > 3 OB > @ 5 "16 1 16 1 4 5 "36 5 67-24 <strong>Chapter</strong> 7: Applications of Vectors
Now we need to calculate @OB > @ to know the lengthof the base.AC > 5 @OB > @ 5 "9 1 1 1 16 5 "26Substituting these results into the equation for area:@OB > @ ? BN 5 6"26 BN 5 6BN 5 6 or about 1.18"2610. a.p > 3 q > 5 (26 2 3, 6 2 3, 1 1 4)( p > 3 q > ) 3 r > 5 (29, 3, 5)5 (0 2 5, 5 1 0, 29 2 3)5 (25, 5, 212)a(1, 22, 3) 1 b(2, 1, 3) 5 (25, 5, 212)Looking at x components:a 1 2b 525;a 525 2 2by components:22a 1 b 5 5Substitute in a:10 1 4b 1 b 5 55b 525b 521Substitute b back into the x components:a 525 1 2;a 523Check in z components:3a 1 3b 521229b. p > 2q > ? r > 3 5212( p > ? r > 5 1 2 2 1 0 521? r > 5)q > 2 12 (q > 1 1? r > 0)p > 5 3521(2, 1, 3) 2 3(1, 22, 3)5 (2, 21, 23) 2 (3, 26, 9)5 (22 2 3, 21 1 6, 23 2 9)5 (25, 5, 212)Review Exercise, pp. 418–4211. a. a > 3 b > 5 (2 2 0, 21 1 1, 0 1 2)b. b > 3 c > 5 (2, 0, 2)5 (0 2 4, 25 1 5, 24 2 0)5 (24, 0, 24)c. 16d. The cross products are parallel, so the originalvectors are in the same plane.2. a.b. @b > 0 a > 0 5 "2 2 1 (21) 2 1 2 2 5 3c. a > @ 52 b > "6 2 1 3 2 1 (22) 2 5 75 (2 2 6, 21 2 3, 2 1 2)@a > 2 b > 5 (24, 24, 4)@ 5 "(24) 2 1 (24) 2 1 4 2 5 4"3d. a > 1 b > 5 (2 1 6, 21 1 3, 2 2 2)@a >e. a > 1 b > 5 (8, 2, 0)? b > @ 5 "8 2 1 2 2 1 0 2 5 2"17f. a > 5 2(6)2 2b > 2 1(3) 1 2(22) 5 55 (2 2 12, 21 2 6, 2 1 4)a > ? (a > 2 2b > 5 (210, 27, 6)) 5 2(210)3. a. If then will be twice thus collinear.b. x > 3 y > y > 2 1(27) 1x > 2(6) 521a 5 6,,5 (3, a, 9) ? (a, 12, 18) 5 03 a 1 12a 1 162 5 015a 52162a 5 2544. cos (u) 5 a> ? b > 5a > ? b > 0 a > 0 @b > @0 a > 5 4(23) 1 5(6) 1 20(22) 5 458@b > 0 5 "4 2 1 5 2 1 20 2 5 21@ 5 "(23) 2 1 6 2 1 22 2 5 23u5cos 21 a 458483 bu 8 18.52°5. a.y42OB OA x–4 –2 0 2 4–2–4b. We can use the dot product of the 2 diagonals tocalculate the angle. The diagonals are the vectorsOA > andOA > 1 OB >OA > 1 OB > OA > 2 OB > .2 OB > 5 (5 2 1, 1 1 4) 5 (4, 5)5 (5 1 1, 1 2cos (u) 5 (OA> 1 OB > 4) 5 (6,) ? (OA > 23)2 OB > )@OA > 1 OB > @@OA > 2 OB > @(OA > 1 OB > ) ? (OA > 2 OB > ) 5 4(6) 1 5(23) 5 9@OA > 1 OB > @ 5 "4 2 1 5 2 5 "41@OA > 2 OB > @ 5 "6 2 1 (23) 2 5 3"59u5cos 21 a3"205 bu 8 77.9°Calculus and Vectors <strong>Solution</strong>s Manual7-25
Now we need to calculate @OB > @ to know the lengthof the base.AC > 5 @OB > @ 5 "9 1 1 1 16 5 "26Substituting these results into the equation for area:@OB > @ ? BN 5 6"26 BN 5 6BN 5 6 or about 1.18"2610. a.p > 3 q > 5 (26 2 3, 6 2 3, 1 1 4)( p > 3 q > ) 3 r > 5 (29, 3, 5)5 (0 2 5, 5 1 0, 29 2 3)5 (25, 5, 212)a(1, 22, 3) 1 b(2, 1, 3) 5 (25, 5, 212)Looking at x components:a 1 2b 525;a 525 2 2by components:22a 1 b 5 5Substitute in a:10 1 4b 1 b 5 55b 525b 521Substitute b back into the x components:a 525 1 2;a 523Check in z components:3a 1 3b 521229b. p > 2q > ? r > 3 5212( p > ? r > 5 1 2 2 1 0 521? r > 5)q > 2 12 (q > 1 1? r > 0)p > 5 3521(2, 1, 3) 2 3(1, 22, 3)5 (2, 21, 23) 2 (3, 26, 9)5 (22 2 3, 21 1 6, 23 2 9)5 (25, 5, 212)Review Exercise, pp. 418–4211. a. a > 3 b > 5 (2 2 0, 21 1 1, 0 1 2)b. b > 3 c > 5 (2, 0, 2)5 (0 2 4, 25 1 5, 24 2 0)5 (24, 0, 24)c. 16d. The cross products are parallel, so the originalvectors are in the same plane.2. a.b. @b > 0 a > 0 5 "2 2 1 (21) 2 1 2 2 5 3c. a > @ 52 b > "6 2 1 3 2 1 (22) 2 5 75 (2 2 6, 21 2 3, 2 1 2)@a > 2 b > 5 (24, 24, 4)@ 5 "(24) 2 1 (24) 2 1 4 2 5 4"3d. a > 1 b > 5 (2 1 6, 21 1 3, 2 2 2)@a >e. a > 1 b > 5 (8, 2, 0)? b > @ 5 "8 2 1 2 2 1 0 2 5 2"17f. a > 5 2(6)2 2b > 2 1(3) 1 2(22) 5 55 (2 2 12, 21 2 6, 2 1 4)a > ? (a > 2 2b > 5 (210, 27, 6)) 5 2(210)3. a. If then will be twice thus collinear.b. x > 3 y > y > 2 1(27) 1x > 2(6) 521a 5 6,,5 (3, a, 9) ? (a, 12, 18) 5 03 a 1 12a 1 162 5 015a 52162a 5 2544. cos (u) 5 a> ? b > 5a > ? b > 0 a > 0 @b > @0 a > 5 4(23) 1 5(6) 1 20(22) 5 458@b > 0 5 "4 2 1 5 2 1 20 2 5 21@ 5 "(23) 2 1 6 2 1 22 2 5 23u5cos 21 a 458483 bu 8 18.52°5. a.y42OB OA x–4 –2 0 2 4–2–4b. We can use the dot product of the 2 diagonals tocalculate the angle. The diagonals are the vectorsOA > andOA > 1 OB >OA > 1 OB > OA > 2 OB > .2 OB > 5 (5 2 1, 1 1 4) 5 (4, 5)5 (5 1 1, 1 2cos (u) 5 (OA> 1 OB > 4) 5 (6,) ? (OA > 23)2 OB > )@OA > 1 OB > @@OA > 2 OB > @(OA > 1 OB > ) ? (OA > 2 OB > ) 5 4(6) 1 5(23) 5 9@OA > 1 OB > @ 5 "4 2 1 5 2 5 "41@OA > 2 OB > @ 5 "6 2 1 (23) 2 5 3"59u5cos 21 a3"205 bu 8 77.9°Calculus and Vectors <strong>Solution</strong>s Manual7-25
6.T130°The vertical components of the tensions must equalthe downward force:T 1 sin (30°) 1 T 2 sin (45°) 5 98 N12 T 1 1 1 "2 T 2 5 98T 1 5 196 2 "2T 2The horizontal components:T 1 cos (30°) 1 T 2 cos (45°) 5 0 N"32 T 1 2 1 "2 T 2 5 0Substitute in T 1 :98"3 2 "62 T 2 5298"32"6 2 "2T2 2 5298"3T 2 8 87.86NSubstitute this back in to get T 1 :T 1 8 71.74N7.x50 km/hx 5 "50 2 1 300 2 8 304.14tan 21 a 50300 b 8 9.46°The resultant velocity is 304.14 km>h, W 9.46° N.8. a.zx98 NyT 245°300 km/hxxyyb. x > 3 y > 5 (215 2 35, 25 2 15, 21 2 3)0 x > 3 y > 5 (250, 220, 18)0 5 "50 2 1 20 2 1 18 2 5 "3224 8 56.789. (0, 3, 25) 3 (2, 3, 1)5 (3 1 15, 210 2 0, 0 2 6) 5 (18, 210, 26)The cross product is perpendicular to the givenvectors, but its magnitude is"18 2 1 (210) 2 1 (26) 2 , or 2"115. A unit vectorperpendicular to the given vectors isa 9!115 , 2 5!115 , 2 3!115 b.10. a. cos (a) 5 AB> ? AC >AB > @AB > @@AC > @AB > AC > 5 (0, 23, 4) 2 (2, 3, 7) 5 (22, 26, 23)? AC > 5 (5, 2, 24) 2 (2, 3, 7) 5 (3, 21, 211)@AB > 522(3) 2 6(21) 2 3(211) 5 33@@AC > 5 "(22) 2 1 (26) 2 1 (23) 2 5 7@ 5 "3 2 1 (21) 2 1a5cos 21 AB> ? AC > (211) 2 5 "131@AB > @@AC > @335 cos 217"1318 65.68°b5cos 21 BA> ? BC >BA >BA > BC > 52AB > @BA > @@BC > @5 (2, 6, 3)? BC > 5 (5 2 0, 2 1 3, 24 2 4, ) 5 (5, 5, 28)@BA > 5 2(5) 1 6(5) 1 3(28) 5 16@@BC > 5 "2 2 1 6 2 1 3 2 5 7@ 5 "5 2 1 5 2 1 (28) 2 5 "144b5cos 21 167"1148 77.64°g5180 2a2b 8 36.68°So b 8 77.64° is the largest angle.b. The area is half the magnitude of the crossproduct of AB >and AC > .12 AB> 3 AC > 5 1 0 (63, 231, 20) 0 8 36.50211. The triangle formed by the two strings and theceiling is similar to a 3-4-5 right triangle, with the30 cm and 40 cm strings as legs. So the angleadjacent to the 30 cm leg satisfiescos u 5 3 57-26 <strong>Chapter</strong> 7: Applications of Vectors
The angle adjacent to the 40 cm leg satisfiescos f 5 4 5Also,sin u 5 4 and sin f 5 3 55 .Let T 1be the tension in the 30 cm string, and T 2bethe tension in the 40 cm string. ThenT 1 cos u2T 2 cos f503T 15 2 T 425 5 0Also,T 1 sin u1T 2 sin f5(10)(9.8) 5 984T 15 2 T 325 5 985 78.4 NSo the tension in the 30 cm string is 78.4 N and thetension in the 40 cm string is 58.8 N.12. a.30 NT 1 5 4 3 T 2a 4 3 T 2b 4 5 1 T 325 5 9853 T 2 5 98T 2 5 58.8 N54 N42 N25 NT 1 5 4 3 (58.8)b. The east- and west-pulling forces result in a forceof 5 N west. The north- and south-pulling forcesresult in a force of 12 N north. The 5 N west and12 N north forces result in a force pulling in thenorth-westerly direction with a force of"5 2 1 12 2 5 13 N,by using the Pythagorean theorem. To find the exactdirection of this force, use the definition of sine.Calculus and Vectors <strong>Solution</strong>s ManualIf u is the angle west of north, thensin u 5 513u 8 22.6°So the resultant is 13 N in a directionN22.6°W. The equilibrant is 13 N in a directionS22.6°E.13. a. Let D be the origin, then:A 5 (2, 0, 0), B 5 (2, 4, 0), C 5 (0, 4, 0),D 5 (0, 0, 0), E 5 (2, 0, 3), F 5 (2, 4, 3),G 5 (0, 4, 3) H 5 (0, 0, 3)b. AF >AF > AC > 5 (0, 4, 3)? AC > 5 (22, 4, 0)@AF > 5 0 1 16 1 0 5 16@@AC > 5 "0 2 1 4 2 1 3 2 5 5@ 5 "(22) 2cos (u) 5 AF> ? AC > 1 4 2 1 0 2 5 2"5@AF > @@AC > @u5cos 21 a 1610"5 bu 8 44.31°c. Scalar projection 5 @AF > @ cos (u)By part b.:5 (5) cos (44.31°)8 3.5814. a > ? b > 5 0 a > 0 @b > @ cos (u) 5 cos (u)cos (u) 52 1 (cosine law)(2a > 2 5b > 25213a > ) ?5213a > ? b > (b > 1? b > 1 6a > 3a > )? a > 2 5b > ? b >1 15213 cos (u) 1 15 7.515. a. The angle to the bank, u, will satisfysin (90° 2u) 5 2 390° 2u8 41.8°u 8 48.2°b. By the Pythagorean theorem, Kayla’s netswimming speed will be"3 2 2 2 2 5 "5 km>h.So since distance 5 rate 3 time, it will take hert 5 0.3"58 0.13 h8 8 min 3 secto swim across.7-27
c. Such a situation would have resulted in a righttriangle where one of the legs is longer than thehypotenuse, which is impossible.16. a. The diagonals are OA > 1 OB >andOA > 2 OB > .OA > 1 OB > 5 (3 2 6, 2 1 6, 26 2 2)OA > 2 OB > 5 (23, 8, 28)5 (3 1 6, 2 2 6, 26 1 2)b. OA > 5? OB > (9, 24, 24)@OA > 5 3(26) 1 2(6) 2 6(22) 5 6@ 5 "3 2 1 2 2 1 (26) 2 5 7@OB > @ 5 "(26) 2 1cos (u) 5 OA> ? OB > 6 2 1 (22) 2 5 2"19@OA > @@OB > @u5cos 21 65 a14"19 b8 84.36°17. a. The z value is double, so if andthe vector will be collinear.b. If p > and q > q >a 5 4b 524,are perpendicular, then their dotproduct will equal 0.p > ? q > 5 2a 2 2b 2 18 5 0c. Let a 5 9, and b then we have a vectorperpendicular to p > 5 0,. Now it must be divided by itsmagnitude to make it a unit vector:0 p > 0 5 "81 1 0 1 324 5 9"5So the unit vector is:a 1 !5 , 0, 218. a. m > !5 b? n >0 m > 5 2"3 2 2"3 1 3 5 30 n > 0 5 "3 1 4 1 9 5 40 5 "4 1cos (u) 5 m> ? n > 3 1 1 5 2"20 m > 00n > 0u5cos 21 a 38"2 b8 74.62°b. Scalar projection 5 0 n > 0cos (u)5 2"2 cos (74.62°)8 0.75c. Scalar projection multiplied with the unit vectorin the direction of m > :5 (0.75) m>0 m > 0("3, 22, 23)5 (0.75)45 (0.1875)("3, 22, 23)d. m > ? k > 523u5cos 21 a 234 b8 138.59°19. a. If the dot product is 0, then the vectors areperpendicular:(1, 0, 0) ? (0, 0, 21) 5 0 1 0 1 0 5 0(1, 0, 0) ? (0, 1, 0) 5 0 1 0 1 0 5 0(0, 0, 21) ? (0, 1, 0) 5 0 1 0 1 0 5 0 specialb. a 1 !2 , 1 21, 0b ? a!2 !3 , 1!3 , 1!3 b52 1 !6 1 1 !6 1 05 0a 1 !2 , 1, 0b ? (0, 0, 21) 5 0 1 0 1 0 5 0!2a 21!3 , 1!3 , 1b ? (0, 0, 21)!35 0 1 0 12 1p > 3 q > !3 52 1 !3not special20. a.5 (22(1) 2 1(21), 1(2) 2 1(1), 1(21) 1 2(2))5 (21,b. p > p > 2 q > 1, 3)1 q > 5 (21, 21, 0)5 (3, 23, 2)(p > 2 q > ) 3 (p > 1 q > ) 5 (22 2 0, 0 1 2, 3 2 (23))c. p > 3 r > 5 (22, 2, 6)5 (4 2 1, 0 1 2, 1 2 0)5 (3, 2, 1)( p > 3d. p > r > )3 q > ? r > 5 0 1 2 2 2 5 05 (22 1 1, 2 2 1, 21 1 4)5 (21, 1, 3)21. Since the angle between the two vectors is 60°,the angle formed when they are placed head-to-tailis 120°. So the resultant, along with these twovectors, forms an isosceles triangle with top angle120° and two equal angles 30°. By the cosine law,the two equal forces satisfy20 2 5 2F 2 2 2F 2 cos 120°F 2 5 4003F 5 20"3822. a > 11.553 b > N5 (2 2 0, 25 2 3, 0 2 10)5 (2, 28, 210)7-28 <strong>Chapter</strong> 7: Applications of Vectors
23. First we need to determine the dot product ofandx > ? y > y > :5 0 x > 00y > 0cos u5 (10) cos (60°)(x > 52 2y > 5) ? (x > 1 3y > )By the distributive property:5 x > ? x > 1 3x > ? y > 2 2x > ? y > 2 6y > ? y >5 4 1 15 2 10 2 1505214124. 0 (2, 2, 1) 0 5 "2 2 1 2 2 1 1 2 5 3Since the magnitude of the scalar projection is 4,the scalar projection itself has value 4 or 24.If it is 4, we get(1, m, 0)?(2, 2, 1)5 432 1 2m 5 12m 5 5If it is 24, we get(1, m, 0)?(2, 2, 1)52432 1 2m 5 212m 5 27So the two possible values for m are 5 and25. a > ?0 a > j > 27.5230 5 "144 1 9 1 16 5 13u5cos 21 a 2313 b8 103.34°26. a.b. CF > C 5 (3, 0, 5), F 5 (0, 4, 0)5 (0, 4, 0) 2 (3, 0, 5) 5 (23, 4, 25)c. @CF > @OP > 5 "9 1 16 1 25 5 5"25CF > @OP > (3, 4, 5)@? OP > 5 "9 1 16 1 25 5 5"2529 1 16 2 25 5218u5cos 21 a 21850 b8 111.1°27.d 130°50°ea. Using properties of parallelograms, we know thatthe other angle is 130 ° (Angles must add up to360 ° , opposite angles are congruent).Using the cosine law,@d > 1 e > @ 2 5 3 2 1 5 2 2 2(3)(5) cos 130°@d > 1 e > @ 8 7.30b. Using the cosine law,@d > 2 e > @ 2 5 3 2 1 5 2 2 2(3)(5) cos 50°@d > 2 e > @ 8 3.84c. is the vector in the opposite direction ofd > e > 22 e > d >, but with the same magnitude. So:@e > 2 d > @ 5 @d > 2(i > e > @ 81 j > 3.84) ? (i > )28. a. Scalar:@i > 5 1@Vector:(i > 1 j > ) ? (j > )b. Scalar:@ j > 5 1@Vector:1a i>1a j>@i > b 5 i >@@ j > b 5 j >@(i > 1 j > ) ? (k > 1 j > )c. Scalar:@k > 1 j > 5 1 @ "21Vector:"2 ? (k> 1 j > )@k > 1 j > 5 1 @ 2 (k> 1 j > )29. a. If its magnitude is 1, it’s a unit vector:0 a > 0 5 " 1 not a unit vector@b > 4 1 1 9 1 136 2 1@ 5 " 1 3 1 1 3 1 1 3 5 1, unit vector0 c > 0 5 " 1 4 1 1 2 1 1 4 5 1, unit vector@d > @ 5 not a unit vectorb. a >"1 1 1 1 1 2 1,is. When dotted with d > , it equals 0.30. 25 ?31. a.b. a > a > sin? b > (30°) ? 0.6 5 7.50 J5 6 2 5 2 1 5 0with the x-axis:0 a > 0 5 "4 1 25 1 1 5 "30cos (a) 5 2a > "30with the y-axis:cos (b) 5 5a > "30with the z-axis:cos (g) 5 21@b > "30@b > 5 "9 1 1 1 1 5 "11with the x-axis:cos (a) 5 3"11x > 7-29Calculus and Vectors <strong>Solution</strong>s Manual
7-30with the y-axis:cos (b) 5 21b > "11with the z-axis:cos (g) 5 1"11c. m > 1 ? m > 2 5 6!330 2 5!330 2 1!330 5 032. Need to show that the magnitudes of thediagonals are equal to show that it is a rectangle.@3i > 1 3j > 1 10k > @ 5 "9 1 9 1 100 5 "118@2i > 1 9j > 2 6k > @ 5 "1 1 81 1 36 5 "11833. a. Direction cosine for x-axis:cos (30°) 5 "32We know the identitycos 2 a1cos 2 b1cos 2 g51.Since a530g, and b5g, we get2 cos 2 b51 2 3cos b5cos g56 142"2cos a5 "32So there are two possibilities, depending uponwhether b5gis acute or obtuse.b. If g is acute, thencos g5 12"2g 8 69.3°If Á is obtuse, thencos g5 12"2g 8 110.7°34. a > ? b > 5 0 a > 0 @b > @ cos (u) 5 1(a >ma > 2 3b >? a > ) ?1 a > (ma >? b > 1 b > 2)2 3ma > 5 0? b > 2 3b > ? b > 5 0m 1 1 2 2 3 2 m 2 6 2 5 035.2 1 2 m 5 5 2m 525a > a > ? b >@a > 1 b > 5 0 2 20 1 12 5281 b > 5 (21, 21, 28)@ 5 "1 1 1 1 64 5 "66a >@a > 2 b >2 b > 5 (1, 9, 24)@ 5 "1 1 81 1 16 5 "9814 @a> 1 b > @ 2 2 136.0 c > c > 5 b > 4 @a> 2 b > @ 2 5 660 2 5 @b > 2 a >4 2 984 5282 a > 2@5 (b >5 b > 25 0 a > ? b > a > ) ?2 a > (b >02 1 @b > ? b > 2 a >1@ 2 2 2a > a > )?? b > a > 2 a > ? b >5 0 a > 0 2 1 @b > @ 2 2 20 a > 0 @b > @ cos u37. AB >@AB > 5 (2, 0, 4)@@AC > 5 "4 1 0 1 16 5 2"5@@AC > 5 (1, 0, 2)@BC > 5 "1 1 0 1 4 5 "5@BC > 5 (21, 0, 22)@ 5 "1 1 0 1 4 5 "5cos A 5 AB> ? AC >@ AB > @@AC > @5 10105 1But this means that angle A 5 0° , so that thistriangle is degenerate. For completeness, though,notice that BC > 52AC > and AB > 5 2 AC >. Thismeans that point C sits at the midpoint of the linesegment joining A and B. So angleC 5 180° and angle B 5 0° . Socos B 5 1;cos C 521.The area of triangle ABC is, of course, 0.<strong>Chapter</strong> 7 Test, p. 4221. a. We use the diagram to calculate a > 3 b > , notinga 1 521, a 2 5 1, a 3 5 1 and b 1 5 2, b 2 5 1,b 3 523.a > b >1 1xx 5 1(23) 2 1(1) 5241 23yy 5 1(2) 2 (21)(23) 52121 2zz 521(1) 2 1(2) 5231 1So, a > 3 b > 5 (24, 21, 23)<strong>Chapter</strong> 7: Applications of Vectors
7-30with the y-axis:cos (b) 5 21b > "11with the z-axis:cos (g) 5 1"11c. m > 1 ? m > 2 5 6!330 2 5!330 2 1!330 5 032. Need to show that the magnitudes of thediagonals are equal to show that it is a rectangle.@3i > 1 3j > 1 10k > @ 5 "9 1 9 1 100 5 "118@2i > 1 9j > 2 6k > @ 5 "1 1 81 1 36 5 "11833. a. Direction cosine for x-axis:cos (30°) 5 "32We know the identitycos 2 a1cos 2 b1cos 2 g51.Since a530g, and b5g, we get2 cos 2 b51 2 3cos b5cos g56 142"2cos a5 "32So there are two possibilities, depending uponwhether b5gis acute or obtuse.b. If g is acute, thencos g5 12"2g 8 69.3°If Á is obtuse, thencos g5 12"2g 8 110.7°34. a > ? b > 5 0 a > 0 @b > @ cos (u) 5 1(a >ma > 2 3b >? a > ) ?1 a > (ma >? b > 1 b > 2)2 3ma > 5 0? b > 2 3b > ? b > 5 0m 1 1 2 2 3 2 m 2 6 2 5 035.2 1 2 m 5 5 2m 525a > a > ? b >@a > 1 b > 5 0 2 20 1 12 5281 b > 5 (21, 21, 28)@ 5 "1 1 1 1 64 5 "66a >@a > 2 b >2 b > 5 (1, 9, 24)@ 5 "1 1 81 1 16 5 "9814 @a> 1 b > @ 2 2 136.0 c > c > 5 b > 4 @a> 2 b > @ 2 5 660 2 5 @b > 2 a >4 2 984 5282 a > 2@5 (b >5 b > 25 0 a > ? b > a > ) ?2 a > (b >02 1 @b > ? b > 2 a >1@ 2 2 2a > a > )?? b > a > 2 a > ? b >5 0 a > 0 2 1 @b > @ 2 2 20 a > 0 @b > @ cos u37. AB >@AB > 5 (2, 0, 4)@@AC > 5 "4 1 0 1 16 5 2"5@@AC > 5 (1, 0, 2)@BC > 5 "1 1 0 1 4 5 "5@BC > 5 (21, 0, 22)@ 5 "1 1 0 1 4 5 "5cos A 5 AB> ? AC >@ AB > @@AC > @5 10105 1But this means that angle A 5 0° , so that thistriangle is degenerate. For completeness, though,notice that BC > 52AC > and AB > 5 2 AC >. Thismeans that point C sits at the midpoint of the linesegment joining A and B. So angleC 5 180° and angle B 5 0° . Socos B 5 1;cos C 521.The area of triangle ABC is, of course, 0.<strong>Chapter</strong> 7 Test, p. 4221. a. We use the diagram to calculate a > 3 b > , notinga 1 521, a 2 5 1, a 3 5 1 and b 1 5 2, b 2 5 1,b 3 523.a > b >1 1xx 5 1(23) 2 1(1) 5241 23yy 5 1(2) 2 (21)(23) 52121 2zz 521(1) 2 1(2) 5231 1So, a > 3 b > 5 (24, 21, 23)<strong>Chapter</strong> 7: Applications of Vectors
. We use the diagram again:c >1 1xx 5 1(27) 2 (23)(1) 52423 2 7yy 523(5) 2 (2)(27) 5212 5zz 5 2(1) 2 1(5) 5231 1So,c. a > b > 3? (b > c > 53 c > (24, 21, 23)) 5 (21, 1, 1) ? (24, 21, 23)5 (21)(24) 1 (1)(21)1 (1)(23)5 0d. We could use the diagram method again, or, wenote that for any vectorsso letting y > 5 x > x > ,we have from the lastequation. Since a > 3 b > x > y > , x >5 from the first twoparts of the problem, (a > b > 3 x > 3 y > 52(y > 3 x > ),,33 b > c > 5 0) 3 (b > 3 c > ) 5 0.2. a. To find the scalar and vector projections ofon we need to calculate anda > b > ,? b > a > ? b >@b > @ 5 "b > a >? b >5 (1, 21, 1) ? (2, 21, 22)5 (1)(2) 1 (21)(21) 1 (1)(22)@b > 5 1@ 5 "2 2 1 (21) 2 1 (22) 2So, @b > 5 3@ 5 3The scalar projection of on isa > >a > b >? b> 5 1 and@b @3,the vector projection of a > on b >is>a a> ? b> 2 bb > 5 1@b @9(2, 21, 22).b. We find the direction cosines for b > :cos (a) 5 b 1@b > 5 2 @ 3a 8 48.2°.cos (b) 5 b 2@b > 5 21@ 3b 8 109.5°.cos (g) 5 b 3@b > 5 22@ 3g 8 131.8°.c. The area of the parallelogram is the magnitude ofthe cross product.a > b >21 21xx 5 (21)(22) 2 1(21) 5 31 22yy 5 1(2) 2 (1)(22) 5 41 2zz 5 (1)(21) 2 (21)(2) 5 121So, a > 21and thus,@a > 3 b >3 b > 5 (3, 4, 1)@ 5 "3 2 1 4 2 1 1 25 "26So the area of the parallelogram formed by a > andis "26 or 5.10 square units.3. We first draw a diagram documenting thesituation:E40 N60°50 N40 N60°120°50 NIn triangle DEF, we use the cosine law:@R > @ 5 "40 2 1 50 2 2 2(40)(50) cos (120°)@R > @ 8 78.10We now use the sine law to find /EDF:sin /EDF sin /DEF@EF > 5@ @R > @sin /EDF sin 120°850 78.10sin /EDF 8 0.5544/EDF 8 33.7°The equilibrant force is equal in magnitude andopposite in direction to the resultant force, so bothforces have a magnitude of 78.10 N. The resultantmakes an angle 33.7°to the 40 N force and 26.3°tothe 50 N force. The equilibrant makes an angle 146.3°to the 40 N force and 153.7°to the 50 N force.DRGb >Fb > 7-31Calculus and Vectors <strong>Solution</strong>s Manual
4. We find the resultant velocity of the airplane.2.5 m/sF1000 km/hPosition diagramESince the airplane’s velocity is perpendicular to thewind, the resultant’s magnitude is given by thePythagorean theorem:@R > @@R > 5 "1000 2 1 100 2@ 8 1004.99The angle is determined using the tangent ratio:tan /EDF 5 1001000/EDF 8 5.7°Thus, the resultant velocity is 1004.99 km>h,N 5.7°W (or W 84.3°N).5. a. The canoeist will travel 200 m across thestream, so the total time he will paddle is:t 5dr canoeistt 5 200 m2.5 m>st 5 80 sThe current is flowing 1.2 m>s downstream, so thedistance that the canoeist travels downstream is:d 5 r current 3 td 5 (1.2 m>s)(80 s)d 5 96 mSo, the canoeist will drift 96 m south.b. In order to arrive directly across stream, thecanoeist must take into account the change in hisvelocity caused by the current. That is, he mustinitially paddle upstream in a direction such thatthe resultant velocity is directed straight acrossthe stream. The resultant velocity:E1.2 m/sRFG DVector diagramSince the resultant velocity is perpendicular to thecurrent, the direction in which the canoeist shouldhead is determined by the sine ratio.sin /EDF 5 1.22.5/EDF 8 28.7°The canoeist should head 28.7° upstream.6. The area of the triangle is exactly:A DABC 5 1 2 @AB> 3 BC > @AB > 5 (2, 1, 3) 2 (21, 3, 5)BC > 5 (3, 22, 22)5 (21, 1, 4) 2 (2, 1, 3)AB > 5 (23,BC > 0, 1)22 x 0x 5 (22)(1) 2 (22)(0) 52222 y 1y 5 (22)(23) 2 (3)(1) 5 33 z 23z 5 (3)(0) 2 (22)(23) 52622 0So, AB > 3 BC > 5 (22, 3, 26) and@AB > 3 BC > @ 5 "(22) 2 1 3 2 1 (26) 2So, A DABC 5 1 2 @AB> 3 BC > @ 5 7 2 .The area of the triangle is 3.50 square units.7.T 125 kgThe system is in equilibrium (i.e. it is not moving),so we know that the horizontal components of T > 1and are equal:T > 24585 "495 7708@T > 1@ sin (45°) 5 @T > 2@ sin (70°)@T > 2@ 5T2sin (45°)sin (70°) @T> 1@DRF7-32 <strong>Chapter</strong> 7: Applications of Vectors
Also, the vertical component of T > 1 1 T > 2 must equalthe gravitational force on the block:@T > 1@ cos 45° 1 @T > 2@ cos 70° 5 (25 kg)(9.8 m>s 2 )Substituting in for , we find that:@T > T > 21@ cos 45° 1sin 45°@T> 1@ cos 70° 5 (25 kg) (9.8 m>s 2 )sin 70°@T > 1@ acos 45° 1So, we can now find@T > sin (45°)2@ 5sin (70°) @T> 1@@T > 2@ 8sin (45°)(254.0 N)sin (70°)@T > 2@ 8 191.1 Nsin 45°cos 70°b 5 245 Nsin 70°@T > 1@ (0.9645) 8 245 N@T > 1@ 8 254.0 NThe direction of the tensions are indicated in thediagram.8. a. We explicitly calculate both sides of theequation. The left side is:x > ? y > 5 (3, 3, 1) ? (21, 2, 23)5 (3)(21) 1 (3)(2) 1 (1)(23)5 0We perform a few computations before computingthe right side:x > 1 y > 5 (3, 3, 1) 1 (21, 2, 23)5 (2, 5, 22)0 x > 1 y > 0 2 5 (x > 1 y > ) ? (x > 1 y > )5 2 2 1 5 2 1 (22) 2x > 2 y > 5 335 (3, 3, 1) 2 (21, 2, 23)0 x > 2 y > 5 (4,0 2 5 (x > 1, 4)2 y > ) ? (x > 2 y > )5 4 2 1 1 2 1 4 25 33Thus, the right side is14 0 x> 1 y > 02 2 1 4 0 x> 2 y > 02 5 1 4 (33) 2 1 4 (33)5 0So, the equation holds for these vectors.b. We now verify that the formula holds in general.We will compute the right side of the equation, butwe first perform some intermediary computations:0 x > 1 y > 0 2 5 (x >5 (x > 10 x > 2 y > 5 (x > ? x > y > ) ? (x >0 2 5 (x > ? x > ) 1 (x > 1?)5 (x > 2? x > y > 1 2(x > y > y > ))) ? (x > ? y > 1 (y > ?) 115 (x > (2y > (x > 2 y > ) 1 (y > x > )? y > 1 (y > ? y > )))? x > ? 2y > ? 2y > ) 1 (2y > ? x > )) 2 2(x > )? y > ) 1 (y > ? y > )So, the right side of the equation is:14 0 x> 1 y > 02 2 1 4 0 x> 2 y > 02 5 1 4 (4(x> ? y > ))5 x > ? y >Thus, the equation holds for arbitrary vectors.Calculus and Vectors <strong>Solution</strong>s Manual7-33