© 2010 Brooks/Cole, Cengage Learning
24 5 3 1
1 K K K KK K K K K K 2 2 2 9 4 1 9 9 4 9 4 2 2 2 2 2 2 2 1 Chapter 11 Vectors and the Geometry of Space 30. A 2, 7, 3 , B 1, 5, 8 , C 4, 6, 1 JJJ JJJ AB 3, 12, 5 BA 3, 12, 5 JJJK JJJ AC 2, 13, 4 CA 2, 13, 4 JJJ JJJ BC 5, 1, 9 CB 5, 1, 9 JJJ JJJK AB ? AC 6 156 20 ! 0 JJJ JJJ BA ? BC 15 12 45 ! 0 JJJ JJJ CA ? CB 10 13 36 ! 0 The triangle has three acute angles, so it is an acute 32. u 5, 3, 1 u 35 cos D 35 cos E 35 cos J 35 cos2 D cos2 E cos2 J 25 9 1 35 35 35 triangle. 33. u 0, 6, 4 , u 52 2 13 31. u i 2 j 2k , u 3 cos D 0 cos D 1 3 cos E 3 13 cos E 3 cos J 3 cos2 D cos2 E cos2 J
1 cos J 13 cos2 D cos2 E cos2 J 0
13 13 1 34. u a, b, c , u a 2 b2 c 2 cos D
cos E
cos J a a 2 b2 c 2 b a 2 b2 c 2 c a 2 b2 c 2 cos2 D cos2 E cos2 J a 2 a b c b 2 a b c c 2 a b 2 c 2 35. u 3, 2, 2 u 17 cos D
cos E
cos J 3 17 2 17 2 17 ? D | 0.7560 or 43.3q
? E | 1.0644 or 61.0q
? y | 2.0772 or 119.0q 36. u 4, 3, 5 u 50 5 2 cos D
cos E 4 5 2 3 5 2 ? D | 2.1721 or 124.4q
? E | 1.1326 or 64.9q cos J 5 5 2 1 2 ? J | S 4 or 45q © 2010 Brooks/Cole, Cengage Learning
25 K 2 2 50 80 800 2 N ¨ © ¹ v 100 ¨ v © ¹ cos D | 230.239 Section 11.3 The Dot Product of Two Vectors 37. u 1, 5, 2 u 30 JJJ 41. OA 0, 10, 10 cos D 1 30 ? D | 1.7544 or 100.5q cos D 0 0 102 102 0 ? D 90q cos E 5 30 ? E | 0.4205 or 24.1q cos E cos J 10 0 102 102 cos J
38. u 2 30
2, 6, 1 ? J | 1.1970 or 68.6q
41 u 42. F1 C1 0, 10, 10 . 1 2 ? E J 45q cos D
cos E
cos J 2 41 6 41 1 41 ? D | 1.8885 or 108.2q
? E | 0.3567 or 20.4q
? J | 1.4140 or 81.0q F1 F1
F2 F3 F 200 C110 2 ? C1 0, 100 2, 100 2
C2 4, 6, 10 C3 4, 6, 10 0, 0, w 10 2 and 39. F1: C1 | 4.3193 F1 F2 : C2 | 5.4183 F2 F F1 F2 | 4.3193 10, 5, 3 5.4183 12, 7, 5 F F1 F2 F3 0 4C2 4C3 0 ? C2 C3
100 2 6C2 6C3 0 ? C2 C3
W 10C2 10C3 100 2 3 25 2 3 108.2126, 59.5246, 14.1336 F | 124.310 lb 43. u 6, 7 , v 1, 4 cos D |
cos E | 108.2126 F 59.5246 F ? D | 29.48q
? E | 61.39q (a) w1 projv u § u ? v · ¨ v 2 ¸ 6 1 7 4 12 42 1, 4 cos J | 14.1336 F ? J | 96.53q (b) w 2 u w1 34 1, 4 2, 8 17 6, 7 2, 8 4, 1 40. F1: C1 300 F1 | 13.0931 44. u 9, 7 , v 1, 3 F2 : C2 | 6.3246 F2 F F1 F2 | 13.0931 20, 10, 5 6.3246 5, 15, 0 (a) w1 projv u § u ? v · ¨ v 2 ¸ 9 1 7 3 1 32 1, 3 230.239, 36.062, 65.4655 F | 242.067 lb
? D | 162.02q F (b) w 2 u w1 30 1, 3 10 9, 7 3, 9
3, 9 6, 2 cos E |
cos J | 36.062 F 65.4655 F ? E | 98.57q
? J | 74.31q © 2010 Brooks/Cole, Cengage Learning
26 ¨ 2 ¸ ¨ v ¸ 5, 1 5 1 , ¨ (a) w1 projv u © ¹ ¨ v © ¹
u ? v . ¨ v © ¹ 1, 1, 1 6 v1 v2 v3 ¨ v © ¹ § u ? v · ¨ v 2 ¸ v ¸ (b) ¨ © ¹ 18 ¨ © ¹ 0, 3, 4 v u S 2 S 2 33 44 8 6 S 2 Chapter 11 Vectors and the Geometry of Space 45. u 2i 3j 2, 3 , v 5i j 5, 1 50. u i 4k 1, 0, 4 (a) w1 projv u v
(b) w 2 u w1 2, 3 ,
46. u 2i 3j 2, 3 , v 3i 2 j § u ? v · © ¹ 2 5 3 1 52 1 13 5 1 5, 1 , 26 2 2 2 2 1 5 2 2
3, 2 v 3i 2k 3, 0, 2
(b) w 2 u w1 1, 0, 4 § u ? v · ¨ v 2 ¸ v 1 3 4 2 32 22 11 3, 0, 2 13 3, 0, 2
33 22 , 0, 13 13 33 22 , 0, 13 13 (a) w1
(b) w 2 projv u
u w1 § u ? v · ¨ v 2 ¸ 2 3 3 2 32 22 0 3, 2 0, 0 2, 3 3, 2 20 30 , 0, 13 13
51. u ? v u1 , u2 , u3 ? v1 , v2 , v3 u1v1 u2v2 u3v3
52. The vectors u and v are orthogonal if u ? v 0. The angle T between u and v is given by cos T u v 47. u 0, 3, 3 , v 1, 1, 1 53. (a) and (b) are defined. (c) and (d) are not defined (a) w1 projv u § u ? v · ¨ v 2 ¸ 0 1 3 1 3 1 1 1 1 1, 1, 1 2, 2, 2 3 because it is not possible to find the dot product of a scalar and a vector or to add a scalar to a vector.
54. See page 786. Direction cosines of v v1 , v2 , v3 are cos D , cos E , cos J . D , E , and J v v v are the direction angles. See Figure 11.26. (b) w 2 u w1 0, 3, 3 2, 2, 2 2, 1, 1 55. See figure 11.29, page 787. 48. u 8, 2, 0 , v
(a) w1 projv u 2, 1, 1 § u ? v · ¨ v 2 ¸ 8 2 2 1 0 1 22 1 1 2, 1, 1 56. (a) ¨ © ¹ § u ? v · ¨ v 2 ¸ v orthogonal. u ? u
0 ? u ? v cv ? u and v are parallel.
0 ? u and v are (b) w 2 u w1 6 2, 1, 1 6, 3, 3
8, 2, 0 6, 3, 3 2, 1, 3 57. Yes, u ? v v 2 v v ? u u 2 u 49. u v 2i j 2k 3j 4k 2, 1, 2 0, 3, 4 u ? v v v 2 1 v ? u
1 u u 2 (a) w1 projv u § u ? v · ¨ v 2 ¸ v 2 0 1 3 2 4 32 42 11 33 44 0, 3, 4 0, , 25 25 25 v u
58. (a) Orthogonal, T
(b) Acute, 0 T (b) w 2 u w1 2,1, 2 0, , 25 25 2, , 25 25 (c) Obtuse, T S © 2010 Brooks/Cole, Cengage Learning
27 v v v K K K
¨ 2 j¸ ¹ u W 1 3 u 0. v W v W F v w Section 11.3 The Dot Product of Two Vectors 59. u 3240, 1450, 2235 1.35, 2.65, 1.85 71. (a) Gravitational Force F
cos 10qi sin 10q j 48,000 j u ? v 3240 1.35 1450 2.65 2235 1.85 $12,351.25 w1 F ? v v 2 v F ? v v This represents the total amount that the restaurant earned on its three products. 48,000 sin 10q v | 8335.1 cos 10qi sin 10q j 60. u 3240, 1450, 2235 w1 | 8335.1 lb 1.35, 2.65, 1.85 Increase prices by 4%: 1.04v New total amount: 1.04 u ? v 1.04 12,351.25 (b) w 2 F w1 48,000 j 8335.1 cos 10qi sin 10q j 8208.5i 46,552.6 j $12,845.30 w 2 | 47,270.8 lb 61. (a)–(c) Programs will vary. JJJ 72. OA 10, 5, 20 , v 0, 0, 1 62. u | 9.165 v | 5.745 T 90q JJJ projv OA JJJ projv OA 20 12 20 0, 0, 1 0, 0, 20 63. Programs will vary.
21 63 42 64. , , 26 26 13 73. F
v W § 1 85¨ i © 10i F ? v 3 · 2 ¸
425 ft-lb 65. Because u and v are parallel, projv u 74. F 25 cos 20qi sin 20q j 66. Because u and v are perpendicular, projv u
67. Answers will vary. Sample answer: 0 v 50i F ? v 1250 cos 20q | 1174.6 ft-lb i j. Want u ? v 4 2 75. F v 1600 cos 25q i sin 25q j 2000i 12i 2 j and v 12i 2 j are orthogonal to u. F ? v 1600 2000 cos 25q 68. Answers will vary. Sample answer: u 9i 4 j. Want u ? v 0. | 2,900,184.9 Newton meters (Joules) | 2900.2 km-N 4i 9 j and v 4i 9 j are orthogonal to u.
69. Answers will vary. Sample answer: JJJK 76. PQ 40i 100 cos 25qi JJJK F ? PQ 4000 cos 25q | 3625.2 Joules u 3, 1, 2 . Want u ? v 0. 77. False. 0, 2, 1 and v 0, 2, 1 are orthogonal to u. For example, let u 1, 1 , v 2, 3 and 70. Answers will vary. Sample answer: 1, 4 . Then u ? v 2 3 5 and u 4, 3, 6 . Want u ? v 0 u ? w 1 4 5. v 0, 6, 3 and v are orthogonal to u. 0, 6, 3 78. True w ? u v w ? u w ? v u v are orthogonal. 0 0 0 so, w and © 2010 Brooks/Cole, Cengage Learning
28 v c c § 1 · © 3 ¹ z . c c 1 y c c 1 x z 2 -1 y x x 6 c c 1 1 6 3 10 10 cos T . 1 3 . c c r cc c c 1 5 1 c 1 1 c y 1 x Chapter 11 Vectors and the Geometry of Space 79. Let s length of a side. 82. (a) The graphs y1 x3 and y2 x1 3 intersect at s, s, s 1, 1 , 0, 0 and 1, 1 . v
cos D s 3
cos E cos J s s 3 1 3 1 (b) y1 3x 2 and y2 . 3x 2 3 At 0, 0 , r 1, 0 is tangent to y1 and r 0, 1 is D E J arcos¨ ¸ | 54.7q tangent to y2 . At 1, 1 , y1 3 and y2 1 3 s v s s r 1, 3 is tangent to y1 , r 10 to y2 . At 1, 1 , y1 3 and y2 1 3 1 10
. 3, 1 is tangent 80. v1 s, s, s r 10 to y2 . 1, 3 is tangent to y1 , r 1 10 3, 1 is tangent v1 v 2 s 3 s, s, 0 y y = x 1/3 v 2
cos T s 2 2 2 2 3 6 3 v1
v2 (s, s, s) 1 (0, 0) -2 -1 (-1, -1) (1, 1)
1 y = x 3 2 (s, s, 0) T arcos | 35.26q 3
81. (a) The graphs y1 x 2 and y2 x1 3 intersect at 0, 0 and 1, 1 . 1 (b) y1 2 x and y2 . 3x 2 3 At 0, 0 , r 1, 0 is tangent to y1 and r 0, 1 is tangent to y2 . -2
(c) At 0, 0 , the vectors are perpendicular 90q . At 1, 1 , 1, 3 ? 3, 1 1 1 10 5 T | 0.9273 or 53.13q By symmetry, the angle is the same at 1, 1 . At 1, 1 , y1 2 and y2 83. (a) The graphs of y1 1 x 2 and y 2 x 2 1 1 5 1, 2 is tangent to y1 , r 1 10 3, 1 is tangent intersect at 1, 0 and 1, 0 . (b) y1 2 x and y2 2 x. to y2 . (c) At 0, 0 , the vectors are perpendicular 90q . At 1, 0 , y1 2 and y2 2. r 1 5 1, 2 is At 1, 1 , tangent to y1 , r 1, 2 is tangent to y2 . cos T
T 1 5
45q 1, 2 ? 1 1 10 3, 1 5 50 1 2 At 1, 0 , y1
tangent to y1 , r 5 2 and y2 2. r 1, 2 is 5 1, 2 is tangent to y2 . 2 y = x 2 (c) At 1, 0 , cos T 1 5 1, 2 ? 1 5 1, 2 3 5 . (1, 1) T | 0.9273 or 53.13q -1 (0, 0) 1 y = x 1/3 2 By symmetry, the angle is the same at 1, 0 . -1 © 2010 Brooks/Cole, Cengage Learning
29 2 2 2 1 y – v z 2 k x k 2 1 1 2
60q © 2 ¹ k k k , , 1 2 K 1 1 5 S
4 1 3 § k · © 2 ¹ 1 90q. 2 . u ? v u v 2 0 Section 11.3 The Dot Product of Two Vectors 84. (a) To find the intersection points, rewrite the second equation as y 1 x3 . Substituting into the first equation y 1 x ? x6 x ? x 0, 1. There are two points of intersection, 0, 1 and 1, 0 , as indicated in the figure.
y = x 3-1 86. If u and v are the sides of the parallelogram, then the diagonals are u v and u v, as indicated in the figure. the parallelogram is a rectangle. ? u ? v 0 ? 2u ? v 2u ? v ? u v ? u v u v ? u v ? u v u v ? The diagonals are equal in length. -1 (0, -1) (1, 0) 1 2 x u u + u v v x = (y +1) 2 -2 87. (a) (b) First equation: y 1 x ? 2 y 1 yc 1 ? yc 1 2 y 1 (k, 0, k) (0, k, k) At 1, 0 , yc 1 2 . k k y Second equation: y 1, 0 , yc 3. x3 1 ? yc 3x 2 . At (k, k, 0)
(b) Length of each edge: k 2 k 2 02 r 2, 1 unit tangent vectors to first curve, 5 r 1, 3 unit tangent vectors to second curve 10 At 0, 1 , the unit tangent vectors to the first curve are r 0, 1 , and the unit tangent vectors to the (c) cos T
T
J (d) r1 k k 2 k 2 § 1 · arccos¨ ¸
k , k , 0 , , 2 2 2 k k k 2 2 2 second curve are r 1, 0 . (c) At 1, 0 ,
cos T 2, 1 ? 1, 3 5 10 50
T | or 45q 4 At 0, 1 the vectors are perpendicular, T 88. u JK k k k k k k r2 0, 0, 0 , , , , 2 2 2 2 2 2 k 2 cos T 2 ¨ ¸ ? 3 T 109.5q cos D , sin D , 0 , v cos E , sin E , 0 85. In a rhombus, u u v. u v ? u v v . The diagonals are u v and
u v ? u u v ? v u ? u v ? u u ? v v ? v The angle between u and v is D E . Assuming that D ! E . Also, cos D E u 2 v So, the diagonals are orthogonal. u – v cos D cos E sin D sin E 1 1 cos D cos E sin D sin E . u u + v v
© 2010 Brooks/Cole, Cengage Learning
30 2 2 u u 2 2 2 u u u 2 i Chapter 11 Vectors and the Geometry of Space 89. u v u v ? u v 91. u v u v ? u v v ? u u v ? v u ? u v ? u u ? v v ? v v ? u u v ? v = u ? u v ? u u ? v v ? v u u ? v u ? v v u 2 2u ? v v u 2 v 2 2u ? v d u 2 2 u v v 2 d
u v
2 90. u ? v u ? v u v cos T v cos T So, u v d u v .
92. Let w1 projvu, as indicated in the figure. Because v cos T w1 is a scalar multiple of v, you can write d u v because cos T d 1. w1 w 2 cv w 2 . Taking the dot product of both sides with v produces u ? v cv w 2 ? v cv ? v w 2 ? v c v 2, because w 2 and v are orthogonol. So, u ? v c v ? c u ? v v 2 and w1 projv u cv u ? v v 2 v. w2 u ? v w1
Section 11.4 The Cross Product of Two Vectors in Space i j k i j k 1. j u i 0 1 0 k 3. j u k 0 1 0 1 0 z
1 0 0 0 z
1 1 k j j x 1 i -1 – k 1 y x 1 i -1 1 y i j k i j k 2. i u j 1 0 0 k 4. k u j 0 0 1 i 0 1 0 z
1 k j 0 z
1 k 1 0
– i j 1 i 1 1 1 x -1 y x -1 y © 2010 Brooks/Cole, Cengage Learning
31 i 1 i 1 j z 1 k x y i 0 1 5 0 5 i 0 Section 11.4 The Cross Product of Two Vectors in Space j k 11. u 12, 3, 0 , v 2, 5, 0 5. i u k 1 0 0 j i j k 0 0 1 u u v 12 3 0 54k 0, 0, 54 z 2 5 0 – 1 – j k u ? u u v 12 0 3 0 0 54 0 ? u A u u v x 1 -1 1 y v ? u u v 2 0 5 0 0 54 0 ? v A u u v 12. u 1, 1, 2 , v 0, 1, 0 i j k i j k 6. k u i 0 0 u u v 1 1 2 2i k 2, 0, 1 1 0 0 0 1 0 u ? u u v 1 2 1 0 2 1 0 ? u A u u v v ? u u v 0 2 1 0 0 1 1 i j 1 0 ? v A u u v -1 13. u 2, 3, 1 , v i j k 1, 2, 1 7. (a) u u v j k 2 4 20i 10 j 16k u u v 2 3 1 2 1 i j k 1, 1, 1 (b) v u u (c) v u v 3 2 u u v
i j k 20i 10 j 16k u ? u u v
v ? u u v 2 1 3 1 1 1 0 ? u A u u v 1 1 2 1 1 1 0 ? v A u u v 8. (a) u u v 3 0 15i 16 j 9k 14. u 10, 0, 6 , v 5, 3, 0 2 3 2 j k (b) v u u (c) v u v u u v 0 15i 16 j 9k u u v 10 0 6 18i 30 j 30k
u ? u u v 10 18 0 30 6 30 5 3 0 18, 30, 30
0 9. (a) u u v i j k 7 3 2 1 1 5 17i 33j 10k ? u A u u v v ? u u v ? v A u u v 5 18 3 30 0 30 (b) v u u (c) v u v u u v 0 17i 33j 10k 15. u i j k , v i j k 2i j k 10. (a) u u v i j k 3 2 2 1 5 1 8i 5 j 17k u u v 1 1 1 2i 3j k 2 1 1 u ? u u v 1 2 1 3 1 1 2, 3, 1 (b) v u u (c) v u v u u v 0 8i 5 j 17k v ? u u v
v u u 0 ? u A u u v 2 2 1 3 1 1 0 ? v A u u v v u u u u v © 2010 Brooks/Cole, Cengage Learning
32 u 6 5
, , , Chapter 11 Vectors and the Geometry of Space 16. u i 6 j, v i 2i j k j k 22. v 8, 6, 4 10, 12, 2 u u v 1 6 0 6i j 13k 2 1 1 u ? u u v 1 6 6 1 0 ? u A u u v u u v u u v u u v 60, 24, 156 1 60, 24, 156 36 22 17. v ? u u v
z 2 6 1 1 1 13 0 ? v A u u v 5 2 13 , , 3 22 3 22 3 22 23. u 3, 2, 5 , v 0.4, 0.8, 0.2 4 3 2 1 4 3 2 1 v
u 4 6 y u u v u u v u u v 3.6, 1.4, 1.6 1.8 4.37 0.7 4.37 , 0.8 4.37 x 18. z 24. u 0, 0, 7 10 , v 3 2 , 0, 31 5 6 5 4 u u v 0, 21 20 ,0 1 3 2 1 v u u v u u v 0, 1, 0 x 4 3 2 u 4 6 y 25. Programs will vary. 19. 6 z 26. u u v 50, 40, 34 u u v | 72.498 5 4 3 v 27. u j x 4 3 2 1 2 1 u 4 6 y v
u u v j k i j k 0 1 0 i 0 1 1 20. 6 z A u u v i 1 5 1 4 3 2 1 v 28. u v i j k j k x 4 3 2 u 4 6 y u u v i j k 1 1 1 j k 0 1 1 21. u u u v 4, 3.5, 7 , v 73.5, 5.5, 44.75 2.5, 9, 3 A u u v j k 2 u u v u u v
2.94 11.8961 0.22 11.8961 1.79 11.8961 29. u v 3, 2, 1 1, 2, 3 i j k u u v 3 2 1 1 2 3 8, 10, 4 A u u v 8, 10, 4 180 6 5 © 2010 Brooks/Cole, Cengage Learning
33 K K K 2 1 5 1 9 5 K K K K K K K K K K 2 K K K 1 cos 40q j sin 40qk z K K KK K K K K 1 1 z F y K K Section 11.4 The Cross Product of Two Vectors in Space 30. u v 2, 1, 0 1, 2, 0 34. A 2, 3, 4 , B 0, 1, 2 , C 1, 2, 0 JJJ JJJK AB 2, 4, 2 , AC 3, 5, 4 u u v i j k 2 1 0 0, 0, 3 JJJ JJJK AB u AC i j k 2 4 2 6i 2 j 2k A 1 u u v 2 0 0, 0, 3 3 A 1 2 3 5 4 JJJ JJJK AB u AC 1 44 11 31. A 0, 3, 2 , B 1, 5, 5 , C 6, 9, 5 , D 5, 7, 2 JJJ AB 1, 2, 3 JJJK DC 1, 2, 3 JJJ BC 5, 4, 0 JJJK AD 5, 4, 0 JJJ JJJK JJJ JJJK Because AB DC and BC AD, the figure ABCD is a parallelogram. JJJ JJJK AB and AD are adjacent sides i j k JJJ JJJK AB u AD 1 2 3 12, 15, 6 5 4 0 JJJ JJJK A AB u AD 144 225 36 35. A 2, 7, 3 , B 1, 5, 8 , C 4, 6, 1 JJJ JJJK AB 3, 12, 5 , AC 2, 13, 4 i j k JJJ JJJK AB u AC 3 12 5 113, 2, 63 2 13 4 JJJ JJJK A 2 AB u AC 1 16,742
36. A 1, 2, 0 , B 2, 1, 0 , C 0, 0, 0 JJJ JJJK AB 3, 1, 0 , AC 1, 2, 0 i j k JJJ JJJK AB u AC 3 1 0 5k 1 2 0 JJJ JJJK A 2 AB u AC 2 32. A 2, 3, 1 , B 6, 5, 1 , C 7, 2, 2 , D 3, 6, 4 JJJ AB 4, 8, 2 JJJK DC 4, 8, 2 JJJ BC 1, 3, 3 JJJK AD 1, 3, 3 JJJ JJJK JJJ JJJK Because AB DC and BC AD, the figure ABCD is a parallelogram. JJJ JJJK AB and AD are adjacent sides 37. F 20k JJJK PQ
JJJK PQ u F 0 cos 40q 2 sin 40q 2
JJJK PQ u F 10 cos 40q | 7.66 ft-lb i j k
PQ 0 0 20 10 cos 40qi i j k JJJ JJJK AB u AD 4 8 2 18, 14, 20 1 3 3 JJJ JJJK A AB u AD 324 196 400 2 230 x 1 2 ft 40° F y 33. A 0, 0, 0 , B 1, 0, 3 , C 3, 2, 0 JJJ JJJK AB 1, 0, 3 , AC 3, 2, 0 38. F 2000 cos 30q j sin 30qk JJJK PQ 0.16k 1000 3j 1000k i j k JJJ JJJK AB u AC 1 0 3 3 2 0 JJJ JJJK A 2 AB u AC 2 6, 9, 2
36 81 4 11 2 JJJK PQ u F
JJJK PQ u F i j k 0 0 0.16 0 1000 3 1000 160 3i 160 3 ft-lb PQ 0.16 ft 60° x
© 2010 Brooks/Cole, Cengage Learning
34 JJJ K K 3 3 3 4 4 K ¬ ¼ ¬ ¼ K ¬ ¼ ¬ ¼ ¨ 2 « © ¬ sin T ¸ 42¨ ¹» ¹ © ¼ K K F 5 K C A K 4 K © 2 ¹ K 4 K Chapter 11 Vectors and the Geometry of Space 39. (a) Place the wrench in the xy-plane, as indicated in the figure. The angle from AB to F is 30q 180q T 210q T JJJ OA 18 inches 1.5 feet y A 30 B JJJ OA
F 1.5ªcos 30q i sin 30q jº i j 56ªcos 210q T i sin 210q T jº O 18 30 in. F x i j k JJJ OA u F 3 3 3 0 4 4 56 cos 210q T 56 sin 210q T 0 100 y = 84 sin ? ª42 3 sin 210q T 42 cos 210q T ºk 0 0 180 ª42 3 sin 210q cos T cos 210q sin T 42 cos 210q cos T sin 210q sin T ºk JJJ OA u F ª § 1 3 «42 3¨ cos T 2 84 sin T , 0 d T d 180q · § 3 1 ·º ¸ ¨ 2 cos T 2 sin T ¸»k 84 sin T k (b) When T JJJ 45q, OA u F 84 2 2 42 2 | 59.40 (c) Let T 84 sin T dT dT 84 cos T 0 when T 90q. This is reasonable. When T 90q, the force is perpendicular to the wrench. 40. (a) AC 15 inches 5 4 feet BC JJJ AB 12 inches j k 4 1 foot B F 180 cos T j sin T k 12 in. (b) JJJ AB u F i j 0 5 0 180 cos T k 1 180 sin T 15 in. JJJ AB u F 225 sin T 180 cos T i 225 sin T 180 cos T (c) When T JJJ 30q, AB u F § 1 · § 3 · 225¨ ¸ 180¨ 2 ¸ | 268.38 © ¹ (d) If T 225 sin T 180 cos T , T 0 for 225 sin T
For 0 T 141.34, T c T 225 cos T 180 sin T 180 cos T ? tan T ? T | 141.34q.
0 ? tan T ? T | 51.34q. AB and F are perpendicular. 4 5 5 JJJ (e) 400 0 180 0 From part (d), the zero is T | 141.34q, when the vectors are parallel.
© 2010 Brooks/Cole, Cengage Learning
35 6 0 2 c V V ¬ ¼ ¬ ¼ ¬ ¼ Section 11.4 The Cross Product of Two Vectors in Space 1 0 0 48. u 0, 4, 0 41. u ? v u w 0 1 0 0 0 1 1 v w 3, 0, 0 1, 1, 5 42. u ? v u w 1 1 1 2 1 0 0 0 1 1 0 4 0 u ? v u w 3 0 0 1 1 5 V u ? v u w 60 4 15 60 43. u ? v u w 2 0 1 0 3 0 0 0 1 49. u u v u ? v 0 ? u and v are parallel. 0 ? u and v are orthogonal. So, u or v (or both) is the zero vector. 44. u ? v u w 2 0 0 1 1 1 0 2 2 50. (a) u ? v u w v u w ? u b w ? u u v u u v ? w c v ? w u u ux w ? v d 1 1 0 45. u ? v u w 0 1 1 1 0 1 V u ? v u w 2 (e) u ? w u v
So, a b v ? w u u w u u ? v h w ? v u u f w ? v u u u u v ? w g d h and e f g 46. u ? v u w 1 3 1 0 6 6 4 0 4 72 51. u u v u1 , u2 , u3 ? v1 , v2 , v3 u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k u ? v u w 72 52. See Theorem 11.8, page 794. 47. u v w 3, 0, 0 0, 5, 1 2, 0, 5 53. The magnitude of the cross product will increase by a factor of 4. 54. From the vectors for two sides of the triangle, and compute their cross product. u ? v u w 3 0 0 0 5 1 2 0 5 75 x2 x1 , y2 y1 , z2 z1 u x3 x1 , y3 y1 , z3 z1
55. False. If the vectors are ordered pairs, then the cross u ? v u w 75 product does not exist.
56. False. In general, u u v v u u 57. False. Let u Then, u u v 1, 0, 0 , v u u w 1, 0, 0 , w 0, but v z w. 1, 0, 0 . 58. True 59. u u1 , u2 , u3 , v i v1 , v2 , v3 , w j w1 , w2 , w3 k u u v w u1 u2 u3 v1 w1 v2 w2 v3 w3 ªu2 v3 w3 u3 v2 w2 º i ªu1 v3 w3 u3 v1 w1 º j ªu1 v2 w2 u2 v1 w1 ºk u2v3 u3v2 i u1v3 u3v1 j u1v2 u2 v1 k u2 w3 u3 w2 i u1w3 u3 w1 j u1w2 u2 w1 k u u v u u w
© 2010 Brooks/Cole, Cengage Learning
36 ¬ ¼ u u ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ Chapter 11 Vectors and the Geometry of Space 60. u u1 , u2 , u3 , v v1 , v2 , v3 , c is a scalar: i j k cu u v cu1 cu2 cu3 v1 v2 v3 cu2v3 cu3v2 i cu1v3 cu3v1 j cu1v2 cu2 v1 k c ª u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k º
61. u = u1 , u2 , u3 c u u v i j k u u u u1 u2 u3 u2u3 u3u2 i u1u3 u3u1 j u1u2 u2u1 k 0 u1 u2 u3 u1 u2 u3 62. u ? v u w v1 w1 v2 w2 v3 w3 w1 w2 w3 u u v ? w w ? u u v u1 v1 u2 v2 u3 v3 w1 u2v3 v2u3 w2 u1v3 v1u3 w3 u1v2 v1u2 u1 v2 w3 w2v3 u2 v1w3 w1v3 u3 v1w2 w1v2 u ? v u w 63. u u v u u v ? u u u v ? v u2v3 u2v3 u2v3 u3v2 i u1v3 u3v1 j u1v2 u2 v1 k u3v2 u1 u3v1 u1v3 u2 u1v2 u2v1 u3 u3v2 v1 u3v1 u1v3 v2 u1v2 u2 v1 v3 0 0 So, u u v A u and u u v A v. 64. If u and v are scalar multiples of each other, u cv for some scalar c. u u v cv u v c v u v c 0 0 If u u v 0, then u v sin T 0. Assume u z 0, v z 0. So, sin T 0, T 0, and u and v are parallel. So, cv for some scalar c. 65. u u v u v sin T If u and v are orthogonal, T S 2 and sin T 1. So, u u v v . 66. u a1 , b1 , c1 , v i a2 , b2 , c2 , w j k a3 , b3 , c3 v u w a2 b2 c2 b2c3 b3c2 i a2c3 a3c2 j a2b3 a3b2 k a3 b3 c3 i j k u u v u w b2c3 a1 b3c2 a3c2 b1 a2 c3 a2b3 c1 a3b2 u u v u w ªb1 a2b3 a3b2 c1 a3c2 a2c3 º i ªa1 a2b3 a3b2 c1 b2c3 b3c2 º j ªa1 a3c2 a2c3 b1 b2c3 b3c2 ºk ªa2 a1a3 b1b3 c1c3 a3 a1a2 b1b2 c1c2 º i ªb2 a1b3 b1b3 c1c3 b3 a1a2 b1b2 c1c2 º j ªc2 a1a3 b1b3 c1c3 c3 a1a2 b1b2 c1c2 ºk a1a3 b1b3 c1c3 a2 , b2 , c2 a1a2 b1b2 c1c2 a3 , b3 , c3 u ? w v u ? v w
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37 t 3 3
3 3
2 5 z 5 5 5 x 3 y 7 1 3 1 7 t 2, JJJK 2. 3 1 3
3 x 0 5 t z 2t 5 y 2 z Section 11.5 Lines and Planes in Space Section 11.5 Lines and Planes in Space 1. x (a) 1 3t , y z 2 t , z 2 5t x
y (b) When t JJJK PQ 0, P 9, 3, 15 1, 2, 2 . When 3, Q 10, 1, 17 . JJJK The components of the vector and the coefficients of t are proportional because the line is parallel to PQ. (c) y 0 when t 2. So, x 7 and z 12. Point: 7, 0, 12 x 0 when t 1. So, y 7 3 and z 1. Point: 0, 7 , 1 0 when t 5 . So, x 1 and y 12 . Point: 1 , 12 , 0 2. x (a) 2 3t , y z 2, z 1 t 4. 2 8 z 2 (a) 7, 23, 0 : Substituting, you have 7 3 2 23 7 8 0 2 x
(b) When t 0, P Q 4, 2, 1 . JJJK PQ 6, 0, 2 y
2, 2, 1 . When 2 2 2 Yes, 7, 23, 0 lies on the line. (b) 1, 1, 3 : Substituting, you have 3 2 2 8 1 1 1 The components of the vector and the coefficients of t are proportional because the line is parallel to PQ. Yes, 1, 1, 3 lies on the line. (c) z 0 when t Point: 1, 2, 0 1. So, x 1 and y 5. Point: 0, 0, 0 Direction vector: 3, 1, 5 x 0 when t 2 . So, y 2 and z Direction numbers: 3, 1, 5 3. x Point: 0, 2, 1
2 t , y
3t , z 4 t (a) Parametric: x x (b) Symmetric: 3 3t , y
y z 5 t , z 5t (a) 0, 6, 6 : For 2 t , you have 6. Point: 0, 0, 0 2. Then y 3 2 6 and 4 2 6. Yes, 0, 6, 6 lies on the line. Direction vector: v 2, , 1 2 (b) 2, 3, 5 : For x t 4. Then y 2 3 4 2 t , you have 12 z 3. No, 2, 3, 5 does Direction numbers: 4, 5, 2 (a) Parametric: x 4t , y 5t , z not lie on the line. (b) Symmetric: x 4 © 2010 Brooks/Cole, Cengage Learning
38 6 , z x 7 y 2 4 y y 2 x 2 y k j x 3 y 5 § 2 2 · © 3 3 ¹ 2 11 y 4 Chapter 11
7. Point: 2, 0, 3 Vectors and the Geometry of Space 13. Points: 7, 2, 6 , 3, 0, 6 Direction vector: v 2, 4, 2 Direction vector: 10, 2, 0 Direction numbers: 2, 4, 2 Direction numbers: 10, 2, 0 (a) Parametric: x 2 2t , y 4t , z 3 2t (a) Parametric: x 7 10t , y 2 2t , z (b) Symmetric:
8. Point: 3, 0, 2 x 2 2 z 3 2 (b) Symmetric: Not possible because the direction number for z is 0. But, you could describe the line as 6. 10 2 Direction vector: v 0, 6, 3 Direction numbers: 0, 2, 1 (a) Parametric: x 3, y
(b) Symmetric: z 2, x 2 2t , z
3 2 t 14. Points: 0, 0, 25 , 10, 10, 0 Direction vector: 10, 10, 25 Direction numbers: 2, 2, 5 (a) Parametric: x 2t , y 2t , z 25 5t 9. Point: 1, 0, 1 Direction vector: v 3i 2 j k Direction numbers: 3, 2, 1 (b) Symmetric:
15. Point: 2, 3, 4 z 25 5 (a) Parametric: x 1 3t , y 2t , z 1 t Direction vector: v (b) Symmetric: x 1 3 y 2 z 1 1 Direction numbers: 0, 0, 1 Parametric: x 2, y 3, z 4 t 10. Point: 3, 5, 4 Directions numbers: 3, 2, 1 16. Point: 4, 5, 2 Direction vector: v (a) Parametric: x 3 3t , y 5 2t , z 4 t Direction numbers: 0, 1, 0 (b) Symmetric: 3 2 11. Points: 5, 3, 2 , ¨ , , 1¸ z 4 Parametric: x 4, y
17. Point: 2, 3, 4 Direction vector: v 5 t , z
3i 2 j k Direction vector: v 17 3 i j 3k 3 Direction numbers: 3, 2, 1 Parametric: x 2 3t , y 3 2t , z 4 t Direction numbers: 17, 11, 9 (a) Parametric: x 5 17t , y 3 11t , z 2 9t 18. Point 4, 5, 2 Direction vector: v i 2 j k (b) Symmetric: x 5 17 y 3 11 z 2 9 Direction numbers: 1, 2, 1 Parametric: x 4 t , y 5 2t , z 2 t 12. Points: 0, 4, 3 , 1, 2, 5 Direction vector: 1, 2, 2 Direction numbers: 1, 2, 2 (a) Parametric: x t , y
(b) Symmetric: x 2 4 2t , z z 3 2 3 2t 19. Point: 5, 3, 4 Direction vector: v 2, 1, 3 Direction numbers: 2, 1, 3 Parametric: x 5 2t , y 3 t , z
20. Point: 1, 4, 3 4 3t Direction vector: v 5i j Direction numbers: 5, 1, 0 Parametric: x 1 5t , y 4 t , z 3 © 2010 Brooks/Cole, Cengage Learning
39 2 8 t v v v 1 2 y y z z 7 s Section 11.5 Lines and Planes in Space 21. Point: 2, 1, 2 30. L1: v 2, 1, 2 3, 2, 2 on line Direction vector: 1, 1, 1 Direction numbers: 1, 1, 1 Parametric: x 2 t , y 1 t , z 2 t L2 : v L3: v L4 : v 4, 2, 4 1, 1 , 1 2, 4, 1 1, 1, 3 on line 2, 1, 3 on line 3, 1, 2 on line 22. Point: 6, 0, 8 Direction vector: 2, 2, 0 Direction numbers: 2, 2, 0 Parametric: x 6 2t , y 2t , z L1 , L2 and L3 have same direction. 3, 2, 2 is not on L2 nor L3 1, 1, 3 is not on L3 So, the three lines are parallel, not identical. 23. Let t
v 0: P 1, 2, 0 3, 1, 2 other answers possible any nonzero multiple of v is correct 31. At the point of intersection, the coordinates for one line equal the corresponding coordinates for the other line. So, (i) 4t 2 2s 2, (ii) 3 2s 3, and 24. Let t
v 0: P 4, 1, 3 0, 5, 4 other answers possible any nonzero multiple of v is correct (iii) t 1 s 1. From (ii), you find that s (iii), t 0. Letting s 0 and consequently, from 0, you see that equation (i) 25. Let each quantity equal 0: is satisfied and so the two lines intersect. Substituting zero for s or for t, you obtain the point 2, 3, 1 . P 7, 6, 2 other answers possible any nonzero multiple of v is correct 4, 2, 1 u 4i k 2i 2 j k First line Second line 26. Let each quantity equal 0: P 3, 0, 3 other answers possible cos T u ? v u v 8 1 17 9 7 3 17 7 17 51 5, 8, 6 any nonzero multiple of v is correct 32. By equating like variables, you have (i) 3t 1 3s 1, (ii) 4t 1 2s 4, and 27. L1: v L2 : v L3: v 3, 2, 4 6, 4, 8 6, 4, 8 6, 2, 5 on line 6, 2, 5 on line 6, 2, 5 not online (iii) 2t 4 s 1. From (i) you have s t and from (iii), t t , and consequently from (ii), 3. The lines do not intersect. L4 : v 6, 4, 6 not parallel to L1 , L2 , nor L3 33. Writing the equations of the lines in parametric form you L1 and L2 are identical. L1
28. L1: v 2, 6, 2 L2 and is parallel to L3 .
3, 0, 1 on line have x x 3t 1 4s 2 t 2 s 1 t 3 3s. L2 : v 2, 1, 3 1, 1, 0 on line For the coordinates to be equal, 3t 1 4s and L3: v 2, 10, 4 1, 3, 1 on line 2 t 2 s. Solving this system yields t 17 7 and L4 : v 2, 1, 3 5, 1, 8 on line 11. When using these values for s and t, the z L2 and L4 are parallel, not identical, because 1, 1, 0 is not on L4 . coordinates are not equal. The lines do not intersect. 29. L1: v L2 : v L3: v L4 : v 4, 2, 3 2, 1, 5 8, 4, 6 2, 1, 1.5 8, 5, 9 on line
8, 5, 9 on line L1 and L3 are identical.
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