11 - CHAPTER 11 Vectors and the Geometry of Space Section...

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Unformatted text preview: CHAPTER 11 Vectors and the Geometry of Space Section 11.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . 469 Section 11.2 Space Coordinates and Vectors in Space . . . . . . . . . . 480 Section 11.3 The Dot Product of Two Vectors . . . . . . . . . . . . . . 489 Section 11.4 The Cross Product of Two Vectors in Space . . . . . . . . 498 Section 11.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . 504 Section 11.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . 517 Section 11.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . 523 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538 CHAPTER 11 Vectors and the Geometry of Space Section 11.1 1. (a) v (b) 5 4 3 y Vectors in the Plane 1 4, 2 2. (a) v (b) −3 −2 −1 −1 −2 5 1, 3 3 3, y 2 4 0, 6 x 1 2 3 (4, 2) 2 1 1 −3 −4 v v x −5 2 3 4 5 −6 (0, − 6) 3. (a) v (b) 4 3, 2 y 2 7, 0 4. (a) v (b) 1 2, 3 y 1 3, 2 4 2 3 (− 3, 2) 2 (− 7, 0) −8 −6 −4 v −2 −2 −4 x v 1 −3 −2 −1 x 5. u v u 5 1 v 3, 6 2 4 2, 4 2, 4 6. u v u 8. u v u 1 7 v 11 25 v 4 ,8 2, 7 0 1 5, 8 5, 8 1 ,8 7. u v u 6 9 v 0, 3, 5 2 3 10 6, 6, 5 5 4, 0, 10 4 13 1 15, 3 15, 3 9. (b) v 5 1, 5 y 2 4, 3 (5, 5) 10. (b) v 3 2, 6 y 12 10 8 6 (1, 12) v 1, 12 (a) and (c). 4 (a) and (c). v 2 (4, 3) 6 4 2 −1 x −4 −6 (3, 6) (1, 2) x 1 2 3 4 5 6 7 2 4 (2, − 6) 469 470 Chapter 11 6 10, Vectors and the Geometry of Space 1 y 6 4 2 −4 2 11. (b) v 2 4, 3 12. (b) v (a) and (c). 5 0, 1 4 y 5, 3 (a) and (c). (− 5, 3) 4 (10, 2) v x 10 v 2 x 2 −2 (6, −1) −6 −4 −2 (− 4, − 3) (− 5, −1) (0, − 4) 13. (b) v 6 6, 6 y 2 0, 4 14. (b) v (a) and (c). 3 7, 1 y 3 2 1 10, 0 (a) and (c). 6 (6, 6) (−10, 0) 1 4 (0, 4) v v 2468 −8 −6 −4 −2 x 2 (6, 2) x 2 4 6 (−3, −1) −2 −3 (7, −1) 15. (b) v 1 2 3 2, 3 4 3 y 1, 5 3 16. (b) v 0.84 0.12, 1.25 y 1.25 1.00 0.75 0.60 0.72, 0.65 (a) and (c). 3 ( 1 , 3( 2 (−1, ( 5 3 (a) and (c). (0.12, 0.60) (0.84, 1.25) 2 v (0.72, 0.65) v x 0.25 0.50 0.75 1.00 1.25 (( 34 , 23 0.50 −2 −1 x 1 2 0.25 17. (a) 2v y 4, 6 (b) 3v 6, y 9 (2, 3) v x 4 6 (4, 6) 2v (2, 3) −8 −4 4 4 − 3v − 4 −8 2 v x 2 4 6 (− 6, − 9) (c) 7 2v y 12 7, 221 (d) 2 3 v y 4 3, 2 (7, 21 ( 2 3 (2, 3) v 8 7 v 2 4 2 ( 4 , 2( 3 2 v 3 x (2, 3) v 4 8 12 x 1 1 2 3 S ection 11.1 18. (a) 4v 4, 20 y Vectors in the Plane 471 (b) 1 2 v (−1, 5) 1 2, y 5 2 (− 4, 20) 20 v 4v 1 (− 1, 5) v −12 − 8 − 4 4 8 12 x −4 −3 −2 −1 −2 −3 x − 1v 2 3 4 5 ( 1, − 2( 2 (c) 0v 0, 0 y 6 (d) 6v 6, y 30 (− 1, 5) (− 1, 5) v −15 −10 − 5 −10 5 10 15 x − 6v v 0v −3 −2 −1 x 1 2 3 −15 −20 −25 −30 (6, − 30) 19. y 20. Twice as long as given vector u. y −u 2u x u x 21. y 22. u y u + 2v 2v −v x u−v u x 23. (a) 2 3u 2 3 4, 9 2, 5 8 3, 6 4, 9 5 2, 5 2, 14 18, 7 24. (a) 2u 3 (b) v (c) 2u 2 3 3, 8 2, 3, 3, 8 16 3 (b) v (c) 2u 25. v 3 2 u 5v 2i j u 5v 8, 25 2 8 5 8, 25 11, 33 34, 109 2 4, 9 3i 3, 3 2j 3 2 26. v 2i 3i y j j i 3, 1 2j y 1 x 2 w 1 v x 2 3 1 2 3 u −1 −2 −3 3 u 2 u −1 472 27. v Chapter 11 2i 4i j 3j 2i 4, 3 Vectors and the Geometry of Space 2j 4 y 28. v 5u 5 2, 3w 1 11 3 1, 2 − 4 −2 2 y x 4 6 8 10 2w 2 7, u + 2w −3w −6 −8 5u v x u −2 4 6 −10 −12 29. u1 u2 4 2 3 1 u1 u2 Q 3 5 3, 5 32. v 144 0 16 30. u1 u2 3 2 4 9 u1 u2 Q 7 7 7, 36 1 7 25 1 5 10 3 unit vector 10 2 61 31. v 34. v 37. u v u u 16 100 32 9 9 122 3, 12 153 5 109 153 3 , 153 12 153 25 4 13 33. v 36. v 52 u u 152 5, 15 5 10 35. v 38. u v 250 1 , 10 17 4 17 , unit vector 17 17 3 2 u u 2 39. u v 5 2 2 34 2 3 , 34 5 34 40. u v u u 6.2 2 3.4 2 50 52 3 2, 5 2 34 2 6.2, 3.4 52 1.24 0.68 , unit vector 2 2 3 34 5 34 , unit vector 34 34 41. u 1, 1 ,v 1 1 v v u u u u (e) v v (f) u u u u v v v v 1 1 5 1 v v 0, 1 1 1, 2 (e) v v (f) u u u u v v 1 4 0, 1 0 1 1, 2 1 1 1 (d) u u v v 1, 2 2 5 42. u 0, 1 , v 0 9 v v u u 3, 1 9 3, 9 0, 1 1 1 3, 32 1 v v 1 1 3, 13 2 3 2 4 13 3 1 32 (a) u (b) v (c) u u (d) (a) u (b) v (c) u u S ection 11.1 1 ,v 2 1 4 v v u u u u (e) v v (f) u u u u v v v v 1 v v 1 2 7 3, 2 85 1 1 2, 3 13 9 3, 7 2 (d) u 9 1 2 1, 2 5 49 4 85 2 u u (e) v v (f) u u u u v v v v Vectors in the Plane 473 43. u (a) u (b) v (c) 1, 2, 3 1 4 5 2 13 44. u 2, 4 ,v 4 25 v v u u 5, 5 16 25 7, 1 49 1 4 52 25 52 (a) u (b) v (c) u u u 1 2, 25 1 1 52 1 v v 5, 5 (d) 1 7, 1 52 1 45. u u v v u u u v v 2, 1 5 5, 4 41 7, 5 −1 y 46. u u v 3, 2 13 1, 5 2, 0 2 v 13 5 u u 2 u u v 1 23 1 3 1, 3 2 2.236 3.606 3 y 2.236 7 6 5 4 3 u u+v −3 −2 −1 2 1 x −1 −2 −3 1 2 3 u+v v u 1 2 3 4 5 6 7 x 6.403 2 1 v u u u v v v 74 8.602 v v≤u 74 ≤ u u 5 v≤u 2≤ 41 u u 4 u u v 1 2 22 47. 4 1 1, 1 2 2 2 1, 1 48. 1, 1 1, 1 49. 3, 3 3, 3 u u v u u 2 2, 2 2 1 0, 3 3 0, 3 2 2, 2 2 50. 3 51. v 3 cos 0 i 3i 3, 0 sin 0 j 52. v 5 cos 120 i 5 i 2 53 j 2 sin 120 j u u v 0, 3 sin 150 j 3, 1 54. v cos 3.5 i 0.9981i sin 3.5 j 0.0610j 0.9981, 0.0610 53. v 2 cos 150 i 3i j 474 55. Chapter 11 u v u v i 32 i 2 2 Vectors and the Geometry of Space 56. 32 j 2 32 j 2 58. u 5 cos 0.5 i 5 sin 0.5 j u v u v 4i i 5i 3j 3j 32 i 2 57. u v u v 2 cos 4 i cos 2 i 2 cos 4 2 sin 4 j sin 2 j cos 2 i 2 sin 4 sin 2 j 5 cos 0.5 i v u v 5 cos 0.5 i 10 cos 0.5 i 5 sin 0.5 j 5 sin 0.5 j 59. A scalar is a real number. A vector is represented by a directed line segment. A vector has both length and direction. 60. See page 764: (ku1, ku2) (u1 + v1, u2 + v2) (u1, u2) u+v ku ku2 u2 u (v1, v2) v v1 u1 v2 u1 ku1 (u1, u2) u2 u 61. (a) Vector. The velocity has both magnitude and direction. (b) Scalar. The price is a number. For Exercises 63–68, au bw ai 2j b bi j a bi 62. (a) Scalar. The temperature is a number. (b) Vector. The weight has magnitude and direction. 2a b j. 0, 2a 1, b b 1. b 3. Solving 3. Solving 63. v 2i j. Therefore, a b 2, 2a simultaneously, we have a 1, b 1. 65. v 3i. Therefore, a b 3, 2a simultaneously, we have a 1, b 67. v i j. Therefore, a b simultaneously, we have a 69. f x (a) m ± 1. Solving 64. v 3j. Therefore, a b simultaneously, we have a 66. v 3i 3j. Therefore, a simultaneously, we have a b 2. 0. Solving b 3, 2a 2, b 1. 1, 2a 2, b 2 3, 1, 2a b b 1. 3 6 37. 1. Solving 68. v i 7j. Therefore, a b Solving simultaneously, we have a b 3. 7. x2, f x 6. Let w w w ± 2x, f 3 1, 6 , w y 10 8 6 4 1 1, 6 unit tangent vectors 37 6, 1 , w 37. −2 (b) m w ± w 1 . Let w 6 1 ± 37 (a) (b) (3, 9) x 2 4 6 8 10 2 6, 1 unit normal vectors S ection 11.1 70. f x (a) m ± Vectors in the Plane 475 x2 5, f x 1, 2 x, f 1 2, w 2 5. 4 3 2 y 2. Let w w w ± 1 1, 5 (a) 2 unit tangent vectors 5. −3 −1 (b) (1, 4) (b) m w ± w 1 . Let w 2 2, 1 , w 1 x −1 1 2 3 1 ± 2, 1 unit normal vectors 5 71. f x (a) m w w (b) m w w x 3, f x 3. Let w ± 1 3. 3x 2 3 at x 1, 3 , then 1. y 1 1, 3 . 10 Let w 1 3, 10 3, 1. 1 , then 2 (a) (1, 1) 1 (b) ± x 1 2 72. f x (a) m w w (b) m w w x3, f x 3x2 12 at x 1, 12 , then 2. y x 2 4 12. Let w ± 1 12 . 1 1, 12 . 145 Let w 1 12, 145 x2 x x2 Let w 4, 3 . 3, 4 , then 3 at x 4 3. 12, 1. 1 , then −6 −4 −2 −4 −6 (a) (b) ± − 10 73. f x fx (a) m w w (b) m w w 74. f x fx (a) m w w (b) m w w 25 25 3 4. 4, 3 , then 4 3 2 y ± 4 3. 1 5 (a) (3, 4) (b) Let w ± 1 −1 x 1 2 3 4 5 1 3, 4 . 5 tan x sec2 x 2 at x 4 . y 2.0 2. Let w ± 1 2. 1, 2 , then 1.5 1.0 1 1, 2 . 5 Let w 2, 1 . 2, 1 , then −π 2 −π 4 (a) (b) π 4 π 2 x 0.5 1 ± 5 − 1.0 476 Chapter 11 2 i 2 2j u v Vectors and the Geometry of Space 2 j 2 75. u u v v 76. u u v v 2 3i 3i u v 2j 3 3j u 3 2 3i 33 2j u 2 i 2 2 j 2 77. Programs will vary. 78. magnitude direction 63.5 8.26 79. F1 F2 F3 R R 2, 3, F1 F2 F3 33 125 110 F3 132.5 1.33 80. F1 F2 F3 R R 2, 4, 3, F1 F1 F1 F2 F3 10 140 200 F2 F3 163.0 180 cos 275 2 2.5, F1 F1 F2 F2 F3 4.09 F2 F3 81. (a) 180 cos 30 i Direction: Magnitude: (c) M 0 455 0 (d) 500 sin 30 j arctan 430.882 30 440.2 11.8 275i 90 430.88 902 60 396.9 23.1 50 430.88 i 0.206 90 j 11.8 (b) M 275 arctan 180 sin 2 180 sin 180 cos 440.18 newtons 90 328.7 33.2 120 241.9 40.1 150 149.3 37.1 180 95 0 (e) M decreases because the forces change from acting in the same direction to acting in the opposite direction as increases from 0 to 180 . 180 0 M α 0 0 180 0 82. F1 F2 500 cos 30 i 250 3 500 sin 30 j 250 2 200 cos 100 2 j 100 2 45 i 200 sin 45 j 100 2 i 100 2 F1 F2 tan 250 3 250 10.7 2 584.6 lb 250 100 2 ⇒ 250 3 100 2 83. F1 F2 F3 75 cos 30 i 75 2 3 F2 F2 F3 75 sin 30 j 125 i 2 228.5 lb 100 cos 45 i 75 2 50 2 100 sin 45 j 125 2 3j 125 cos 120 i 125 sin 120 j 50 2 F3 71.3 R R F1 F1 S ection 11.1 84. F1 F2 F3 400 cos 200 3 R R Vectors in the Plane sin 135 j 477 30 i 140 2 sin 30 j 280 cos 45 i 200 315 2 39.6 140 2 2 sin 45 j 175 2 j 350 cos 135 i 175 2 i 2 200 3 arctan 35 2 200 385.2483 newtons 200 315 2 200 3 35 2 0.6908 85. (a) The forces act along the same direction. (b) The forces cancel out each other. 180 . 0. (c) No, the magnitude of the resultant can not be greater than the sum. 86. F1 20, 0 , F2 F2 10 cos , sin 20 400 500 (b) 40 (a) F1 10 cos , 10 sin 400 cos 400 cos (c) The range is 10 ≤ F1 F2 ≤ 30. 0 and 100 cos2 100 sin2 The maximum is 30, which occur at 2. 0 0 2 The minimum is 10 at . (d) The minimum of the resultant is 10. 87. 4, 1 , 6, 5 , 10, 3 y 8 6 4 2 2 4 8 y 8 y (8, 4) (1, 2) (3, 1) 6 8 x 6 4 2 −4 −2 −2 −4 2 (6, 5) (1, 2) (3, 1) 4 6 (8, 4) 6 4 2 x (1, 2) (8, 4) (10, 3) x (− 4, −1) −4 8 −2 −2 −4 (3, 1) 4 6 8 10 88. u 1 u 3 P1 P2 7 2, 1 1, 2 1, 2 → CB → CA 1, 5 2 6, 3 2, 1 2 2, 1 u cos 30 i v cos 130 i 3, 3 5, 4 sin 30 j sin 130 j v sin 130 v cos 130 2000 0 A 50° v C 89. u v y 130° 30° B u x Vertical components: u sin 30 Horizontal components: u cos 30 Solving this system, you obtain u 1305.5 pounds and v 30° 1758.8 pounds. 478 Chapter 11 24 20 24 arctan 10 arctan u cos v cos 1 2 Vectors and the Geometry of Space y 90. 1 0.8761 or 50.2 A θ2 v C u B 2 1.9656 or 112.6 sin sin 1 2 u v i i j j 1 1 θ1 x Vertical components: u sin v sin 2 5000 2 Horizontal components: u cos Solving this system, you obtain u 2169.4 and v v cos 0 3611.2. v cos 1200 cos 6 1200 sin 6 1193.43 ft sec 125.43 ft sec 91. Horizontal component Vertical component v sin 92. To lift the weight vertically, the sum of the vertical components of u and v must be 100 and the sum of the horizontal components must be 0. u v u cos 60 i v cos 110 i sin 60 j sin 110 j 100, or 100. 0 or 0. 3 and adding to the first equation gives 100 ⇒ v 0 gives 65.27 lb. 20° 30° u v Thus, u sin 60 u 3 2 v sin 110 v sin 110 v cos 110 v cos 110 100 lb And u cos 60 u 1 2 Multiplying the last equation by u sin 110 Then, u u 1 2 3 cos 110 65.27 cos 110 44.65 lb. 44.65 lb, v 38.67 lb 61.33 lb 65.27 lb (a) The tension in each rope: u (b) Vertical components: u sin 60 v sin 110 93. u v u v 900 cos 148 i 100 cos 45 i 900 cos 148 692.53 i arctan u v sin 148 j sin 45 j 100 cos 45 i 547.64 j 38.34 ; 547.64 2 900 sin 148 100 sin 45 j 547.64 692.53 2 38.34 North of West 692.53 882.9 km hr S ection 11.1 94. u v u tan v 400i plane 50 cos 135 i 400 25 2 i sin 135 j 25 2 j 5.54 N 84.46 E 25 2 i 364.64 i 25 2 j wind 35.36 j Vectors in the Plane 479 35.36 ⇒ 364.64 Direction North of East: Speed: 95. True 98. False a 101. u v b 0 cos2 sin2 sin2 cos 2 336.35 mph 96. True 99. False ai 1, 1 y 97. True 100. True bj 2a 102. Let the triangle have vertices at 0, 0 , a, 0 , and b, c . Let u be the vector joining 0, 0 and b, c , as indicated in the figure. Then v, the vector joining the midpoints, is v a 2 b i 2 c j 2 b a c j i 2 2 1 bi cj 2 (b, c) c (a + b , 2( 2 v u 1 u. 2 x (0, 0) a ( 2 , 0( (a, 0) 103. Let u and v be the vectors that determine the parallelogram, as indicated in the figure. The two diagonals are u v and v u. Therefore, r x u v , s y v u . But, u r xu Therefore, x 104. w uv s v y vu vi u s u r yv u y x yu x y v. y 1 2. v 1 and x 0. Solving we have x u v cos v sin v vj u v u cos v ui u u sin v uj u uv v cos u cos v i sin u sin v j 2 u v cos sin tan w u 2 v cos u 2 v i sin 2 cos 2 j 2 u v cos cos 2 u v tan u v cos Thus, w u 2 2 v 2 2 and w bisects the angle between u and v. 105. The set is a circle of radius 5, centered at the origin. u x, y x2 y2 5 ⇒ x2 y2 25 480 Chapter 11 Vectors and the Geometry of Space 12 gt . 2 1 x g 2 v0 cos 2 106. Let x t v0 t cos and y v0 t sin x v0 cos y x v0 cos ⇒y v0 sin x tan x tan v02 2g v02 2g v02 2g (x, y) g2 x sec 2 2v02 g x2 1 2v02 g x2 2v02 g x2 2v02 gx2 2v02 tan 2 x tan 2 tan v02 gx 2 α x g x2 tan2 2v02 g x2 tan2 2v02 gx2 tan 2v02 v02 2g v02 gx v04 g 2x 2 If y ≤ y v02 2g v0 t gx2 , then 2v02 12 gt 2 v02 2g can be chosen to hit the point x, y . To hit 0, y : Let v02 g t 2g v0 1 , and you need y ≤ v02 2g gx2 2v02 2 90 . Then v02 . 2g The set H is given by 0 ≤ x, 0 < y and y ≤ Note: The parabola y v02 2g gx2 is called the “parabola of safety”. 2v02 Section 11.2 1. 6 5 4 3 Space Coordinates and Vectors in Space 2. 8 z z 3. (5, − 2, 2) − 2 4 3 2 −2 −3 3 2 1 1 z (3, −2, 5) 6 (2, 1, 3) 1 4 x 3 2 (−1, 2, 1) 6 y x 4 2 12 3 y 23 4 x 6 y ( z 8 6 3 , 2 4, −2 ( (5, − 2, − 2) 4. 5. A 2, 3, 4 B 1, 2, 2 6. A 2, B 3, 3, 1, 4 1 (4, 0, 5) 2 6 x −2 −4 −6 6 y (0, 4, − 5) 7. x 3, y 4, z 5: 3, 4, 5 8. x 7, 7, y 2, 1 2, z 1: 9. y z 0, x 10: 10, 0, 0 10. x 0, y 3, z 2: 0, 3, 2 11. The z-coordinate is 0. 12. The x-coordinate is 0. 13. The point is 6 units above the xy-plane. 14. The point is 2 units in front of the xz-plane. S ection 11.2 15. The point is on the plane parallel to the yz-plane that passes through x 4. 17. The point is to the left of the xz-plane. 19. The point is on or between the planes y 3 and y 3. Space Coordinates and Vectors in Space 3. 481 16. The point is on the plane z 18. The point is behind the yz-plane. 20. The point is in front of the plane x 4. 21. The point x, y, z is 3 units below the xy-plane, and below either quadrant I or III. 23. The point could be above the xy-plane and thus above quadrants II or IV, or below the xy-plane, and thus below quadrants I or III. 25. d 5 25 27. d 6 25 0 4 1 0 2 2 22. The point x, y, z is 4 units above the xy-plane, and above either quadrant II or IV. 24. The point could be above the xy-plane, and thus above quadrants I or III, or below the xy-plane, and thus below quadrants II or IV. 26. d 2 16 64 2 49 2 2 36 2 36 0 2 6 0 2 2 2 5 96 5 2 62 2 3 2 2 2 2 65 2 61 4, 4 3 16 9 6 35 2 16 46 6 3 2 2 4 2 28. d 4 4 9 29. A 0, 0, 0 , B 2, 2, 1 , C 2, AB AC BC BC 2 30. A 5, 3, 4 , B 7, 1, 3 , C 3, 5, 3 AB AC BC Since AB 4 4 16 4 4 16 1 1 0 3 3 42 4 4 0 AB 2 4 16 36 1 AC 2 AC , the triangle is isosceles. Right triangle 31. A 1, AB AC BC Since AB 3, 2 , B 5, 16 4 36 4 16 4 1, 2 , C 16 16 0 6 6 2 10 1, 1, 2 32. A 5, 0, 0 , B 0, 2, 0 , C 0, 0, AB AC BC Neither 34. The y-coordinate is changed by 3 units: 5, 6, 4 , 7, 4, 3 , 3, 8, 3 3 , 2 4 2 80 , 8 2 6 2 20 25 25 0 4 4 0 9 0 9 3 29 34 13 AC , the triangle is isosceles. 33. The z-coordinate is changed by 5 units: 0, 0, 5 , 2, 2, 6 , 2, 5 2 2 9 2 37 , 4, 9 3 2 35. , 3, 5 36. , 6, 4, 7 37. Center: 0, 2, 5 Radius: 2 x x 2 2 2 2 38. Center: 4, Radius: 5 2 1, 1 2 2 0 y z2 2 4y z 10z 5 4 0 x 2 x y2 4 y 8x 1 z 2z 1 25 0 y2 25 z2 2y 7 482 Chapter 11 2, 0, 0 2 10 y y2 x2 x2 2x 1 y2 x Center: 1, Radius: 5 3, 4 3 2 2 Vectors and the Geometry of Space 0, 6, 0 39. Center: Radius: x 1 1, 3, 0 40. Center: r 3 3, 2, 4 z 2x y2 6y 1 2 0 2 10 0 2x 6y z2 8z 8z z 1 16 4 2 (tangent to yz-plane) x 3 2 x2 41. z2 6y z2 9 y y 2 2 z 4 2 9 0 1 25 1 9 16 3 2 42. x2 9x 81 4 x2 y2 y2 x z2 2y 9 2 2 9x 1 y 2y z2 1 2 10z 10z z 19 25 5 2 0 19 109 4 81 4 1 25 Center: Radius: 9 , 1, 2 109 2 9y2 x2 y2 1 9 1 3 1 , 3 2 5 43. 9x2 9z2 z2 y2 y 6x 2 x 3 2y 1 2 18y 2y 1 z 1 1 9 z2 0 2 0 0 1 9 1 1 9 1 x2 2 x 3 x Center: 1, 0 Radius: 1 44. 4x2 4y2 x2 x2 x 1 4 y2 x Center: 1 , 4, 2 1 y2 4z2 z2 8y 1 2 2 4x x 16 y 32y 8y z2 4 2 8z 2z 2z z 33 33 4 1 1 2 0 0 33 4 9 1 4 16 1 Radius: 3 S ection 11.2 45. x2 y2 z2 ≤ 36 46. x2 Space Coordinates and Vectors in Space y2 z2 > 4 483 Solid ball of radius 6 centered at origin. Set of all points in space outside the ball of radius 2 centered at the origin. y2 8z z z2 < 4x 16 < 4 4 2 47. x2 4x 4 y2 x 6y 2 2 x2 9 y z2 3 2 6y 9 8z 16 13 13 < 16 Interior of sphere of radius 4 centered at 2, 48. x2 4x 4 y2 x 6y 2 2 3, 4 . y2 z2 > 16 > 4 2 x2 9 y z2 3 2 4x 13 6y 4 8z 9 13 16 8z z > 16 Set of all points in space outside the ball of radius 4 centered at 49. (a) v 2 2i (b) 5 4 3 2 −2 1 2 3 x 1 1 −3 z 2, 3, 4. 4 4i 0i 5j z 8 4i 2j 4 2k 2j 3 2, 2, 2 1k 50. (a) v 0 2k 5j 4, 3 5, 2 1k (b) − 2, 2, 2 〈4, − 5, 2〉 6 4 2 6 4 2 2 2 4 3 4 y x 6 y 51. (a) v 0 3i 3i 3k − 3, 0, 3 5 4 3 2 1 −3 3 3j 3, 0, 3 3 0k 52. (a) v 2 4k 2i 0, 0, 4 z 4 3 2 1 3 3j 4 0k (b) z (b) 〈 0, 0, 4 〉 −2 1 2 3 x 1 3 2 1 1 2 2 3 4 y x 3 y 53. 4 1, 3, 1 1, 6 2, 6 1 1, 0 1 1, 36 1, 6 38 1 , 38 1 , 38 1, 0, 2 6 38 1 54. 1 4, 7 5 5, 3 25 2 144 25 5, 12, 5 5, 12, Unit vector: 194 12 , 194 6 5 194 Unit vector: 1, 6 38 3, 0 1 1 , 0, 2 1 1 2 1 5, 12, 5 194 2, 1 1 , 73 2 36 6 , 73 4 36 6 73 5 , 194 1, 6, 73 55. 5 1, 0, 4 ,3 1 56. 2 1, 4 6 1, 6, Unit vector: Unit vector: 484 Chapter 11 3 4i (a) and (c). z 5 3 Vectors and the Geometry of Space 3 k 2j 4, 1, 1 4 3k 58. (b) v 4 6i (a) and (c). z 12 9 6 3 9 57. (b) v 1i j 2i 4j 3 9k 1j 7 2k 6, 4, 9 (3, 3, 4) 4 (− 1, 2, 3) (0, 0, 0) 2 −2 (− 4, 3, 7) (− 6, 4, 9) v 2 4 y x (4, 1, 1) 2 4 x 9 y (2, −1, − 2) 59. q1, q2, q3 Q 3, 1, 8 0, 6, 2 3, 5, 6 60. q1, q2, q3 Q (b) v 1, 4 3, 0, 2, 5 2 3 2, z 3 2 −2 −3 1, 21 3, 2 61. (a) 2v 2, 4, 4 z 5 4 3 2 −2 1 2 3 1 2 y 1, 2 2, 4, 4 −3 − 1, − 2, − 2 2 3 x −2 1 −2 −3 2 3 4 x (c) 3 2 v 3 2, 3, 3 z (d) 0v 0, 0, 0 z 3 3 2 −3 −2 2 3 x −2 −3 3 , 2 2 −3 −2 1 2 3 y x −2 −3 1 0, 0, 0 −1 −2 −3 1 2 3 y 3, 3 1 −2 −3 62. (a) v z 4 3 2 1 3 x 2, 2, 1 (b) 2v 4, 4, 2 z 8 6 〈4, − 4, 2〉 4 2 6 3 〈− 2, 2, −1〉 y x 6 y (c) 1 2 v 1, 1, 1 2 z 2 (d) 5 2v 5, 5, 5 2 z 8 6 〈1, −1, 〈 1 2 x 1 5 〈5, −5, 2 〈 y x 4 2 6 y S ection 11.2 63. z 64. z 65. z 66. z 67. 2z 2z1 2z2 2z3 z 68. 2u 9 7 2, Space Coordinates and Vectors in Space 485 u v 1, 2, 3 2, 2, 1 1, 0, 4 7, 0, 4 2 4 u 2u 5u 3u v 4v 3v 2w w 1 2 1, 2, 3 2, 4, 6 5, 10, 15 2, 2, 8, 8, 1 4 3 8, 0, 4, 0, 8 6, 12, 6 3, 4, 20 w 6, 6, 4, 0, 2, 0, 4 2 z1, z2, z3 3 6 4 ⇒ z1 0 ⇒ z2 4 ⇒ z3 5 2 7 2 3 1, 2, 3 3 5 2 3, v w 3z 2 1, 2, 3 0, 0, 0 2, 2, 1 4, 0, 4 3 z1, z2, z3 0, 0, 0 0, 6, 9 0 6 9 z 3z1 3z2 3z3 0, 3z1, 3z2, 3z3 0 ⇒ z1 0 ⇒ z2 0 ⇒ z3 2, 3 0 2 3 69. (a) and (b) are parallel since 70. (b) and (d) are parallel since 71. z 3i 4j 2k 6i 8j i 6, 4 3 4, 10 j 3 2 2 3, 2, k 2 1i 2 5 and 2, 4 , 3 2 3 10 3 2 3 3, 2, 9 8 5. 31 22 j 3 4k and 3 i 4 j 7, k i 2 3 j 3 4 k. 72. z 4k 2z. 8, 3 zz 14, 16, 6. (a) is parallel since 73. P 0, 2, \ (b) is parallel since 5 , Q 3, 4, 4 , R 2, 2, 1 3, 6, 9 2, 4, 6 3 2 74. P 4, \ 2, 7 , Q 6, 2, 3, 1, 2 \ 2, 0, 3 , R 7, 4 3, 9 PQ \ PQ \ PR 3, 6, 9 PR 3, \ 1, 2 1 2 2, 4, 6 \ 6, 2, \ 4 Therefore, PQ and PR are parallel. The points are collinear. 75. P 1, 2, 4 , Q 2, 5, 0 , R 0, 1, 5 \ Therefore, PQ and PR are parallel. The points are collinear. 76. P 0, 0, 0 , Q 1, 3, \ 2 , R 2, 6, 4 PQ \ 1, 3, 1, \ 4 1, 1 \ PQ \ 1, 3, 2, \ 2 6, 4 \ PR PR Since PQ and PR are not parallel, the points are not collinear. Since PQ and PR are not parallel, the points are not collinear. 486 Chapter 11 Vectors and the Geometry of Space 78. A 1, 1, 3 , B 9, \ 77. A 2, 9, 1 , B 3, 11, 4 , C 0, 10, 2 , D 1, 12, 5 \ 1, 5 5 7 7 \ 2 , C 11, 2, 9 , D 3, 4, 4 AB \ 1, 2, 3 1, 2, 3 2, 1, 1 2, 1, 1 \ \ \ \ AB \ 8, 8, 2, 3, 2, 3, \ 2, 2, CD \ DC \ AC \ AD \ BD BC Since AB CD and AC BD , the given points form the vertices of a parallelogram. 79. v 0 80. v 1 0 9 \ \ Since AB DC and AD BC , the given points form the vertices of a parallelogram. 10 81. v v 1, 1 1, 3, 1 3, 2, 9 u u u u 2, 4 2 9 5 4 25 5 5 38 4 14 3 9 14 82. v v 4, 3, 7 16 2, 4 u u u u 9 1, 2 1 1 2, 3 4 1, 2 1 2, 3 1, 2 3 49 74 83. v v 0, 3, 0 5 9 25 34 84. v v 85. u u (a) (b) 86. u u (a) (b) u u 6, 0, 8 36 0 64 10 87. u u (a) (b) 1 6, 0, 8 10 u u 1 6, 0, 8 10 1 3, 2, 38 1 3, 2, 38 v v 4, 7.5, 8.732 2 88. u u (a) (b) u u 8, 0, 0 8 1, 0, 0 u u 1, 0, 0 89. Programs will vary. 90. (a) u (b) u (c) u (d) v 5.099 9.014 91. cv cv 9c 2 2c, 2c, 4c 2 25 ± c 4c 2 c2 5 92. cu cu 14c 2 c c, 2c, 3c c2 9 ± 93. v 9c 2 3 10 u u 10 0, 0, 1 1 , 2 2 4c 2 10 10 , 2 2 c 5 3 3 14 14 3u 2u 32 21 , , 23 3 3 1 2 94. v 3 u u 3 1 1 1 , , 3 3 3 2 , 14 3 , 14 3 3 3 , , 3 3 3 1 14 95. v 1, 1, 96. v 5 u u 5 97. v 2 cos ± 30 j 3j ± k 0, sin ± 30 k 3, ± 1 −2 −1 1 −1 −2 2 1 z 70 3 70 70 , , 7 14 14 −2 0, 3, 1 2 x y 0, 3, −1 S ection 11.2 52 i 2 52 2 Space Coordinates and Vectors in Space 487 98. v v 5 cos 45 i 5 cos 135 i sin 45 k sin 135 k k or i k 52 (i + k) 2 8 6 4 2 6 x z 6 y 99. v 2 3v 3, 2, 2, 6, 3 4, 2 1, 2 100. v 2 3v 5, 6, 10 3, 13 3, 3 2 4, 6, 3 4, 3, 0 101. (a) 2, z 4, 2 1, 2, 5 (b) w a au 10 3, 4, bv b 2 ai 0, b a 0 bj bk 0 0, a 1 Thus, a and b are both zero. v 1 y 1 x u (c) ai a w a 1, b u bj 1 v bk i 2j k (d) a i a a 1, a bj b bk 2, b i 3 2j 3k Not possible 102. A sphere of radius 4 centered at x1, y1, z1 . v x x x 104. d x1 2 103. x0 is directed distance to yz-plane. y0 is directed distance to xz-plane. x2, y x1 y x1 2 2 y1, z y y1 2 z1 y1 z y1 2 z z1 2 z1 2 4 z0 is directed distance to xy-plane. 16 sphere z2 z1 2 x2 y2 2 105. x cv 107. x0 2 y y0 2 z z0 2 r2 106. Two nonzero vectors u and v are parallel if u for some scalar c. B C A → BC → Hence, AB 108. r r0 x x 1 1 2 2 → AB → AC → BC → CA → AC → CA 0 y y 1 1 2 2 z z 1 1 2 2 2 4 This is a sphere of radius 2 and center 1, 1, 1 . 488 Chapter 11 Vectors and the Geometry of Space L2 182. 109. (a) The height of the right triangle is h The vector PQ is given by \ \ Q (0, 0, h) PQ 0, 18, h . L The tension vector T in each wire is 24 T c 0, 18, h where ch 3 8 Hence, T 0, 18, h and h T (b) L T (c) 30 8. (0, 0, 0) 18 (0, 18, 0) P T 20 18.4 L = 18 8 h 182 25 11.5 h2 30 10 8 L2 35 9.3 182 40 9.0 182 45 8.7 L2 50 8.6 182 8L . L2 182 T=8 0 0 100 x 18 is a vertical asymptote and y horizontal asymptote. (d) lim L → 18 8 is a 8L L2 182 8L L2 182 L→ L→ lim lim 8 1 18 L 2 8 (e) From the table, T 10 implies L 30 inches. 110. As in Exercise 109(c), x 111. Let v v cos v z 0.6 0.4 0.2 0.2 0.4 0.6 x 0.4 y a will be a vertical asymptote. Hence, lim T r0 → a . be the angle between v and the coordinate axes. cos i cos 1 3 3 j k 3 1, 1, 1 3 j cos k 112. 550 302,500 c2 c F c 75i 18,125c 2 50j 100k 3 cos 1 3 3 i 3 16.689655 4.085 4.085 75i 306i 204j 50j 100k 409k ( 3 , 3 3 , 3 3 3 ( S ection 11.3 \ The Dot Product of Two Vectors 489 113. AB \ 0, 70, 115 , F1 60, 0, 115 , F2 45, F1 F2 C1 0, 70, 115 C2 60, 0, 115 65, 115 114. Let A lie on the y-axis and the wall on the x-axis. Then A → AB AB 0, 10, 0 , B 8, 8, 0, 6 , C 10, 2 59 → AC 650 → AC 297.0, 178.2 423.1, 253.9 720.1, 432.1 10, 0, 6 and 10, 6 . → 10, 6 , AC AC \ AD F 65, 115 , F3 F3 C3 45, 0, 0, 500 10 2, AC Thus: 60C2 70C1 115 C1 C2 45C3 65C3 C3 0 0 500 104 ,C 69 2 28 , and 23 F F F1 Thus, F1 → AB 420 → , F2 AB F2 237.6, 423.1, 185.5, Solving this system yields C1 112 C3 . Thus: 69 F1 F2 F3 115. d AP x2 x2 y2 202.919N 157.909N 226.521N 2d BP y z2 1 2 860.0 lb z 2z 1 2 2 4 x2 3x2 x2 x x 1 y2 3y2 8 x 3 4 3 2 2 y z2 3z2 16 9 y 3 2 2 2x 8x 2 z2 4y 18y 6y z 1 3 5 2z 9 2 2y 2 0 18 z2 2 z 3 1 9 6 16 9 9 1 9 44 9 y2 Sphere; center: 4 , 3, 3 1 2 11 , radius: 3 3 Section 11.3 1. u (a) u (b) u (c) u (d) u (e) u 2 The Dot Product of Two Vectors 2, 4 44 3 3 25 6 2. u (a) u (b) u (c) u 6 2, 2u v 3 2 6 12, 18 12 (d) u (e) u 12 1 1 17 26 2 3, 4 , v v u 32 33 25 vv 2v 4, 10 , v v u 4 44 116 vv 2v 22 2u 4, 8 4, 8 u vv 2v u 0v 2u 2 2, 3 10 3 10 10 22 116 2, 3 v 6, 3 4, 8 80 0 v 20 2 22 44, 66 44 46 4 83 4 0 88 80 3. (a) u v (b) u u (c) u (d) u (e) u 2 5, 5, u u 1 1 26 17 2u v 5, 3, 2 1 5 55 3 4. (a) u v (b) u u (c) u 2 vv 2v 3, 2 2 17 51, 34 34 (d) u (e) u 0 490 5. u Chapter 11 2, v u 2 Vectors and the Geometry of Space 0, 6, 5 36 3 3 45 44 2 29 6. u i, v v u 2 3, 4 , v 20 22 29 vv 2v j v u i 1 1 1 vv 2v j i 2u 2k, v 21 22 9 vv 2v 5i 2u v 3j 2 2k 5 5i 10 15j 10k 1 11 v i 3 2 2 3j 2k 22 2 9 5 (a) u (b) u (c) u (d) u (e) u 7. u 2i (a) u (b) u (c) u 2 0, 6, 5 2u k, v 21 22 v i 0, 12, 10 22 k 4 (d) u (e) u 8. u 2i (a) u (b) u (c) u (d) u (e) u uv uv u v 2 10 1 1 1 1 11 1 6 (a) u (b) u (c) u 2 v u 6 vv 2v v 2u i v k 2 (d) u (e) u uv uv u v 9. cos 8 5 cos 20 10. cos 40 25 cos 5 6 1 5 10 5 1 2 500 3 3 2 11. u cos 1, 1 , v uv uv 2 2, 12. u 0 cos 3, 1 , v uv uv 4 2, 0 28 13. u cos 3i j, v uv uv arccos 2i 4j 1 52 14. u v cos cos cos 6 i sin sin 6 j 3 i 2 1 j 2 2 i 2 2 j 2 2 10 20 1 52 3 i 4 3 j 4 98.1 uv uv 3 2 arccos 2 2 2 1 4 k, v 32 2i 1 2 3 2 2 2 1 4 105 3 15. u cos 1, 1, 1 , v uv uv arcos 2 3 2, 1, 2 36 61.9 1 2 3 16. u cos 3i 2j uv uv 2 3j 23 uv 0 0 S ection 11.3 17. u cos 3i 4j, v uv uv arccos 2j 8 5 13 8 13 65 3k 8 13 65 116.3 18. u cos 2i 3j uv uv The Dot Product of Two Vectors k, v i 2j k 491 9 14 6 arccos 9 2 21 3 21 14 10.9 3 21 14 19. u u u v 4, 0 , v 4 1, 1 0 ⇒ not orthogonal 20. u u u 2, 18 , v c v ⇒ not parallel 3 , 2 1 6 21. u u u 4, 3 , v 1 , 2 2 3 c v ⇒ not parallel v 0 ⇒ orthogonal c v ⇒ not parallel v 0 ⇒ orthogonal Neither 22. u u 1 3 1 6 i 2j , v 2i 4j 23. u u uv j 6k, v 8 i 2j k 24. u u u v 2i 3j k, v 2i j k v ⇒ parallel c v ⇒ not parallel 0 ⇒ not orthogonal c v ⇒ not parallel 0 ⇒ orthogonal Neither 25. u u u v 2, 3, 1 , v 1, 1, 1 26. u v u u v cos , sin , sin , 1, c v ⇒ not parallel 0 ⇒ orthogonal cos , 0 c v ⇒ not parallel 0 ⇒ orthogonal 27. The vector 1, 2, 0 joining 1, 2, 0 and 0, 0, 0 is perpendicular to the vector 2, 1, 0 joining 2, 1, 0 and 0, 0, 0 : 1, 2, 0 2, 1, 0 0 3, 0, 0 joining 0, 0, 0 and 28. Consider the vector 3, 0, 0 , and the vector 1, 2, 3 joining 0, 0, 0 and 1, 2, 3 : 3, 0, 0 1, 2, 3 3<0 The triangle is a right triangle. 29. The vectors forming the sides of the triangle are: u v w u u v v w w 0 1 1 6 2 4 2, 1 2, 2 0, 2 9 4 5 8 4 8 3, 2 3, 0 1, 0 4 4 2 2, 4, 3, 5, 1, 1, 2 4 2 The triangle has an obtuse angle. 30. The vectors forming the sides of the triangle are: u v w u w 4 4 1 2, 5 2, 6 1, 6 15 7, 5, 12 7, 8 1 1 45 3 3 8 3, 12, 5 2, 13, 5, 1, 48 < 0 4 9 23 > 0 10 > 0 17 > 0 The triangle has an obtuse angle. The triangle has three acute angles. 31. u cos cos cos cos2 i 2j 1 3 2 3 2 3 cos2 cos2 1 9 4 9 4 9 1 2k, u 3 32. u cos cos cos cos2 5, 3, 1 5 35 3 35 1 35 cos2 cos2 25 35 9 35 1 35 1 u 35 492 33. u cos cos cos Chapter 11 0, 6, 0 3 13 2 13 cos2 4, u Vectors and the Geometry of Space 52 2 13 cos2 cos2 0 9 13 c2 4 13 1 34. u cos cos cos cos2 a, b, c , u a2 a2 a2 cos2 a b2 b b2 c b2 a2 c2 c2 c2 cos2 b2 a2 17 a2 b2 c2 a2 b2 b2 c2 a2 c2 b2 c2 1 35. u cos cos cos 3, 2, 2 3 ⇒ 17 2 ⇒ 17 2 ⇒ 17 u 36. u cos cos cos 4, 3, 5 4 52 3 52 5 52 2, 6, 1 u ⇒ ⇒ 1 ⇒ 2 u 50 52 2.1721 or 124.4 1.1326 or 64.9 or 45 0.7560 or 43.3 1.0644 or 61.0 2.0772 or 119.0 4 37. u cos cos cos 1, 5, 2 u 30 1.7544 or 100.5 0.4205 or 24.1 1.1970 or 68.6 38. u cos cos cos 41 1.8885 or 108.2 0.3567 or 20.4 1.4140 or 81.0 1 ⇒ 30 5 ⇒ 30 2 ⇒ 30 50 F1 80 F2 F2 2 ⇒ 41 6 ⇒ 41 1 ⇒ 41 300 F1 100 F2 F2 20, 39. F1: C1 F2: C2 F F1 4.3193 5.4183 40. F1: C1 F2: C2 F 5.4183 12, 7, 14.1336 F 29.48 61.39 96.53 cos cos cos 5 F1 13.0931 6.3246 4.3193 10, 5, 3 13.0931 10, 5 6.3246 5, 15, 0 108.2126, 59.5246, F cos cos cos 124.310 lb 108.2126 ⇒ F 59.5246 ⇒ F 14.1336 ⇒ F 230.239, 242.067 lb 36.062, 65.4655 230.239 ⇒ F 36.062 ⇒ F 65.4655 ⇒ F 162.02 98.57 74.31 S ection 11.3 \ The Dot Product of Two Vectors 200 C1 10 2 ⇒ C1 493 10 2 41. OA cos cos 0, 10, 10 02 cos 1 ⇒ 2 0 102 02 102 10 102 45 0⇒ 90 42. F1 C1 0, 10, 10 . F1 and F1 F2 C2 C3 4, 0, 100 2, 100 2 4, 6, 10 6, 10 102 F3 F F 0, 0, w F1 4C2 F2 F3 4C3 6C2 0 0 ⇒ C2 6C3 C3 C3 25 2 N 3 100 2 W 43. w2 45. w2 47. u u u w1 w1 6, 7 0, 3, 3 5, 1 v v u 2 0 ⇒ C2 100 2 3, 9 6, 3, 10C2 u u 2, w1 w1 10C3 9, 7 800 2 3 6, 3 2 2, 1, 3 2, 8 4, 2, 2, 2 1 2, 1, 1 44. w2 46. w2 48. u 8, 2, 0 3, 2 v v 0v 2, 3 2, 3 , v u 3 ,v u v u 2 (a) w1 (b) w2 v 13 5, 1 26 15 , 22 0, 3, 4 51 , 22 (a) w1 (b) w2 0, 0 w1 w1 49. u 2, 1, 2 , v u v v 2 50. u 1, 0, 4 , v u v u v 2 3, 0, 2 v 11 3, 0, 2 13 1, 0, 4 20 30 , 0, 13 13 33 22 , 0, 13 13 33 22 , 0, 13 13 (a) w1 v 0, 2, 33 44 , 25 25 86 , 25 25 u1v1 u2v2 u3v3 (a) w1 (b) w2 11 0, 3, 4 25 (b) w2 u w1 w1 51. u v u1, u2, u3 v1, v2, v3 52. The vectors u and v are orthogonal if u The angle cos between u and v is given by uv . uv v 0. 53. (a) Orthogonal, (b) Acute, 0 < (c) Obtuse, < < 2 2 54. (a) and (b) are defined. 2 < 55. See page 784. Direction cosines of v cos , , and v1 , cos v v2 , cos v v1, v2, v3 are v3 . v 56. See figure 11.29, page 785. are the direction angles. See Figure 11.26. 494 Chapter 11 u v u v 2 Vectors and the Geometry of Space u v u v 2 57. (a) (b) v u⇒u c v ⇒ u and v are parallel. v 0 ⇒ u and v 58. Yes, v v v2 1 v u v 1 u v v u u u 2 u u u2 v v 0⇒u v2 are orthogonal. v 59. u v u v 3240, 1450, 2235 1.35, 2.65, 1.85 3240 1.35 $12,351.25 1450 2.65 2235 1.85 60. u v 3240, 1450, 2235 1.35, 2.65, 1.85 Increase prices by 4%: 1.04v New total amount: 1.04 u v 1.04 12,351.25 $12,845.30 This represents the total amount that the restaurant earned on its three products. 61. Programs will vary. 62. u v 9.165 5.745 90 63. Programs will vary. 64. 21 63 42 ,, 26 26 13 65. Because u appears to be perpendicular to v, the projection of u onto v is 0. Analytically, projv u u v v 2 v 2, 3 6, 4 6, 4 6, 4 2 0 6, 4 0. 66. Because u appears to be a multiple of v, the projection of u onto v is u. Analytically, projv u u v v 2 v 3, 2 6, 4 6, 4 6, 4 6, 4 3, 2 u. 26 6, 4 52 1 i 2 8i 2 j. Want u 3 6j and v 67. u v v 8i 0. 6j are orthogonal to u. 68. u v 3i 8i 3j. Want u 8j and v v 3i 0. 8j are orthogonal to u. 69. u v 3, 1, 2 . Want u v v 0, 0. 2, 1 are orthogonal to u. 70. u v (b) w2 0, 3, 6 . Want u v v 0, 0. 6, 3 are orthogonal to u. 8335.1 cos 10 i sin 10 j 0, 2, 1 and 0, 6, 3 and F w1 71. (a) Gravitational Force F v w1 cos 10 i F v v 2 48,000 j 48,000j 46,552.6j sin 10 j F vv 48,000 sin 10 v 8335.1 cos 10 i sin 10 j w2 8208.5i v 47,270.8 lb w1 8335.1 lb S ection 11.3 \ The Dot Product of Two Vectors 495 72. OA 10, 5, 20 , v \ 0, 0, 1 0, 0, 20 73. F v W 85 10 i F 1 i 2 v 3 j 2 425 ft lb 74. F v W 25 cos 20 i 50 i F v sin 20 j projvOA \ 20 0, 0, 1 12 20 1250 cos 20 lb projvOA 75. False. Let u uv 1174.6 ft 76. True 2, 4 , v 1, 7 and w 2 28 30 and u w 5, 5 . Then 10 20 30. w u v w 0 u 0 w v 0⇒w and u 77. Let s v v cos length of a side. s, s, s s3 v s z v are orthogonal. s, s, s s3 s, s, 0 s2 s2 s3 arccos 6 3 6 3 35.26 v2 x z 78. v1 v1 v2 v1 (s, s, s) y cos cos s s3 1 3 1 3 s x y s v2 cos (s, s, 0) arccos 54.7 79. (a) The graphs y1 y 1 x 2 and y2 2 x1 3 intersect at 0, 0 and 1, 1 . 2 y y = x2 2 x and y 1 . 3x 2 3 1 . 3 −1 1 (1, 1) y = x 1 /3 (0, 0) 1 x 2 At 0, 0 , ± 1, 0 is tangent to y1 and ± 0, 1 is tangent to y2. At 1, 1 , y ± 1 2 and y 2 −1 1 1 1, 2 is tangent to y1, ± 3, 1 is tangent to y2. 5 10 (b) At 0, 0 , the vectors are perpendicular 90 . 1 1, 2 5 11 45 . 80. (a) The graphs y1 y 1 At 1, 1 , cos 1 3, 1 10 5 50 1 2 x 3 and y2 2 x1 3 intersect at 1, 1 , 0, 0 and 1, 1 . 2 y 3x 2 and y 1 . 3x 2 3 (−1, −1) −2 −1 y = x 1/3 (1, 1) y = x3 x 1 2 1 At 0, 0 , ± 1, 0 is tangent to y1 and ± 0, 1 is tangent to y2. At 1, 1 , y ± 1 (0, 0) −1 −2 3 and y 2 1 . 3 1 1 1, 3 is tangent to y1, ± 3, 1 is tangent to y2. 10 10 1,y 1 At 1, ± 3 and y 2 1 . 3 1 1 1, 3 is tangent to y1, ± 3, 1 is tangent to y2. 10 10 —CONTINUED— 496 Chapter 11 Vectors and the Geometry of Space 80. —CONTINUED— (b) At 0, 0 , the vectors are perpendicular 90 . 1 1, 3 10 11 0.9273 or 53.13 By symmetry, the angle is the same at 81. (a) The graphs of y1 y 1 At 1, 1 , cos 1 3, 1 10 6 10 3 . 5 1, 1. 1, 0 . 1 2 x 2 and y 2 2 x. 2 x2 1 intersect at 1, 0 and 2 x and y 1 At 1, 0 , y ± 2 and y 2. 1 1, 2 is tangent to y2. 5 1 1, 5 1 2 is tangent to y1, ± 2 and y 2 At 1, 0 , y ± 2. 2 is tangent to y2. 2 3 . 5 1 1 1, 2 is tangent to y1, ± 1, 5 5 1 1, 5 2 1 1, 5 (b) At 1, 0 , cos 0.9273 or 53.13 By symmetry, the angle is the same at 1, 0 . 1 x3. Substituting into the first equation y 82. (a) To find the intersection points, rewrite the second equation as y y 1 2 x ⇒ x6 x⇒x 0, 1. 1 and 1, 0 , as indicated in the figure. 1y 1⇒y 1 2y 1 −1 1 y = x 3− 1 There are two points of intersection, 0, First equation: y At 1, 0 , y Second equation: y ± (1, 0) 1 2 1 1 . 2 2 x⇒2y x (0, −1) x = ( y + 1) 2 −2 x3 1⇒y 3x2. At 1, 0 , y 3. 1 1 2, 1 unit tangent vectors to first curve, ± 1, 3 unit tangent vectors to second curve 5 10 At 0, 1 , the unit tangent vectors to the first curve are ± 0, 1 , and the unit tangent vectors to the second curve are ± 1, 0 . (b) At 1, 0 , cos 1 2, 1 5 4 At 0, or 45 90 . v and u v u 0 v v v v v. u u+v v 1 1, 3 10 5 50 1 . 2 1 the vectors are perpendicular, v . The diagonals are u v u u u 2 83. In a rhombus, u u v u u−v v u v v u u 2 u Therefore, the diagonals are orthogonal. S ection 11.3 84. If u and v are the sides of the parallelogram, then the diagonals are u as indicated in the figure. The parallelogram is a rectangle. ⇔u ⇔ 2u ⇔u ⇔u v v v v 2 The Dot Product of Two Vectors v, u v 497 v and u u+ u − v v 0 2u u u v v v 2 u v u v ⇔ The diagonals are equal in length. \ 85. (a) (k, 0, k) k z (d) r1 (0, k, k) \ k, k, 0 0, 0, 0 k2 4 k 2 2 kkk ,, 222 kkk ,, 222 1 3 kk ,, 22 k , 2 k 2 k , 2 k 2 r2 k x k y cos (k, k, 0) 3 (b) Length of each edge: k2 (c) cos k2 02 k2 1 2 60 109.5 k2 k 2k 2 arccos 1 2 86. u cos , sin , 0 , v cos , sin , 0 . Assuming that cos cos 11 v u u v u 2u u v v v v v v v 2 The angle between u and v is cos uv uv 2 > . Also, cos cos sin sin . sin sin 87. u v u u u u u 2 2 v v u u v v v u 2 2 2 u u v 88. u u v v u v cos u v cos u v cos ≤ u v since cos ≤ 1. v 89. u v 2 u u u u ≤ ≤ u u v 2u u v v u u v 2 v v 2 v v v u u 2u v v 2 v Therefore, u v≤u v. 498 Chapter 11 Vectors and the Geometry of Space 90. Let w1 u projv u, as indicated in the figure. Because w1 is a scalar multiple of v, you can write w1 v w2 cv cv 2, cv w2. w2 u Taking the dot product of both sides with v produces u w2 v cv v w2 v θv since w2 and v are orthogonol. 2 w1 Thus, u v cv ⇒c u v v 2 and w1 projv u cv u v v 2 v. Section 11.4 ijk 010 100 z The Cross Product of Two Vectors in Space ijk 100 010 z 1. j i k 2. i j k 3. j k ijk 010 001 z i 1 1 1 k j 1 x k j j 1 y x i −1 −k 1 y x 1 i −1 1 i −1 1 y 4. k j ijk 001 010 z i 5. i k ijk 100 001 z j 6. k i ijk 001 100 z j 1 1 1 k −i j −1 −j k k j 1 x −1 1 y x 1 i −1 1 y x 1 i −1 1 y 7. (a) u (b) v (c) v v u v ijk 234 372 u v 22, 0 22, 16, 16, 23 23 8. (a) u (b) v (c) v v u v i 3 2 u 0 j 0 3 v k 5 2 15, 15, 16, 9 16, 9 ijk 372 372 i 7 1 u 0 j 3 1 v k 2 5 9. (a) u (b) v (c) v v u v 17, 33, 10 10. (a) u (b) v (c) v v u v i 3 1 u 0 j 2 5 v k 2 1 8, 8, 5, 5, 17 17 17, 33, 10 S ection 11.4 11. u u u v 12. u u u v u v v u u 2, 3, 1 , v i 2 1 v v j 3 2 2 1 k 1 1 1 1 1, 2, 1 i 3 2 0, 1, 0 k 2 0 2 u 2i 10 v 0 1 v k 2 2, 0, 1 1 j 1 1 k 1 1 1, 1 1 1, 1 u u v v The Cross Product of Two Vectors in Space 499 0⇒u 0⇒v 1, 1, 2 , v i 1 0 j 1 1 1 13. u u u v 12, 3, 0 , v i 12 2 v j 3 5 12 0 0⇒u k 0 0 2, 5, 0 54k 30 u 50 u j 2i 13 u 13 u u v v v 1 1 v k 3j 1 1 k 2, 3, 1 v 0 54 0, 0, 54 0 54 u 0⇒u v u v 0 2 0⇒v 14. u u u v u v 10, 0, 6 , v i 10 7 j 0 0 10 0 0⇒u v u v 70 0⇒v 10 u 7, 0, 0 k 6 0 42j 0 42 u 0 42 u v v v u v 20 0⇒v 15. u 0, 42, 0 60 u u i v u j i 1 2 v k, v j 1 1 1 2i k 1 1 2 0⇒u 00 v u v 2 2 0⇒v v u v u 16. u u v 17. v u u z 6 5 4 3 2 1 1 4 3 2 ijk 160 211 v v 16 26 6 6i 1 1 1 j 13k 0⇒u 1 13 u v u v 19. 6 5 4 3 2 1 1 2 0⇒v z 6 5 4 3 2 1 1 18. z 20. 6 5 4 3 2 1 1 2 z v v v v u 4 6 y x 4 3 2 u 4 6 y x 4 3 u 4 6 y x 4 3 u 4 6 y x 21. u v u u u v 4, 3.5, 7 1, 8, 4 70, 23, 57 2 46 , 24,965 57 24,965 v v 140 , 24,965 500 22. u v u u u Chapter 11 8, 10, v v v 6, 4 12, 2 Vectors and the Geometry of Space 23. u v u u u 1 i 2 v v v 3i 2j 3 j 4 71 , 20 20 7602 5k 1 k 10 11 5 , 54 71 , 20 11 5 , 54 44 , 7602 25 7602 60, 24, 156 1 36 22 60, 24, 156 5 2 13 , , 3 22 3 22 3 22 71 , 7602 2 k 3 1 i 2 6k 24. u v u u u v v v v v 25. Programs will vary. 1 0, , 0 3 0, 1, 0 26. u u 50, 40, 72.498 34 27. u v u A j j v u k ijk 010 011 v i 1 i 28. u v u A i j v u j k k ijk 111 011 v j k j k 2 29. u v u A v 3, 2, 1, 2, 3 i 3 1 u v 1 30. u v j 2 2 8, k 1 3 8, 10, 4 10, 4 180 65 u A 32. A 2, \ 2, 1, 0 1, 2, 0 v u v i 2 1 j 1 2 k 0 0 0, 0, 3 3 6, 4 , D 7, 2, 2 \ 0, 0, 3 31. A 1, 1, 1, , B 2, 3, 4 , C 6, 5, 2 , D 7, 7, 5 \ \ \ 3, 1 , B 6, 5, \ 1 , C 3, 1, \ AB BD \ 1, 2, 3 , AC 5, 4, 1 \ \ \ 5, 4, 1 , CD \ \ 1, 2, 3 , AB BD \ 4, 8, 2 , AC 1, 3, 3 \ \ \ 3, 3 , CD \ 4, 8, 2, Since AB CD and AC BD , the figure is a parallelogram. AB and AC are adjacent sides and \ \ \ Since AB CD and AC BD , the figure is a parallelogram. AB and AC are adjacent sides and \ \ \ AB A AC \ ijk 123 541 \ 10i 2 83 14j 6k. AB Area AC \ i 4 1 \ j 8 3 k 2 3 920 18, 2 230 14, 20 . AB AC 332 AB AC S ection 11.4 33. A 0, 0, 0 , B 1, 2, 3 , C \ \ The Cross Product of Two Vectors in Space 3, 4 , B 0, 1, 2 , C \ 501 3, 0, 0 3, 0, 0 34. A 2, \ 1, 2, 0 3, 5, k 2 4 44 6i 4 2j 2k AB \ 1, 2, 3 , AC \ AB \ \ 2, 4, AC 1 AB 2 2 , AC i 2 3 j 4 5 1 2 AB AC 1 AB 2 i 1 3 \ \ j 2 0 1 2 k 3 0 117 9j 3 2 1 4 113, 6k AB \ \ A AC 13 A AC 11 35. A 2, \ 7, 3 , B 1, 5, 8 , C 4, 6, \ 36. A 1, 2, 0 , B \ 2, 1, 0 , C 0, 0, 0 \ AB \ \ 3, 12, 5 , AC AC 1 AB 2 20k i j 3 12 2 13 \ \ 2, 13, k 5 4 1 2 AB 2, 63 \ \ 3, AC 1 AB 2 1, 0 , AC i 3 1 j 1 2 5 2 k 0 0 1, 5k 2, 0 AB AB Area AC 16,742 \ \ A AC 37. F \ 38. F \ 2000 cos 30 j 0.16 k F \ sin 30 k 1000 3 j 1000k PQ \ PQ \ 1 cos 40 j sin 40 k 2 i j k 0 cos 40 2 sin 40 2 F 0 0 20 F z PQ \ 10 cos 40 i PQ PQ i 0 0 j 0 1000 3 lb k 0.16 1000 160 3 i PQ 10 cos 40 7.66 ft lb F z 160 3 ft PQ PQ 1 ft 2 0.16 ft 60° F 40° F y y x x 39. (a) Place the wrench in the xy-plane, as indicated in the figure. \ y The angle from AB to F is 30 \ 180 210 . A 18 in. OA \ 18 inches 1.5 cos 30 i 60 cos 210 1.5 feet sin 30 j i 33 i 4 j k 0 0 k 45 cos 210 cos 3 cos 2 1 sin 2 3 j 4 30 O 30 B F x AB F \ sin 210 OA F i 3 34 60 cos 210 45 3 sin 210 45 3 sin 210 cos 45 3 90 sin \ j 34 60 sin 210 45 cos 210 cos 210 sin 3 sin 2 45 sin 210 sin k 140 k 1 cos 2 k 90 sin . y = 90 sin Hence, OA F 0 180 0 —CONTINUED— 502 Chapter 11 Vectors and the Geometry of Space 39. —CONTINUED— \ (b) When (c) Let T 45 : OA 90 sin . F 90 2 2 45 2 63.64. dT 90 cos 0 when 90 . d This is what we expected. When 90 the pipe wrench is horizontal. 15 12 \ 40. (a) B is 5 4 5 4 to the left of A, and one foot upwards: k sin k k 1 200 sin i \ (d) If T dT d AB F, 8 sin 0 ⇒ tan ⇒ 5 4 51.34 . AB F \ j 25 10 cos 200 cos j F i 0 0 (b) AB j 54 200 cos The vectors are orthogonal. \ 250 sin \ 200 cos 200 cos 8 cos (e) The zero is to F. 400 141.34 , the angle making AB parallel AB F 250 sin 25 10 sin (c) For \ 30 , F 25 10 25 5 1 2 43 8 3 2 298.2. 0 180 AB − 300 41. u v w 1 0 0 2 1 0 0 1 0 0 1 2 0 0 1 0 1 2 1 42. u v w 1 2 0 1 0 1 w 1 1 0 1 1 0 2 1 0 1 0 1 1 1 43. u v w 2 0 0 1 0 4 w 0 3 0 1 0 1 3 6 0 72 1 6 4 6 44. u v w 0 45. u V v u w v 2 46. u V v u w v 72 47. u v w u V 49. u v 3, 0, 0 0, 5, 1 2, 0, 5 w v 3 0 2 w 0 5 0 0 1 5 75 v1, v2, v3 u 2v3 u 3v2 i u 1v3 75 48. u v w u V u 3v1 j v 1, 1, 0 1, 0, 2 0, 1, 1 w v 1 1 0 w u 2v1 k 1 0 1 3 0 2 1 3 u v u u 1v2 u1, u2, u3 50. See Theorem 11.8, page 792. 51. The magnitude of the cross product will increase by a factor of 4. 52. Form the vectors for two sides of the triangle, and compute their cross product: x2 x1, y2 y1, z 2 z1 x3 x1, y3 y1, z3 z1 S ection 11.4 53. If the vectors are ordered pairs, then the cross product does not exist. False. The Cross Product of Two Vectors in Space 1, 0, 0 , v u w 1, 0, 0 , w w. 1, 0, 0 . 503 54. False, let u Then, u v 0, but v 55. True 56. u u v u u1, u2, u3 , v u1i w v u2 j v2w3 w u3 k v3w2 i u1 v2w3 v1w3 v3w2 v3w1 j u2 v1w3 v1w2 v3w1 v2w1 k u3 v1w2 v2w1 u1 u2 u3 v1 v2 v3 w1 w2 w3 v1, v2, v3 , w w1, w2, w3 57. u u u1, u2, u3 , v v w v1 u2 v3 u2v3 i u1 v1, v2, v3 , w j u2 w1 v2 w3 u3v2 i u3w1 j u w w2 u3 v2 u1v3 w1, w2, w3 k u3 v3 w3 u1 v3 u1v2 w3 u3 v1 w1 j u2w3 u1 v2 w2 u2 v1 w1 k w2 i u3v1 j u2v1 k u3w2 i u1w3 u 58. u cu u1, u2, u3 , v v i j cu1 cu2 v1 v2 cu2v3 c u2v3 59. u u u u1, u2, u3 ij u1 u2 u1 u2 k u3 u3 v u1w2 u2w1 k v1, v2, v3 , c is a scalar. k cu3 v3 cu3v2 i u3v2 i cu1v3 u1v3 cu3v1 j u3v1 j cu1v2 u1v2 cu2v1 k cu v u2v1 k u2u3 u3u2 i u1u3 u3u1 j u1u2 u2u1 k 0 60. u v w u1 u2 v1 v2 w1 w2 w v2 u 3 w2v3 w u v u3 v3 w3 w1 w2 u1 u2 v1 v2 v1u 3 w1v3 w3 u3 v3 w3 u 1v2 u 3 v1w2 v1u 2 w1v2 u v w w1 u 2v3 u 1 v2w3 u v w2 u 1v3 u 2 v1w3 504 61. u u Chapter 11 u v v v u v v u2v3 u2v3 u2v3 Vectors and the Geometry of Space u3v2 i u3v2 u1 u3v2 v1 v v. c v for some scalar c. u1v3 u3v1 u3v1 u3v1 j u1v3 u2 u1v3 v2 u1v2 u1v2 u1v2 u2v1 k u2v1 u3 u2v1 v3 0 0 Thus, u u and u 62. If u and v are scalar multiples of each other, u u If u u v v cv v cv v c0 0 0, then u v sin c v for some scalar c. 0. Assume u 0, v 0. Thus, sin 0, 0, and u and v are parallel. Therefore, 63. u v u v sin If u and v are orthogonal, 64. u a1, b1, c1 , v v w 2 and sin a3, b3, c3 b3c2 i 1. Therefore, u v u v. a2, b2, c2 , w k c2 c3 b2c3 j b1 b3c2 a3b2 a2c3 b1b3 b1b3 b1b3 a3c2 ij a2 b2 a3 b3 i a1 b2c3 b1 a2b3 a1 a3c2 a2 a1a3 b2 a1a3 c2 a1a3 a1a3 u a2c3 k c1 a3c2 j a2b3 a3b2 k u u v v w w a2c3 a2b3 a2c3 i b3c2 k a3b2 a1 a2b3 a3b2 c1 b2c3 b3c2 j c1 a3c2 b1 b2c3 c1c3 c1c3 c1c3 a3 a1a2 b3 a1a2 c3 a1a2 b1b2 b1b2 b1b2 c1c2 i c1c2 j c1c2 k c1c2 a3, b3, c3 b1b3 u c1c3 a2, b2, c2 vw a1a2 b1b2 wv Section 11.5 1. x (a) 1 3t, y z Lines and Planes in Space 2 t, z 2 5t (b) When t 0 we have P Q 10, 1, 17 . \ 1, 2, 2 . When t 3 we have PQ 9, 3, 15 \ The components of the vector and the coefficients of t are proportional since the line is parallel to PQ . x y (c) y 0 when t 2. Thus, x Point: 7, 0, 12 x z 0 when t 0 when t 1 . Point: 3 2 . Point: 5 7 and z 71 0, , 33 1 12 , ,0 55 12. S ection 11.5 2. x (a) 2 3t, y z Lines and Planes in Space 505 2, z 1 t (b) When t 0 we have P Q 4, 2, 1 . \ 2, 2, 1 . When t 2 we have PQ 6, 0, 2 \ x y The components of the vector and the coefficients of t are proportional since the line is parallel to PQ . (c) z 0 when t 1, 2, 0 2 . Point: 3 0, 2, 1 3 1. Thus, x 1 and y 2. Point: x 0 when t 3. Point: (0, 0, 0 Direction vector: v 1, 2, 3 4. Point: 0, 0, 0 Direction vector: v 5 2, , 1 2 4, 5, 2 4t, y x 4 y 5 z 2 5t, z 2t Direction numbers: 1, 2, 3 (a) Parametric: x (b) Symmetric: x t, y y 2 z 3 2t, z 3t Direction numbers: (a) Parametric: x (b) Symmetric: 5. Point: 2, 0, 3 2, 4, 2 2t, y y 4 z 4t, z 3 2 3 2t 2 6. Point: 3, 0, 2 0, 6, 3 Direction vector: v Direction vector: v Direction numbers: 2, 4, (a) Parametric: x (b) Symmetric: x 2 2 2 Direction numbers: 0, 2, 1 (a) Parametric: x (b) Symmetric: y 2 z 3, y 2, x 2t, z 3 2 t 7. Point: 1, 0, 1 Direction vector: v Direction numbers: 3, (a) Parametric: x (b) Symmetric: x 3 1 1 3i 2j 2, 1 3t, y y 2 22 , ,1 33 17 i 3 11, 17t, y y 3 11 11 j 3 9 3 z 11t, z 2 9 2 9t 3k z 1 2t, z 1 1 t k 8. Point: 3, 5, 4 2, 1 3 3 3 y 3t, y 5 2 5 z 2t, z 4 4 t Directions numbers: 3, (a) Parametric: x (b) Symmetric: x 9. Points: 5, 3, 2, 10. Points: 2, 0, 2 , 1, 4, Direction vector: 1, Direction numbers: 1, (a) Parametric: x (b) Symmetric: x 2 2 3 4, 5 4, 5 t, y y 4 z 5 4t, z 2 2 5t Direction vector: v Direction numbers: 17, (a) Parametric: x (b) Symmetric: x 17 5 5 506 Chapter 11 Vectors and the Geometry of Space 12. Points: 0, 0, 25 , 10, 10, 0 Direction vector: 10, 10, Direction numbers: 2, 2, 3 3 5 z 12 5t, z 12t (a) Parametric: x (b) Symmetric: x 2 2t, y y 2 z 25 5 2t, z 25 5 25 5t 11. Points: 2, 3, 0 , 10, 8, 12 Direction vector: 8, 5, 12 Direction numbers: 8, 5, 12 (a) Parametric: x (b) Symmetric: x 8 2 2 8t, y y 13. Point: 2, 3, 4 Direction vector: v k 14. Point: 4, 5, 2 j Direction vector: v Direction numbers: 0, 0, 1 Parametric: x 15. Point: 2, 3, 4 Direction vector: v 3i 2j 1 3 2t, z 4 t k 2, y 3, z 4 t Direction numbers: 0, 1, 0 Parametric: x 16. Point: 4, 5, 2 i 1, 2, 1 4 3 5i j 1, 0 5t, y 4 t, z 3 t, y 5 2t, z 2 t 2j k 4, y 5 t, z 2 Direction vector: v Direction numbers: Parametric: x 18. Point: 1, 4, Direction numbers: 3, 2, Parametric: x 2 3t, y 17. Point: 5, 3, 4 2, 1, 3 Direction vector: v Direction numbers: 2, Parametric: x 19. Point: 2, 1, 2 Direction vector: Direction numbers: Parametric: x 21. Let t v 0: P 1, 2, 0 2 3, 5 Direction vector: v Direction numbers: 5, 1, 3 3 t, z 4 3t 2t, y Parametric: x 1 20. Point: 1, 1, 1 1, 1, 1 t, y 1, 1 t, z 2 t 6, 0, 8 2, 2, 0 2, 2, 0 6 2t, y 2t, z 8 Direction vector: Direction numbers: Parametric: x 22. Let t v 4, 0: P 1, 3 2 other answers possible 0, 5, 4 other answers possible any nonzero multiple of v is correct 3, 0, 3 other answers any nonzero multiple of v is correct 7, 6, 2 other 23. Let each quantity equal 0: P answers possible v 25. L1: v L 2: v L 3: v L 4: v 6, 4, 2, 1 24. Let each quantity equal 0: P possible v 26. L1: v L 2: v L 3: v L 4: v 5, 8, 6 4, any nonzero multiple of v is correct 3, 2, 4 4, 6, 4, 8 8 6, 6, 6, 2, 5 on line 2, 5 on line 2, 5 not on line any nonzero multiple of v is correct 2, 3 8, 5, 9 on line 2, 1, 5 8, 4, 6 8, 5, 9 on line 6, 4, 6 not parallel to L1, L 2, nor L 3 2, 1, 1.5 Hence, L1 and L 2 are identical. L1 L 2 and L 3 are parallel. L1 and L 2 are identical. S ection 11.5 Lines and Planes in Space 507 27. At the point of intersection, the coordinates for one line equal the corresponding coordinates for the other line. Thus, (i) 4t 2 2s 2, (ii) 3 2s 3, and (iii) t 1 s 1. From (ii), we find that s 0 and consequently, from (iii), t 0. Letting s t 0, we see that equation (i) is satisfied and therefore the two lines intersect. Substituting zero for s or for t, we obtain the point (2, 3, 1 . u v cos 4i 2i k 2j k 81 17 9 (First line) (Second line) 7 3 17 7 17 51 uv uv 28. By equating like variables, we have (i) 3t 1 3s 1, (ii) 4t 1 2s 4, and (iii) 2t 1 2 4 s 1. 3. The lines do not intersect. From (i) we have s t, and consequently from (ii), t and from (iii), t 29. Writing the equations of the lines in parametric form we have x x 3t 1 4s y y 2 2 t s z z 1 3 t 3s. 17 7 2 s. Solving this system yields t For the coordinates to be equal, 3t 1 4s and 2 t When using these values for s and t, the z coordinates are not equal. The lines do not intersect. 30. Writing the equations of the lines in parametric form we have x x 2 3 3t 2s y y 2 5 6t s z z 3t 3 2 3 t 4s. 2s, 2 6t 5 s, 3 t 2 and s 11 7. By equating like variables, we have 2 point of intersection is 5, 4, 2 . u v cos 3, 6, 1 2, 1, 4 uv uv (First line) 4s. Thus, t 1, s 1 and the (Second line) 4 46 21 2s s 2s 8 1 1 7 4 966 2 966 483 31. x y z 2t 5t t 3 2 1 x y z 32. x y z 2t 4t t 1 10 x y z 3s 5s 12 11 2s 4 Point of intersection: 7, 8, Note: t 4 2 −2 4 6 8 Point of intersection: 3, 2, 2 x = 2t − 1 y = − 4t + 10 z=t 3 2 −2 −3 2 and s z 0 z x = − 5s − 12 y = 3s + 11 z = − 2s − 4 10 x 8 6 4 10 y x 3 2 2 −8 (7, 8, − 1) (3, 2, 2) 3 y 508 33. 4x Chapter 11 3y \ Vectors and the Geometry of Space 6 1 ,Q \ 6z 0, 0, 0, \ 34. 2x 0, 2, 0 , R 3, 4, 0 k 1 0 4, 3, 6 3, 4, 1 P 3y \ 4z 4 2, 0, 0 , R \ (a) P PQ \ 0, 0, 1 , Q 2, 0, \ 3, 2, 3, 2, 3 2, 3, 4 2 2, 1 , PR i 0 3 j 2 4 (a) PQ \ 1 , PR i 2 3 j 0 2 k 1 3 (b) PQ PR (b) PQ PR The components of the cross product are proportional to the coefficients of the variables in the equation. The cross product is parallel to the normal vector. The components of the cross product are proportional (for this choice of P, Q, and R, they are the same) to the coefficients of the variables in the equation. The cross product is parallel to the normal vector. 36. Point: 1, 0, n k 1 3 0 3 0 1z 3 0 35. Point: 2, 1, 2 n 1x x 2 i 2 0 1, 0, 0 0y 1 0z 2 0 0, 0, 1 0y 0x z 37. Point: 3, 2, 2 Normal vector: n 2x 2x 3 3y 3y z 10 2 2i 3j 1z k 2 0 38. Point: (0, 0, 0 Normal vector: n 3x 3x 0 2z 0y 0 0 3i 2k 2z 0 0 39. Point: 0, 0, 6 Normal vector: n 1x x x y y 2z 0 2z 1y 12 12 0 0 i j 2z 2k 6 0 40. Point: 3, 2, 2 Normal vector: v 4x 4x 3 y 3z y 8 2 4i 3z j 3k 2 0 41. Let u be the vector from 0, 0, 0 to 1, 2, 3 : u i 2j 3k Let v be the vector from 0, 0, 0 to v 2i 3j 3k Normal vector: u v i 1 2 3i 3x 3x 9y 0 7z 9y 0 2i 2k 2i 3j 4k 0 7z 0 j 2 3 2, 3, 3 : k 3 3 9j 0 7k 42. Let u be vector from 2, 3, Let v be vector from 2, 3, 1, 1, 0 : 1, 4, 2 . Normal vector: u v 2 to 3, 4, 2 : 1, 1, 4 . 2 to i 1 1 1z j 1 4 2 z k 4 2 0 8 18, 3 6, 3 6, 2, 1 6x 2 2y 3 6x 2y 43. Let u be the vector from 1, 2, 3 to 3, 2, 1 : u Let v be the vector from 1, 2, 3 to Normal vector: 4x 4x 1 3y 3y 4z 1 2u 1, j 0 4 0 2, 2 : v k 1 1 4i 4j k v 4z 3 i 1 2 2 10 S ection 11.5 44. 1, 2, 3 , Normal vector: v i, 1 x 1 0, x 1 Lines and Planes in Space k, 1 z 3 0, z 2i j 3 509 45. 1, 2, 3 , Normal vector: v 46. The plane passes through the three points 0, 0, 0 , 0, 1, 0 3, 0, 1 . The vector from 0, 0, 0 to 0, 1, 0 : u The vector from 0, 0, 0 to Normal vector: u x 3z 0 v i 0 3 3, 0, 1 : v j 1 0 k 0 1 i j 3i k 3k 47. The direction vectors for the lines are u v 3i 4j k. Normal vector: u v i 2 3 j 1 4 k 1 1 1, 5, 1 0 k, 5i j k Point of intersection of the lines: x x 1 y z y 5 5 z 1 48. The direction of the line is u 2i j k. Choose any point on the line, 0, 4, 0 , for example , and let v be the vector from 0, 4, 0 to the given point 2, 2, 1 : v 2i 2j k v 0 i 2 2 j 1 2 k 1 1 i 2k Normal vector: u x x 2 2z 2z 0 1 49. Let v be the vector from 1, 1, 1 to 2, 2, 1 : v 3y z 3i 3: n j 2i 2k 3j k Let n be a vector normal to the plane 2x Since v and n both lie in the plane P, the normal vector to P is v 7x 7x 2 y n i 3 2 1y 11z 5 5: 6i 7j 2k 51. Let u i and let v be the vector from 1, 2, 5, 6 : v i 7j 7k 2, 1 to j 1 3 2 k 2 1 11 z 7i 1 j 11k 0 50. Let v be the vector from 3, 2, 1 to 3, 1, v j 6k Let n be the normal to the given plane: n Since v and n both lie in the plane P, the normal vector to P is: v n i 0 6 j 1 7 k 6 2 40i 2 20i 20 x 20x 52. Let u 3 18y 18 y 3z 2 27 3, 5, 7 : v 3z 1 36j 18j 0 6k Since u and v both lie in the plane P, the normal vector to P is: u y v 2 z 1 ijk 100 177 z 1 7j 0 7k 7j k 3k y k and let v be the vector from 4, 2, 1 to 7i 3j 6k Since u and v both lie in the plane P, the normal vector to P is: u 3x 3x 4 7y v 7y 26 i 0 7 2 j 0 3 k 1 6 0 3i 7j 3i 7j 510 Chapter 11 Vectors and the Geometry of Space 0. 4⇒x 1 24 7 and 2 x 53. xy-plane: Let z Then 0 y 4 2 z −8 6 4 4− 6 2 2 68 10 y t⇒t 34 0. 3t ⇒ t 2 3 (− 7, 10, 0) 10. Intersection: 7, 10, 0 xz -plane: Let y Then 0 z 2 4 2 3 ⇒x 1 1 3, 2 2 3 10 3 1 3 and (0, − 1 , − 7 ) 2 2 ( −1 , 3 0, − 10 3 ) 10 3. Intersection: 0, yz -plane: Let x Then 0 z 1 4 0. 1 2 7 2. 2t ⇒ t 1 2 ⇒y 2 3 1 2, 1 2 7 2 1 2 and Intersection: 0, 2 3t, y 1 54. Parametric equations: x xy-plane: Let z Then 3 y 2t 1 0. 0⇒t 3 2 t, z 3 2t z (0 , − 5 , 5 ) 33 (5, 0, 5) 6 4 3 2 5 2. ⇒x 2 3 5 2, 3 2 5 2, 5 2 and −4 ( − 5 , − 5 , 0) 2 2 −1 2 −2 y −3 −3 Intersection: 0 x 5 4 −2 −1 2 1 xz -plane: Let y Then t z 1⇒x 3 21 0. 2 31 5 and 3 5. Intersection: 5, 0, 5 0. 2 3 5 3. yz -plane: Let x Then 2 z 3t 3 0⇒t 2 2 3 ⇒y 1 2 3 55 3, 3 5 3 and Intersection: 0, 55. Let x, y, z be equidistant from 2, 2, 0 and 0, 2, 2 . x x2 2 4x 2 y 4 y2 2 2 z 4 4x x 0 2 x x2 4z 0 y2 0 2 y 4y 4 2 2 z 4z 2 2 4y z2 8 z z2 4 8 Plane 2, 4 . x x2 6 2 56. Let x, y, z be equidistant from x x2 6x 9 3 y2 2 3, 1, 2 and 6, 2 y 2y 6x 18x 9x 1 1 z 4z 4z 4z 2z 2 2 y y2 8z 2 2 z 4y 4 4 2 z2 2y 6y 3y 4 14 42 21 12x 36 12x 4y z2 8z 16 56 0 0 Plane 58. The normal vectors to the planes are n1 n2 n1 n2 0. n1 Since n2 3, 1, 4 , n2 9, 3, 12 . 57. The normal vectors to the planes are n1 Thus, 5, 3, 1 , n2 1, 4, 7 , cos 3n1, the planes are parallel, but not equal. 2 and the planes are orthogonal. S ection 11.5 59. The normal vectors to the planes are n1 cos Therefore, i 3j 6k, n2 5 5i 3 j 6 k, 4 138 414 2 138 . 207 Lines and Planes in Space 511 60. The normal vectors to the planes are n1 cos Therefore, 3i 2j n1 n2 n1 n2 arccos k, n2 3 1 6 i 8 4j 2 2k, 1 . 6 n1 n2 n1 n2 arccos 46 27 2 138 207 14 21 65.9 . 83.5 . 61. The normal vectors to the planes are n1 1, 5, 1 and n2 5, 25, 5 . Since n2 5n1, the planes are parallel, but not equal. 62. The normal vectors to the planes are n1 cos Thus, 2, 0, 1 , n2 0. 4, 1, 8 , n1 n2 n1 n2 2 and the planes are orthogonal. 63. 4x 2y 6z z 6 4 12 64. 3x 6y z 3 2z 6 65. 2x y 3z z 3 4 (0, 0, 2) 4 (0, − 4, 0) −4 ( 0, 0, 4 2 3 ( 6 x (0, 6, 0) (3, 0, 0) 3 6 y x x 2 2 3 y 3 −1 y (2, 0, 0) 66. 2x y z z 4 4 67. y z 5 z 6 68. x 2y 4 z 4 (0, 0, 5) 2 −4 1 6 1 3 x x y x (0, 5, 0) 6 y 4 3 y 69. x 5 z 3 70. z 8 z 8 71. 2x y z z 2 6 6 x 4 2 4 6 y 5 x 5 y −6 Generated by Maple (5, 0, 0) 5 x 5 y 512 72. x Chapter 11 3z 3 z 2 1 Vectors and the Geometry of Space 73. 5x −2 4y 6z z 8 0 74. 2.1x z 4.7y z 3 0 3 2 3 x 2 1 1 2 y 2 x −1 1 y 1 x 2 y 1 Generated by Mathematica Generated by Maple Generated by Mathematica 75. P1: n P2: n P3: n P4: n 3, 2, 5 6, 4, 3, 2, 5 10 1, 1, 1, 1 on plane 1, 1 not on plane 75, 50, 125 1, 1, 1 on plane P1 and P4 are identical. P1 76. P1: n P2: n P3: n P4: n 6, P4 is parallel to P2. 60, 90, 30 or 9, 3 or 2, 3, 1 2, 3, 1 2, 3, 1 2, 3, 1 9 0, 0, 10 on plane 0, 0, 0, 0, 5 6 2 3 on plane 20, 30, 10 or 12, 18, 6 or on plane P1, P2, and P3 are parallel. 77. Each plane passes through the points c, 0, 0 , 0, c, 0 , and 0, 0, c . 79. If c If c 0, z 0, cy 0 is xy-plane. z 0⇒y 78. x y c Each plane is parallel to the z-axis. 80. x If c If c cz 0 0, x 0, x 0 is the yz-plane. cz 0 is a plane parallel to the y-axis. 1 z is a plane parallel to c x-axis and passing through the points 0, 0, 0 and 0, 1, c . 81. The normals to the planes are n1 3i 2j k and n2 i 4j 2k. The direction vector for the line is n2 n1 i 1 3 2z 2z x j 4 2 14 0 14 2 k 2 1 7j 2k . 82. The normals to the planes are n1 and n2 1, 1, 5 . The direction vector for the line is n1 n2 i 6 1 j 3 1 k 1 5 6, 3, 1 16, 31, 3 . Now find a point of intersection of the planes. 6x x 7x 4y 4y Now find a point of intersection of the planes. 6x x Let y x 3y y 9, z 4 z 5z 5⇒ 5⇒ 6x 6x 4⇒ 9 31t, z 3y 6y 3y 4, 2 z 30z 31z 9, 2 . 3t 5 30 35 Substituting 2 for x in the second equation, we have 4y 2z 2 or z 2y 1. Letting y 1, a point of intersection is 2, 1, 1 . x 2, y 1 t, z 1 2t 2⇒x 16t, y S ection 11.5 83. Writing the equation of the line in parametric form and substituting into the equation of the plane we have: x 1 2 2 1 2 t t, y 3 2 t t, z 1 1 2t 2t 12, t 3 2 Lines and Planes in Space 513 84. Writing the equation of the line in parametric form and substituting into the equation of the plane we have: x 21 1 4t 4t, y 3 2t 2t, z 3 5, t 6t 1 2 3 2 2 Substituting t 3 2 into the parametric equations for the line we have the point of intersection 2, 3, 2 . The line does not lie in the plane. 85. Writing the equation of the line in parametric form and substituting into the equation of the plane we have: x 21 1 3t 3t, y 3 1 1 2t 2t, z 10, 3 1 t 10, contradiction 1 Substituting t 2 into the parametric equations for the line we have the point of intersection 1, 1, 0 . The line does not lie in the plane. 86. Writing the equation of the line in parametric form and substituting into the equation of the plane we have: x 54 4 2t 2t, y 3 1 1 3t 3t, z 17, t 2 0 5t Therefore, the line does not intersect the plane. Substituting t 0 into the parametric equations for the line we have the point of intersection 4, 1, 2 . The line does not lie in the plane. 89. Point: Q 2, 8, 4 z 8 8, 4, 1 Plane: 2x y z 5 2, 1, 1 87. Point: Q 0, 0, 0 Plane: 2x 3y z 12 0 Normal to plane: n \ 88. Point: Q 0, 0, 0 Plane: 8x 4y 2, 3, 1 Point in plane: P 6, 0, 0 Vector PQ \ Normal to plane: n Normal to plane: n \ 6, 0 0 12 14 6 14 7 Point in plane: P 1, 0, 0 \ Point in plane: P 0, 0, 5 Vector: PQ \ D PQ n n Vector: PQ \ 1, 0, 0 8 81 8 9 2, 8, 11 6 1 11 6 6 D PQ n n D PQ n n 90. Point: Q 3, 2, 1 Plane: x y 2z 4 1, 1, 2 91. The normal vectors to the planes are n1 1, 3, 4 and n2 1, 3, 4 . Since n1 n2, the planes are parallel. Choose a point in each plane. P Q \ Normal to plane: n Point in plane: P 4, 0, 0 \ 10, 0, 0 is a point in x 6, 0, 0 is a point in x \ 3y 3y 4z 4z 4 26 10. 6. 2 26 13 Vector: PQ \ 1, 2, 1 1 6 1 6 6 6 PQ 4, 0, 0 , D D PQ n n PQ n1 n1 4, 4, 9 and 92. The normal vectors to the planes are n1 n2 4, 4, 9 . Since n1 n2, the planes are parallel. Choose a point in each plane. P Q \ 3, 6, 7 and 93. The normal vectors to the planes are n1 n2 6, 12, 14 . Since n2 2n1, the planes are parallel. Choose a point in each plane. P Q \ 5,0, 3 is a point in 4x 0, 0, 2 is a point in 4x 5, 0, \ 4y 4y 9z 9z 7. 18. 0, 1, 1 is a point in 3x 6y 12y 7z 14z 1. 25. PQ D 1 11 113 11 113 113 PQ D 25 , 0, 0 is a point in 6x 6 25 , 1, 6 \ PQ n1 n1 1 27 2 94 27 2 94 27 94 188 PQ n1 n1 514 Chapter 11 Vectors and the Geometry of Space 4, 0, 1 is the direction vector for the line. 95. u Q 1, 5, 2 is the given point, and P 2, 3, 1 is on the line. Hence, PQ 3, 2, 3 and \ \ 2, 0, 4 94. The normal vectors to the planes are n1 and n2 2, 0, 4 . Since n1 n2, the planes are parallel. Choose a point in each plane. P Q \ 2, 0, 0 is a point in 2x 5, 0, 0 is a point in 2x \ 4z 4z 4. 10. 6 20 35 5 PQ \ u PQ 3, 0, 0 , D PQ n1 n1 i 3 4 j 2 0 149 17 k 3 1 2, 2533 17 9, 8. D PQ u u 1, 1, 96. u P \ 2, 1, 2 is the direction vector for the line. 0, 3, 2 is a point on the line let t 1, 1, 2 u \ 97. u 2 is the direction vector for the line. 0. PQ \ Q 2, 1, 3 is the given point, and P 1, 2, 0 is on the line let t 0 in the parametric equations for the line . \ PQ ijk 112 212 5 9 Hence, PQ 0, 2, 5 3 1 \ 3, i 3 1 j 1 1 1 1, 3 and k 3 2 1 81 1 1, 16 4 9, 4. 98 6 7 3 73 3 PQ u \ D PQ u u D PQ u u 98. u 0, 3, 1 is the direction vector for the line. \ 99. The direction vector for L1 is v1 The direction vector for L 2 is v2 Since v2 3, 1, 2, 1 . 6, 3. Q 4, 1, 5 is the given point, and P on the line. Hence, PQ 1, 2, 4 and \ 3, 1, 1 is 3v1, the lines are parallel. \ PQ u \ i 1 0 j 2 3 k 4 1 142 9 515 5 1 14, 9 1 1, 3 . 206 10 2, 3, 4 to be a point on L1 and P Let Q 2, 0, 0 . point on L 2. PQ u v2 is the direction vector for L 2. \ 0, 1, 4 a D PQ u u 103 5 PQ v2 \ i 2 3 j 2 6 k 0 3 36 9 6, 6, 36 36 324 9 18 396 54 D PQ v2 v2 22 3 66 3 100. The direction vector for L1 is v1 The direction vector for L 2 is v2 Since v1 3 2 v2, 6, 9, 4, 6, 12 . 8. 101. The parametric equations of a line L parallel to v a, b, c, and passing through the point P x1, y1, z1 are x 1, 3, 0 x1 at, y y1 bt, z z1 ct. the lines are parallel. \ 3, 2, 1 to be a point on L1 and P Let Q 4, 5, 1 . a point on L 2. PQ u v2 is the direction vector for L 2. \ The symmetric equations are x a x1 y b y1 z c z1 . PQ v2 \ i 4 4 j 5 6 k 1 8 342 16 31813 29 34, 36, 44 36 2 36 442 64 4388 116 D PQ v2 v2 1097 29 S ection 11.5 102. The equation of the plane containing P x1, y1, z1 and having normal vector n a, b, c is ax x1 by y1 cz z1 0. Lines and Planes in Space 515 103. Solve the two linear equations representing the planes to find two points of intersection. Then find the line determined by the two points. You need n and P to find the equation. 104. x y z a: plane parallel to yz-plane containing a, 0, 0 b: plane parallel to xz-plane containing 0, b, 0 c: plane parallel to xy-plane containing 0, 0, c 105. (a) The planes are parallel if their normal vectors are parallel: a1, b1, c1 t a2, b2, c2 , t 0 (b) The planes are perpendicular if their normal vectors are perpendicular: a1, b1, c1 106. Yes. If v1 and v2 are the direction vectors for the lines L1 and L 2, then v v1 v2 is perpendicular to both L1 and L 2. a2, b2, c2 0 107. An equation for the plane is y z x 1. abc For example, letting y a, 0, 0 . z 0, the x-intercept is 108. (a) Sphere x x2 109. 0.04x (a) 3 y2 0.64y Year x y z Model, z 2 (b) Parallel planes y z2 z 2 6x 2 z 4y 5 2 16 22 0 0.04x 1997 6.6 7.7 8.2 8.06 0.64y 1998 6.5 7.4 7.8 7.88 1999 6.3 7.3 7.9 7.82 4x 3y z 10 ± 4 n 10 ± 4 26 10z 3.4 3.4 ⇒ z 1995 6.2 8.2 8.4 8.40 1994 5.8 8.7 8.8 8.74 1996 6.4 8.0 8.4 8.26 2000 6.1 7.1 7.8 7.70 The approximations are close to the actual values. (b) According to the model, if x and z decrease, then so will y, the consumption of reduced-fat milk. 110. On one side we have the points 0, 0, 0 , 6, 0, 0 , and n1 i 6 1 j 0 1 k 0 8 48j 6k 1, 1, 8 . (− 1, − 1, 8) 6 4 2 z 1, 1, 8 . On the adjacent side we have the points 0, 0, 0 , 0, 6, 0 , and n2 i 0 1 j 6 1 k 0 8 48i 36 2340 89.1 6k 1 65 cos n1 n2 n1 n2 arccos 1 65 6 x (0, 0, 0) 4 (6, 0, 0) (0, 6, 0) 6 y 516 Chapter 11 6 1 t, y1 t, y2 Vectors and the Geometry of Space 8 2 t, z1 t, z2 3 2t 6, 8, 3 and the second insect is at P2 2 y1 2t 2 2 111. L1: x1 L 2: x 2 (a) At t t 0, the first insect is at P1 6 x1 52 5t 2 1 2 2 1, 2, 0 . Distance (b) Distance 8 3 2 0 z1 2 70 z2 2 8.37 inches 15 x2 6 y2 3 t 2 30t 70, 0 ≤ t ≤ 10 0 0 15 (c) The distance is never zero. (d) Using a graphing utility, the minimum distance is 5 inches when t 112. First find the distance D from the point Q normal to the plane. \ 3 minutes. 4, 0, 0 be on the plane. n 18 29 29 3 2 3, 2, 4 to the plane. Let P 14 8 29 12 18 29 2, 4, 3 is the D PQ n n 7, 2, 4 2, 4, 3 4 16 9 The equation of the sphere with center 3, 2, 4 and radius 18 29 29 is x y 2 2 z 4 2 324 . 29 113. The direction vector v of the line is the normal to the plane, v The parametric equations of the line are x 35 3t 4 t 4 3 4t 26t t Point of intersection is 5 3 4 ,4 13 1, 7 8 4 13 4 , 13 3 4 4 13 5 3t, y 4 t, z 3, 1, 4 . 3 4t. To find the point of intersection, solve for t in the following equation: 77 48 ,, 13 13 23 13 2, 4, 0 of the line because 2, 1, 4 be on the line and 114. The normal to the plane, n 2, 2, 1, 3 2, 4, 0 0. \ 3 is perpendicular to the direction vector v Hence, the plane is parallel to the line. To find the distance between them, let Q P 2, 0, 0 on the plane. PQ 4, 1, 4 . \ D PQ n n 4, 1, 4 4 2, 1 9 1, 3 19 14 19 14 14 115. The direction vector of the line L through 1, The parametric equations for L are x 1 2t 3 t 1 t 4t t 1 2 3 3 . 4 3, 1 and 3, 3 t, z 4, 2 is v 1 t. 2, 1, 1 . 2t, y Substituting these equations into the equation of the plane gives Point of intersection is 1 2 3 , 4 3 3 ,1 4 3 4 1 , 2 91 , 44 116. The unknown line L is perpendicular to the normal vector n u 1, 1, 1 . Hence, the direction vector of L is v i 1 1 j 1 1 k 1 1 2, 2, 0 . 1, 1, 1 of the plane, and perpendicular to the direction vector The parametric equations for L are x 1 2t, y 2t, z 2. S ection 11.6 117. True 118. False. They may be skew lines. (See Section Project) t, z 1 are both parallel to 119. True Surfaces in Space 517 120. False. The lines x t, y 0, z 1 and x 0, y the plane z 0, but the lines are not parallel. Section 11.6 1. Ellipsoid Matches graph (c) 4. Elliptic cone Matches graph (b) 7. z 3 Surfaces in Space 2. Hyperboloid of two sheets Matches graph (e) 5. Elliptic paraboloid Matches graph (d) 8. x 4 3. Hyperboloid of one sheet Matches graph (f) 6. Hyperbolic paraboloid Matches graph (a) 9. y2 z2 9 Plane parallel to the xy-coordinate plane z Plane parallel to the yz-coordinate plane z The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a circle. z 4 4 2 2 3 x 2 y x 4 2 4 y x 76 4 y 10. x2 z2 25 11. y x2 12. z 4 y2 The y-coordinate is missing so we have a cylindrical surface with rulings parallel to the y-axis. The generating curve is a circle. z 6 4 The z-coordinate is missing so we have a cylindrical surface with rulings parallel to the z-axis. The generating curve is a parabola. z 4 The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a parabola. z 8 4 8 x 8 y x 4 3 2 3 4 y x 8 4 8 12 y 13. 4x2 x 1 2 y2 y 4 2 4 1 z 3 14. y 2 y2 4 z2 z 4 2 4 1 z 5 The z-coordinate is missing so we have a cylindrical surface with rulings parallel to the z-axis. The generating curve is an ellipse. −3 2 3 x 2 3 y The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a hyperbola. 5 x 5 y 518 15. z Chapter 11 sin y Vectors and the Geometry of Space 16. z z 2 1 y ey z 20 15 10 5 3 x 1 The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is the sine curve. x 3 3 4 The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is the exponential curve. 2 3 4 y 17. z x2 y2 (c) You are viewing the paraboloid from the z-axis: 0, 0, 20 (d) You are viewing the paraboloid from the y-axis: 0, 20, 0 (a) You are viewing the paraboloid from the x-axis: 20, 0, 0 (b) You are viewing the paraboloid from above, but not on the z-axis: 10, 10, 20 18. y2 z2 4 (a) From 10, 0, 0 : z (b) From 0, 10, 0 : z (c) From 10, 10, 10 : z 3 y y 3 x 3 y 19. x2 1 y2 4 x 1 z2 1 2 1 2 z 20. x2 16 y2 25 x2 16 x2 16 z2 25 y2 25 z2 25 z2 1 5 4 3 2 1 4 3 2 1 z Ellipsoid xy-trace: y2 4 z2 z2 1 1 ellipse 2 Ellipsoid xy-trace: 1 ellipse 5 xz-trace: x 2 yz-trace: y2 4 1 circle 1 ellipse x −2 2 y 12 34 5 y xz-trace: 1 ellipse 25 circle x yz-trace: y2 21. 16x 2 4x 2 y2 y 4 2 16z 2 4z 2 1 4 3 2 z 22. z 2 −2 −3 Hyperboloid on one sheet xy-trace: 4x 2 xz-trace: 4 x yz-trace: 2 y2 1 4 Hyperboloid of two sheets xy-trace: none x2 xz-trace: z 2 x2 y2 4 x2 9 1 hyperbola 1 hyperbola y2 36 1 ellipse 5 x z 5 5 y y2 4 z 2 2 1 hyperbola 1 circle 2 3 x −2 −3 3 y yz-trace: z2 z ± y2 4 z2 0 4z 1 hyperbola 10 : 23. x2 y z 3 2 1 −3 24. z x2 4y2 4 z Elliptic paraboloid xy-trace: y xz-trace: x2 yz-trace: y y 1: x2 x2 z2 z2 z2 1 0, 3 x 2 1 Elliptic paraboloid xy-trace: point 0, 0, 0 3 point 0, 0, 0 xz-trace: z 4 y x2 parabola 4y2 parabola 3 x −2 −3 yz-trace: z 2 1 1 2 y S ection 11.6 y2 x2 Surfaces in Space 519 25. x2 y2 z 0 3 z 26. 3z z 28 24 20 Hyperbolic paraboloid xy-trace: y xz-trace: z yz-trace: z y ± 1: z ±x Hyperbolic paraboloid xy-trace: y ±x 12 3x 12 3y 10 x 23 y x2 y2 1 x2 x 32 xz-trace: z yz-trace: z 10 y 27. z 2 y2 4 Elliptic Cone x2 xy-trace: point 0, 0, 0 xz-trace: z yz-trace: z z ± 1: x2 ±x ±1 −2 x 2 z 28. x2 2 1 2y 2 2z 2 5 z Elliptic Cone xy-trace: x 2 y ± ± 2y 5 xz-trace: x 2z x 5 y 2 y2 4 9y2 1 16 x y 1 −2 yz-trace: point: 0, 0, 0 29. 16x2 16 x2 2x 16z2 9 y2 1 x 1 2 32x 4y 9y 36y 4 2 2 2 36 16z2 16z 2 z2 1 0 36 16 2 z 16 36 1 x 2 1 1 2 y2 16 9 2 −2 4 y 1 Ellipsoid with center 1, 2, 0 . 30. 9 x2 6x 9x2 9 y2 y2 9x 3 9z2 4y 2 54x 4 y 3. 4y 9 z2 2 2 54z 6z 9z 4 9 3 2 0 81 0 x z 81 15 10 10 5 Elliptic cone with center 3, 2, 15 20 25 y −15 31. z 2 sin x z 32. z x2 0.5y 2 z 4 33. z 2 z x2 ± 4y 2 x2 z 5 4y2 3 π x 3 y −2 1 2 x 1 2 y x −2 −1 1 2 y 520 Chapter 11 Vectors and the Geometry of Space 2 z 4 z2 z 4 2 34. z 2 z 4y ± x2 4y z 35. x2 x2 y −8 y2 ± 36. x2 x2 −4 y2 ln x2 e y2 z 4 z z 8 4 4 8 x −3 −3 −4 4 −4 −8 3 y 4 x 4 y x 4 y 37. z 4 z 5 xy 38. z 8 x x2 z y2 39. 4x 2 z ± y2 4z 2 y2 4 z 8 6 4 −6 −8 16 x2 4 4 2 5 x 4 3 3 4 5 y x 4 2 2 4 y 8 x 6 4 −2 −2 −4 −6 −8 −4 2 y 40. 9x 2 z ± 4y 2 8z 2 92 x 8 72 12 y 2 9 41. z z 2 x2 2 y2 y2 z y2 42. z y 4 4 0, y 5 x2 x2 0, z z 2 x2 x2 2 1 x 0 z 20 10 3 10 10 x 20 20 y 2 −2 −2 2 x 1 2 y x 43 4 y 43. x2 x z y2 z 0 2 1 4 3 2 z 44. z y z 2z 0 4 x2 y2 3 z −3 3 3 x y 3 x 2 3 y 45. x2 z2 x2 z2 ry 2 and z ry ±2 y ; therefore, 46. x2 z2 x2 z2 ry 2 and z ry 3y; therefore, 4y. z ; therefore, 2 z 2. 9y 2. 1 2 4y 2 47. x2 y2 x2 y2 rz 2 and y rz 4y 2 48. y2 z2 y2 z2 rx 2 and z rx 4 4z 2 x2 ; therefore, 4. z2 2 , 4x 4 1 4 4 x2 , x 2 S ection 11.6 2 ; therefore, x 4 . x2 52. x 2 Surfaces in Space 521 49. y 2 z2 y2 z2 rx 2 and y rx z2 50. x2 y2 x2 y2 rz 2 and y rz ez; therefore, 22 2 ,y x 0 2z 2 e2z. 51. x 2 x2 y2 y2 2z z2 cos2 y cos y or z cos y Equation of generating curve: x 2z or x 2z Equation of generating curve: y 53. Let C be a curve in a plane and let L be a line not in a parallel plane. The set of all lines parallel to L and intersecting C is called a cylinder. 55. See pages 812 and 813. 54. The trace of a surface is the intersection of the surface with a plane. You find a trace by setting one variable equal to a constant, such as x 0 or z 2. 56. In the xz-plane, z In three-space, z x 2 is a parabola. x 2 is a cylinder. 4 z 57. V 2 0 x 4x 4x3 3 x4 4 x2 4 0 dx 128 3 4 3 2 1 58. V 2 0 y sin y dy 1.0 z 2 2 h ( x) x 1 2 3 4 sin y y cos y 0 2 2 0.5 p ( x) π 2 π y 59. z x2 2 y2 4 2 we have 2 42 4 4, c 2 x2 2 y2 , or 1 4 x2 4 y2 8 (b) When z 8 we have 8 x2 2 y2 , or 1 4 x2 16 y2 . 32 (a) When z Major axis: 2 8 Minor axis: 2 4 c2 a2 b2, c2 Foci: 0, ± 2, 2 x2 2 y2 4 Major axis: 2 32 Minor axis: 2 16 c2 32 16 Foci: 0, ± 4, 8 82 8 4 16, c 60. z 61. If x, y, z is on the surface, then 4 we have z 0, 4, 9 2 x2 2 1 4, 4 z 2 y 4 x. 2 2 2 x2 x2 8y y y2 2)2 4y z2 4 z2 (a) When y Focus: (b) When x z 2 y2 4y x2 4 z2 Elliptic paraboloid 2 we have y2 ,4 z 4 2 y 2. Traces parallel to xz-plane are circles. Focus: 2, 0, 3 522 Chapter 11 Vectors and the Geometry of Space x2 39632 y2 39632 z 62. If x, y, z is on the surface, then z2 z2 8z x2 x 2 63. z2 3950 2 1 y2 y 2 z z 2 4 2 8z 16 x 8 2 4000 x2 y2 16 ⇒ z y2 8 2 4000 4000 x y Elliptic paraboloid shifted up 2 units. Traces parallel to xy-plane are circles. 2 64. (a) x2 y2 rz 2 (b) V 1 2 2 0 2 x3 2x 0 12 x 2 13 x dx 2 x4 8 2 2 1 dx (c) V 2 12 2 x3 2x 12 12 x 2 13 x dx 2 2 12 1 dx 2z x2 y2 z 4 3 2z 2 0 2 2 4 y 2 2 x2 x2 0 x4 8 31 64 12.6 cm3 y 4 225 64 11.04 cm 3 −2 2 x 2 2 1 1 x 1 2 3 1 2 3 y 3 3 x 1 2 3 65. z y2 b2 x2 ,z a2 bx bx ay a4b2 4 a2b 2 a2 y 2 ay y2 b2 12 y b2 y x2 a2 ab2y ab2 2 b2 ± 2 66. Equating twice the first equation with the second equation, 2x2 6y2 4z2 4y 4y a2b4 4 3x 8 8 4y 2x2 3x 6y2 2 4z2 3x 2 12 x a2 a2 bx x 6, a plane b x a a2b 2 ab2 2 Letting x at, you obtain the two intersecting lines x at, y bt, z 0 and x at, y bt ab2 z 2abt a2b2. 68. False. For example, the surface x 2 z 2 e 2y can be formed by revolving the graph of x e y about the y-axis, as the graph of z e y about the y-axis. 67. True. A sphere is a special case of an ellipsoid (centered at origin, for example) x2 a2 having a y2 b2 b z2 c2 c. 1 69. The Klein bottle does not have both an “inside” and an “outside.” It is formed by inserting the small open end through the side of the bottle and making it contiguous with the top of the bottle. S ection 11.7 Cylindrical and Spherical Coordinates 523 Section 11.7 1. 5, 0, 2 , cylindrical x y z 5 cos 0 5 sin 0 2 5 0 Cylindrical and Spherical Coordinates 2. x y z 0, 4, 4, , 2 2 , cylindrical 0 4 3. x y z 2 , rectangular 1, 2, , 2 , cylindrical 3 2 cos 2 sin 2 3, 2 , rectangular 3 , 1 , cylindrical 2 cos sin 1 1, 1 , rectangular 3 2 3 2 0 1 3 3 1 3 4 cos 4 sin 2 2 2 5, 0, 2 , rectangular 4. x y z 6, 4 , 2 , cylindrical 32 32 5. x y z 4, 7 , 3 , cylindrical 6 4 cos 4 sin 3 2 3, 2, 3 , rectangular 7 6 7 6 23 2 6. x y z 1, 6 cos 6 sin 2 4 4 0, 3 2, 3 2, 2 8. 2 2, 5 r 2 2, 4 , rectangular 22 arctan z 4, 4 4 2, 22 arctan z 4 2 2, 4 , 4 , cylindrical 1 , 4 , cylindrical 2 7. 0, 5, 1 , rectangular r 0 2 9. 1, 4 r 3, 4 , rectangular 12 arctan 3 3 2 5 2 22 4 2 2 5 arctan 0 z 1 2 1 3 z 4 2, , 4 , cylindrical 3 5, , 1 , cylindrical 2 10. 2 3, r 2, 6 , rectangular 12 arctan z 4, 13. z 6 4 4 1 3 5 6 11. 2, r 4 , rectangular 2 2 12. r 3, 2, 1 , rectangular 3 2 22 4 22 2 3 13 arctan 2 3 arctan z 1 13, 6 , 6 , cylindrical 2 arctan , 3 1 , cylindrical 5 is the equation in cylindrical coordinates. 14. x 4, 4 r 4 sec , y2 2, 8x, 8r cos 8 cos , 2, rectangular coordinates r cos cylindrical coordinates rectangular equation cylindrical equation rectangular equation 15. x2 y2 r2 z2 z2 10, 10, rectangular equation cylindrical equation rectangular equation 16. z z 18. x2 x2 r2 y2 r2 r 17. y r sin sin x2, r cos r cos2 r sec tan , 2 cylindrical equation cylindrical equation 524 19. Chapter 11 y2 r sin r2 sin 2 2 Vectors and the Geometry of Space 10 10 10, z 2, rectangular coordinates z 2 20. x2 y2 r 2 z2 z 2 3z 3z 0, rectangular coordinates 0 cylindrical coordinates z2 cylindrical coordinates 21. x2 x2 r y2 y2 z 3 2 2 2 4 22. z 2 23. tan 6 6 y x y x 3y 0 z 2 1 −2 x 2 1 2 y −2 Same z 3 1 3 x 1 3 −3 3 x 3 2 −2 −3 2 3 y x 2 y x 3y 24. x2 x2 y2 z 4 r y2 z2 4 z 2 z 2 0 25. x2 x2 x2 y2 y z r r2 y2 2y 1 2 2 sin 2r sin 2y 0 1 26. x2 x2 x y2 1 2 z r r2 y2 2x y2 2 cos 2r cos 2x 0 1 2 −2 2 x 2 y −2 2 x −1 −2 x 1 −2 1 2 y 2 −2 2 3 y 27. x2 r2 y2 z2 z2 4 4 −2 x 2 1 −1 2 1 z 28. z z −2 2 y r 2 cos2 9 z x2 1 32 x 12 34 56 y 29. 4, 0, 0 , rectangular 42 arctan 0 arccos 0 4, 0, 02 0 2 02 4 30. 1, 1, 1 , rectangular 12 arctan 1 1 arccos 3 3, , arccos 4 1 , spherical 3 12 4 12 3 2 , spherical S ection 11.7 31. 2, 2 3, 4 , rectangular 2 arctan arccos 4 2, 1 2 4 2 Cylindrical and Spherical Coordinates 525 32. 2, 2, 4 2 , rectangular 4 2 23 3 2 42 22 arctan 1 arccos 22 4 2 5 42 2 2 10 2 3 2 , , spherical 34 2 10, , arccos 4 2 , spherical 5 33. 3, 1, 2 3 , rectangular 3 1 12 6 6 4 1 arctan 3 3 arccos 2 34. 4, 0, 0 , rectangular 4 2 35. 4 x 02 02 4, , , spherical 64 4 sin 4 sin 4 cos 6, 4 4 4 cos sin 6 6 22 6 2 arccos 0 4, , 2 y z 4, , , spherical 66 2 , spherical 2, 2 2 , rectangular 36. x y z 12, 3 , , spherical 49 12 sin 12 sin 12 cos 9 9 9 cos sin 3 4 3 4 11.276 2.902 2.902 37. x y z 12, 4 , 0 , spherical 0 0 12 sin 0 cos 12 sin 0 sin 12 cos 0 4 4 12 0, 0, 12 , rectangular 2.902, 2.902, 11.276 , rectangular 3 5, , , spherical 44 x y z 5 sin 5 sin 3 cos 4 4 3 sin 4 4 5 2 5 2 52 2 38. x y z 9, , 4 9 sin 9 sin 9 cos , spherical cos sin 0 0 9 39. 40. x y z 6, , 2 , spherical cos sin 0 0 6 4 4 6 sin 6 sin 6 cos 2 2 2 3 5 cos 4 55 ,, 22 0, 0, 9 , rectangular 52 , rectangular 2 42. cos z 2, 2 2 sec , 6, 0, 0 , rectangular 41. sin sin y 3, 3 3 csc rectangular equation rectangular equation csc , spherical equation pherical equation 526 43. x2 Chapter 11 y2 z2 2 Vectors and the Geometry of Space 36, rectangular equation 36 6, spherical equation 44. x2 x2 y2 y2 3z2 z2 2 0, 4z2 4 2 rectangular equation cos2 1 cos 4 cos2 1 2 3 , (cone) spherical equation 45. 2 x2 sin2 cos2 2 y2 9, 9 9 3 rectangular equation sin2 2 sin2 sin2 sin 3 csc , spherical equation 46. sin cos x 10, 10 10 csc sec , rectangular equation spherical equation 48. x 2 2 47. x 2 2 2 2 y2 sin2 sin2 sin2 2 2 ± 2 z rectangular coordinates cos2 cos2 2 2 2 2 y2 2 z2 9 cos 9z 0 0 9 cos rectangular equation sin2 sin2 2 2 2 cos2 cos2 sin2 cos2 spherical equation sin2 cos2 tan2 tan 2 spherical equation 3 4 tan −2 2 1 −1 2 y 49. x2 2 y2 z2 4 −2 x 2 1 z 50. z 3 y x y x y 0 −3 3 x −3 y 1 x 51. cos 3 2 3 4 3x 2 z 6 2 52. cos z 2 x z y2 2 x2 x2 z2 y2 z y2 z y2 z2 z2 z2 z2 x 2 −2 −1 1 −1 1 2 y −1 −2 z y2 z2 3 z 2 −3 2 −3 0 z 0 x2 3 x 3 −2 −3 y x2 xy-plane 0, z ≥ 0 3y 2 S ection 11.7 53. 4 cos x2 x2 x2 y2 y2 y2 z2 z z2 4z 2 2 Cylindrical and Spherical Coordinates 527 z 54. cos z −2 −3 2 sec 2 3 z x 0 2 4z y2 5 z 2 4 3 2 1 3 2 2 3 x 4, z ≥ 0 x 3 2 1 2 y 3 y 55. sin x2 x2 y2 y2 z 2 csc 1 1 1 56. 4 csc sin sin x 4 z 6 sec 57. 4, , 0 , cylindrical 4 42 4 arccos 0 2 02 4 4 cos cos 4 1 −2 x 2 1 −1 −2 x 1 2 −2 y 4 6 4 4, , , spherical 42 6 y 58. 3, 4 , 0 , cylindrical 02 3 59. 4, , 4 , cylindrical 2 42 2 arccos 4 42 4 42 42 60. 2, 2 , 3 22 2 3 2 , cylindrical 2 2 32 4 22 0 arccos 9 3, arccos 2 2, 1 2 3 4 2 , , spherical 42 , 6 , cylindrical 42 6 arccos 3 13 , arccos 3 , spherical 13 65. r 62 2 13 62. 4 2, , , spherical 24 23 , , spherical 34 61. 4, 6 4, , 4 , cylindrical 3 4 3 arccos 1 2 4 2 63. 12, , 5 , cylindrical 122 52 5 13 5 , spherical 13 13 42 42 arccos 13, , arccos 2 13, 6 4 2, , , spherical 34 64. 4, , 3 , cylindrical 2 42 2 arccos 3 5 32 5 10, , , spherical 62 10 sin 2 10 66. r 4, , , spherical 18 2 4 sin 2 4 6 z 10 cos 2 0 z 4, 18 4 cos 2 0 3 5, , arccos , spherical 2 5 10, , 0 , cylindrical 6 18 , 0 , cylindrical 528 Chapter 11 Vectors and the Geometry of Space 67. r 36, , 2 , spherical 36 sin 36 68. r 18, , , spherical 33 sin 18 sin 3 9 69. r 6, , , spherical 63 6 sin 3 33 sin 2 3 z cos 36 cos 2 0 z cos 18 cos 3 93 3 3, 9, , 9 3 , cylindrical 3 5 , 6 5 sin 5 6 z 0, 5 cos 5 , 6 5 5 , cylindrical z 4, 7 , , spherical 66 8 sin 7 6 8 cos 6 83 2 z 6 4 z 6 6 cos 3 6 3 36, , 0 , cylindrical , 3 , cylindrical 70. r 5, , spherical 0 71. r 8, 72. r 3 7, , , spherical 44 7 sin 3 4 72 2 4 7 cos 72 ,, 24 Spherical 7.810, 0.983, 1.177 3 7.000, 0.322, 2.014 3 4 72 2 72 , cylindrical 2 7 , 4 3 , cylindrical 6 Rectangular 73. 4, 6, 3 74. 6, 2, 3 Cylindrical 7.211, 0.983, 3 6.325, 5, , 8 9 10, 0.75, 6 0.322, 75. 4.698, 1.710, 8 76. 7.317, 77. 6.816, 6 9.434, 0.349, 0.559 11.662, 20, 2 , 34 0.750, 1.030 7.071, 12.247, 14.142 14.142, 2.094, 14.142 6.311, 0.25, 4.052 3.606, 0.588, 2 3 1.5 78. 6.115, 1.561, 4.052 79. 3, 2, 2 3 7.5, 0.25, 1 4.123, 0.588, 1.064 80. 3 2, 3 2, 81. 54 3 ,, 23 2 5, 4 6, 0.785, 6.708, 0.785, 2.034 3.206, 0.490, 2.058 6.403, 1.571, 0.896 2.833, 0.490, 5, 5 5, 1.571, 4 3 , 4 2, 5 82. 0, 83. 84. 3.536, 3.536, 1.732, 1, 3 7.071, 2.356, 2.356 3.606, 2.618, 0.588 11 ,3 6 Note: Use the cylindrical coordinate 2, 85. 2.804, 2.095, 6 5 ,3 6 6.946, 5.642, 0.528 3.5, 2.5, 6 [Note: Use the cylindrical coordinates 3.5, 5.642, 6 86. 2.207, 7.949, 4 8.25, 1.3, 4 9.169, 1.3, 2.022 S ection 11.7 Cylindrical and Spherical Coordinates 529 87. r 5 88. Cylinder Matches graph (d) 4 Plane Matches graph (e) 89. 5 Sphere Matches graph (c) 90. Cone 4 Matches graph (a) 91. r 2 z, x 2 y2 z 92. 4 sec , z Plane Matches graph (b) cos 4 Paraboloid Matches graph (f) 93. Rectangular to cylindrical: r 2 tan z Cylindrical to rectangular: x y z 95. Rectangular to spherical: 2 x2 y x z r cos r sin z x2 y x arccos y2 94. r a Cylinder with z-axis symmetry b Plane perpendicular to xy-plane z c Plane parallel to xy-plane y2 z2 Spherical to rectangular: x y z y2 z z2 97. x2 (a) (b) 99. x2 y2 r2 2 sin sin cos cos sin tan x2 96. a Sphere b Vertical half-plane c Half-cone z2 z2 16, z2 z2 16 16 4 2z 2z 0 0, r 2 0, z 1 2 98. 4 x2 (a) 4r 2 (b) 4 y2 z2 z 2, 2r z 2 y2 (a) r 2 sin2 sin2 1 , 4 1 2 101. x2 (a) y2 r2 2 2 1 0, 2 sin2 cos2 cos2 , (b) 2 2 cos 2 cos 2 cos 4 sin2 tan 1 , 2 z z sin2 cos 2 , tan2 arctan 100. x2 y2 4y 4r sin , r 4 sin sin , 4 sin 4 sin csc 0, (a) r 2 (b) 2 cos , csc sin2 cot cos , (b) sin2 4 sin , sin 4 sin sin cos , sin2 sin 102. x2 y2 16 16, r 4 16, 4 2 (a) r 2 (b) 2 sin2 sin sin2 4 16 0, 0, 4 csc sin 530 103. x2 Chapter 11 y2 9 Vectors and the Geometry of Space (b) r 2 sin2 9 sin2 9, 2 sin2 sin2 cos2 cos2 2 sin2 sin2 , sin2 9, (a) r 2 cos2 r2 cos2 2 9 2 9 csc2 cos2 sin2 z 104. y 4 4, r sin 4 csc 4, 4 csc csc 105. 0 ≤ ≤ 2 5 (a) r sin (b) sin 0≤r≤2 0≤z≤4 2 3 x 3 2 1 2 3 y 106. 2 ≤ ≤ 2 107. 0 ≤ ≤2 108. 0 ≤ z2 ≤ ≤2 r2 z 4 3 109. 0 ≤ 6r 8 0≤ 0≤ ≤2 ≤ 6 0≤r≤3 0 ≤ z ≤ r cos z 4 3 0≤r≤a r≤z≤a z a 2≤r≤4 ≤ a sec z a −a −4 4 4 y x a a −a −5 5 y x 5 x y 30° x y 110. 0 ≤ 4 ≤ ≤2 ≤ 2 111. 0 ≤ 0≤ 0≤ ≤ ≤ 2 2 112. 0 ≤ 0≤ 1≤ ≤ ≤ 2 113. Rectangular 0 ≤ x ≤ 10 0 ≤ y ≤ 10 0 ≤ z ≤ 10 z 10 0≤ ≤1 z ≤2 z ≤3 z 3 2 2 −2 −2 2 −3 y −2 4 −1 1 2 10 x x 10 y 2 x 2 y 2 y 3 2 x 114. Cylindrical: 0.75 ≤ r ≤ 1.25 0≤z≤8 115. Spherical 4≤ ≤6 z 8 116. Cylindrical 1 2 ≤r≤3 ≤2 9 r2 0≤ 9 4 z r2 ≤ z ≤ −8 8 x −4 −8 x −4 4 y y R eview Exercises for Chapter 11 117. Cylindrical coordinates: r2 z 2 ≤ 9, ≤2 531 118. Spherical coordinates: ≥2 ≤3 3 z r ≤ 3 cos , 0 ≤ 2 0≤ ≤ 4 −2 −1 2 x 1 1 2 y 119. False. c represents a half-plane. 120. True. They both represent spheres of radius 2 centered at the origin. 122. True (except for the orgin). 121. False. r, , z 0, 0, 1 and r, , z 0, , 1 represent the same point x, y, z 0, 0, 1 . 123. z z sin , r sin y r 1 y 1 y 124. 2 sec 4 sphere ⇒ cos 2⇒z 2 plane The intersection of the plane and the sphere is a circle. The curve of intersection is the ellipse formed by the intersection of the plane z y and the cylinder r 1. Review Exercises for Chapter 11 1. P (a) u v (b) v (c) 2u 3. v v 1, 2 , Q \ 4, 1 , R 3, 4, 2 1 4i 25 3i 5, 4 j, 2j 2. P (a) u (b) v (c) 2u 10, 0 10i 8 sin 120 j 4. v 2, \ 1, Q 7, 0 42 v 14i 52 5, 1R 7i, v 41 PR 2, 4 \ PQ \ PQ 4, 5 4i 5j PR 42 22 6, 2 4i 5j 18i 5j 4, 2 v cos i v sin j 8 cos 120 i 4i 4 3j v cos i v sin j 1 cos 225 i 2 2 i 4 2 j 4 1 sin 225 j 2 5. z 0, y 4, x 5: 5, 4, 0 6. x z 0, y 7: 0, 7, 0 7. Looking down from the positive x-axis towards the yz-plane, the point is either in the first quadrant y > 0, z > 0 or in the third quadrant y < 0, z < 0 . The x-coordinate can be any number. 15 2 2 8. Looking towards the xy-plane from the positive z-axis. The point is either in the second quadrant x < 0, y > 0 or in the fourth quadrant x > 0, y < 0 . The z-coordinate can be any number. 9. x 3 2 y 2 2 z 6 2 10. Center: Radius: x 2 2 0 2 2 y 40 , 0 3 2 2 2 64 , 3 z 0 2 0 2 2 2 2, 3, 2 2 17 4 2 4 9 4 17 532 11. x2 x Chapter 11 4x 2 2 Vectors and the Geometry of Space y2 3 2 4 y 6y z 2 9 9 z2 z 4 3 2 4 4 9 12. x2 10x 25 y2 z2 6y 4z z 2 2 9 4 4 6 4 2 y 34 25 z 9 4 Center: 2, 3, 0 Radius: 3 6 x 5 4 3 x 45 6 y 5 2 y 3 2 Center: 5, Radius: 2 3, 2 2 4 6 8 x 13. v 4 z 2, 4 1, 7 3 2, 5, 10 14. v 3 6, z 8 7 6 5 4 3 2 1 1 2 1 3 2, 8 0 3, 5, 8 (2, − 1, 3) 3 2 1 12 3 5 (3, −3, 8) 5 x 4 3 −2 y v 3 y 6 5 4 (4, 4, − 7) x (6, 2, 0) 15. v w Since 5 1 3, 6 3, 3 4, 9 4, 6 1 1 4, 2, 10 2, 1, 5 16. v w 8 11 5, 5 4, 5 4, 3 7 7 3, 1, 4 2 5, 6 6, 10, 2w v, the points lie in a straight line. Since v and w are not parallel, the points do not lie in a straight line. 6, 3, 2 49 2, 8 6, 7 48 , 7 24 16 , 77 5, 5, 0 2i 3i 66 9 54 16, 12, 24 6j 2 6j 3k 3 36 2k, 17. Unit vector: u u 2, 3, 5 38 4, 4, 0 , R 1, 4, 0 3, 0, 6 1 3 36 45 1, 4, 5 40 i 3i 2 , 38 3 , 38 5 38 18. 8 3, 2 19. P (a) u v (b) u (c) v 21. u 5, 0, 0 , Q \ 2, 0, 6 4j, 6k 06 3 20. P (a) u v (b) u (c) v 22. u 1, 3 , Q \ 0, 5, 1 , R 2, 6, 2 3 PQ \ PQ \ PR v v 7, 9 PR v v 4, 3, 9 3, 6, 23 36 6,v 2, 3 , v v Since u 0, the vectors are orthogonal. 3 j 4 2 j 3 3 v 2 5 22 1 52 2 4 6 15 3 3 4 2 4 or, 2 3 12 or 15 6 52 2 i Since v 4u, the vectors are parallel. 23. u v u u cos 5 cos 2 cos v 5 3 i 4 2 i 3 sin sin i 3j j 24. u u v 4, 1, 5 , v 3, 2, 2 0 ⇒ is orthogonal to v. 2 52 1 2 uv uv arccos R eview Exercises for Chapter 11 25. u 10, 5, 15 , v 2, 1, 3 26. u v u u v cos 3 uv uv 83.9 27. There are many correct answers. For example: v ± 6, 5, 0 . \ \ 533 1, 0, 2, v 3 2, 1 1 u 5v ⇒ u is parallel to v and in the opposite direction. 10 1 3 10 28. W F PQ F PQ cos 75 8 cos 30 300 3 ft lb In Exercises 29–38, u < 2 14 2 3, 2, 1 , v > < 2, 4, 3,w > < 1, 2, 2 . 11 14 29 11 14 29 56.9 u u w 2 > 29. u u 33 14 2 u 2 11 30. cos uv uv arccos 31. projuw u 2, 1 5 14 5 14 5 3, 14 15 10 ,, 14 14 15 5 ,, 14 7 i 2 1 j 4 2 k 3 2 32. Work u w 3 4 2 5 33. n n n n v w 5 1 5 2i 2i j j , unit vector or 1 2i 5 j 34. u v i 3 2 i 2 3 v j 2 4 j 4 2 v k 1 3 k 3 1 u. 10i 11j 8k 35. V u 3, v w 2, 1, 0 4 4 2, 1 v u 10i 11j 8k Thus, u 36. u v w i 3 2 i 3 1 u j 2 4 3, 2, 1 k 1 3 1, 2, 1 i 3 1 j 2 2 k 1 1 4i 4j 4k u v 10i k 1 2 11j 8k u u w v j 2 2 w 4i 6i 4j 4k 7j u 4k v w 534 Chapter 11 Vectors and the Geometry of Space u v 10, 11, 285 1 v 2 1 2 5 (See Exercise 33) 2 z 37. Area parallelogram 8 102 112 8 2 (See Exercises 34, 36) 38. Area triangle w 2 2 1 2 39. F \ c cos 20 j 2k F \ sin 20 k PQ PQ \ PQ 200 c F F i 0 0 F j k 0 2 c cos 20 c sin 20 2c cos 20 2 ft 70° F 2c cos 20 i y PQ 100 cos 20 x 100 cos 20 j cos 20 100 1 tan2 20 2 0 0 sin 20 k 100 j tan 20 k 100 sec 20 1 2 1 0 1 2 106.4 lb 40. V u v w 25 10 41. v 9 3, 11 0, 6 2 3 3 6 6, 11, 4 6t, y y 11 z 4 11t, z 2 2 4t (a) Parametric equations: x (b) Symmetric equations: x 42. v 8 1, 10 4, 5 3 9, 6, 2 1 1 9 y 6 9t, y 4 4 z 2 44. Direction numbers: 1, 1, 1 6t, z 3 3 2t (a) Parametric equations: x (b) Symmetric equations: x 43. v j 1, y 2 t, z 3 (a) x (a) x (b) x 1 1 t, y y i 2 3 2 j 5 1 2 z t, z 3 k 1 4 3 t (b) None 45. 3x 3y 7z 4, x y 2z 3 46. u v 21 i 11j 13k Solving simultaneously, we have z 1. Substituting z 1 into the second equation we have y x 1. Substituting for x in this equation we obtain two points on the line of intersection, 0, 1, 1 , 1, 0, 1 . The direction vector of the line of intersection is v i j. (a) x (b) x t, y y 1, z 1 1 t, z 1 Direction numbers: 21, 11, 13 (a) x (b) x 21 21t, y y 11 1 1 z 13 11t, z 4 4 13t R eview Exercises for Chapter 11 47. P \ 535 3, 0, 8, \ 4, 2 , Q \ 3, 4, 1 , R 4, 5, j 8 5 4 27x k 1 4 32 z 4y 2 32z 4 27i 1, 1, 2 48. n 3x 3x 3i 2 y j k 1y 3 0 1z 1 0 PQ n 1 , PR \ PQ 27 x 3 PR i 0 4 4y z 8 4j 32k 0 33 5 2, 1 2, 3 1 3, 3, 2 be the 50. Let v direction vector for the line through the two points. Let n 2, 1, 1 be the normal vector to the plane. Then v n i 3 2 j 3 1 k 2 1 5, 7, 3 49. The two lines are parallel as they have the same direction numbers, 2, 1, 1. Therefore, a vector parallel to the plane is v 2i j k. A point on the first line is 1, 0, 1 and a point on the second line is 1, 1, 2 . The vector u 2i j 3k connecting these two points is also parallel to the plane. Therefore, a normal to the plane is v u i 2 2 2i j 1 1 4j k 1 3 2i 1 x 51. Q 1, 0, 2 point 2x 3y \ is the normal to the unknown plane. 5x 2j . 2y 2y 0 1 52. Q 3, 2, 4 point 5x 5 7y 7y 3z 1 27 0 3z 3 0 Equation of the plane: x 6z 6 P 5, 0, 0 point on plane n \ A point P on the plane is 3, 0, 0 . PQ n D 2, \ 2, 5, 1 normal to plane 2, \ 2, 0, 2 3, 6 normal to plane 8 7 PQ D 2, 4 10 30 30 3 PQ n n PQ n n 53. The normal vectors to the planes are the same, n 5, 3, 1 . 54. Q u P \ 5, 1, 3 point 1, 2, 1 direction vector 0, 0, 2 . Choose a Choose a point in the first plane, P point in the second plane, Q 0, 0, 3 . \ 1, 3, 5 point on line 6, u \ PQ \ 2, i 6 1 2 j 2 2 k 2 1 264 6 2, 8, 14 PQ D 0, 0, \ 5 5 35 5 35 35 7 PQ PQ n n D PQ u u 2 11 536 Chapter 11 Vectors and the Geometry of Space 1 z 2 55. x 2y 3z 6 56. y z2 57. y Plane Intercepts: 6, 0, 0 , 0, 3, 0 , 0, 0, 2 z 3 Since the x-coordinate is missing, we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a parabola in the yz-coordinate plane. z 2 Plane with rulings parallel to the x-axis z 2 (0, 0, 2) 3 6 x (0, 3, 0) (6, 0, 0) y 1 6 2 x 3 4 y x 2 y 58. y cos z 59. x2 16 y2 9 z2 1 60. 16x 2 Cone 16y 2 9z 2 0 Since the x-coordinate is missing, we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is y cos z. z Ellipsoid xy-trace: xz-trace: x2 16 x2 16 y2 9 2 y2 9 z2 z2 xy-trace: point 0,0, 0 1 xz-trace: z 1 yz-trace: z 1 z 4, x 2 z 4 ± 4x 3 4y 3 9 4 ± yz-trace: −2 x 2 2 y −4 y2 4 x 5 −2 y −3 x 3 −3 23 y 61. x2 16 y2 9 y2 9 x2 16 z z2 z2 1 1 −2 5 x 2 62. 5 y x2 25 y2 4 z2 100 z 1 12 Hyperboloid of one sheet x2 xy-trace: 25 xz-trace: yz-trace: x2 25 y2 4 y2 4 z2 100 z2 100 1 1 1 x 5 −5 y Hyperboloid of two sheets xy-trace: y2 9 x2 16 1 xz-trace: None y2 yz-trace: 9 63. x 2 z2 z 2 2 z2 1 4. Cylinder of radius 2 about y-axis 64. y 2 z2 z 16. Cylinder of radius 4 about x-axis 2 x y −2 2 x y −2 −2 R eview Exercises for Chapter 11 65. Let y r x the x-axis. 2 x and revolve the curve about 66. z 2 z x2 x2 67. 2 2, 2 2, 2 , rectangular (a) r (b) 22 22 2 537 2y revolved about y-axis ± 2y ry 2y 2 z2 z2 2y 22 22 2 4, 2 2 arctan 2 5, 1 3 ,z 4 3 , 4 2, arccos 4, 3 , 2 , cylindrical 4 2 arccos 1 , 5 2 5, 5 3 , arccos , spherical 4 5 2 2 25 68. 333 3 ,, , rectangular 442 (a) r 3 4 3 4 2 3 4 3 4 2 3 , 2 33 2 2 arctan 3 30 , 2 3 ,z 33 , 2 arccos 33 3 ,, , cylindrical 222 3 , 10 30 , , arccos 23 3 , spherical 10 2 2 (b) 3 , 69. 100, 6 , 50 , cylindrical 502 50 5 70. 81, 5 , 27 3 , cylindrical 6 6561 5 6 2187 54 3 1002 6 arccos 50 5, 50 50 5 arccos 1 5 63.4 or 1.107 , 1.1071 arccos 54 3, 27 3 54 3 arccos 1 2 3 6 , 63.4 , spherical or 50 5, 6 5 , , spherical 63 71. 25, r2 3 , , spherical 44 25 sin 3 4 2 72. 25 2 2 12, r2 2 , , spherical 23 12 sin 2 3 2 ⇒r ⇒r 63 4 z 25 cos 2 , 2 y2 , 25 cos 3 4 25 2 2 z 2 cos 6 3, , 12 cos 2 3 6 4 25 2 , cylindrical 2 2 6 , cylindrical 73. x2 2z r 2 sin2 cos2 2 (a) Cylindrical: r 2 cos2 (b) Spherical: 74. x 2 y2 z2 2 2z, r 2 cos 2 sin2 sin2 2z 2 cos , sin2 cos 2 2 cos 0, 2 sec 2 cos csc2 sin2 16 z2 16 (b) Spherical: 4 (a) Cylindrical: r 2 538 75. r r2 x2 x2 x2 Chapter 11 4 sin 4r sin y2 y2 y 4y 4y 2 2 Vectors and the Geometry of Space cylindrical coordinates 3 z 21 4 4 4 rectangular coordinates x 4 y 76. z z 4 4 cylindrical coordinates rectangular coordinates z 77. tan y x y 4 tan 1 x, x ≥ 0 z spherical coordinates 1 4 4 rectangular coordinates half-plane −2 2 x 2 y 4 x 3 2 1 3 −3 4 y 78. 2 cos Because cos x2 y2 z2 x r 2 x2 x x2 x y2 spherical coordinates y2 , we have rectangular equation Problem Solving for Chapter 11 1. b b a b a a a a b c b x b b b b c c c c 0 0 0 Then, sin A a b c c a abc ab abc sin C . c b b c sin A a b sin C The other case, sin A a sin B is similar. b 2. f x 0 t4 y 1 dt (b) f x f0 x 2 4 (a) 4 2 −4 −2 −2 −4 x4 1 4 tan 1 (c) ± 2 , 2 2 2 x: x t, y t. (d) The line is y u 1 i 2 j 2 2 , 2 2 P roblem Solving for Chapter 11 3. Label the figure as indicated. From the figure, you see that \ 539 Q SP \ 1 a 2 \ 1 b 2 \ \ RQ \ and \ SR 1 a 2 1 b 2 a \ P PQ . 1 2 R Since SP RQ and SR PQ , PSRQ is a parallelogram. a− 1b S 2 b 1 2 a+ 1b 2 4. Label the figure as indicated. \ S R PR \ a b b b a b a b 2 SQ a a a a 2 0, because P b Q b in a rhombus. \ \ 0, 1, 1 direction vector of line determined by 5. (a) u P1 and P2. \ 6. n PP0 n PP0 D P1Q u u 2, 0, 1, 1 2 2, 2 2 3 2 32 2 4 x 3 2 z 6 5 4 3 Figure is a square. → n and the points P form a circle of radius Thus, PP0 n in the plane with center at P0. n + PP0 n n − PP0 n 0, 1, 1 1 P1 23 4 P2 Q y P0 P (b) The shortest distance to the line segment is P1Q 1 2, 0, 1 z2 2 5. 1 0 7. (a) V 0 z dz 1 base altitude 2 y2 b2 x2 ca 2 2 1 2 1 1 2 r 8. (a) V 2 0 r2 x2 dx d > 0, 2 r 2x x3 3 r 0 43 r 3 Note: x2 (b) 2 a 1 2 c) (b) At height z z : (slice at z y2 cb 2 x2 a2 y2 b2 x2 a2 d2 c2 y2 b2 d2 1 1 1. d2 c2 c2 c2 d2 1 x2 a2 c2 c2 k 0 y2 d2 b2 c2 c2 a2 c2 d 2 c2 ab 2 c c2 c At z c, figure is ellipse of area ca k cb dc abc. abc2 2 abk2 2 Area V 0 abc 1 2 abk k b2 c2 d 2 c2 (c) V 1 area of base height 2 V 2 0 d2 d 2 dd d3 3 c 0 ab 2 c c2 2 ab 2 cd c2 4 abc 3 540 9. (a) Chapter 11 2 sin Torus Vectors and the Geometry of Space z (b) 2 −3 2 cos Sphere −3 −2 1 2 3 x z 1 −2 3 x 3 −2 y 2 3 y 10. (a) r 2 cos (b) z z2 r 2 cos 2 x2 y2 11. From Exercise 64, Section 11.4, u v w z u v zw u v w z. Cylinder Hyperbolic paraboloid 1 t 2 12. x (a) u P t 3, y 1, z 2t 1; Q 4, 3, s 2, 1, 4 direction vector for line 3, 1, \ 1 point on line 1 i 1 2 j 2s 1 7 s k 1 4 2 PQ \ 1, 2, s u \ PQ 7 si 6 2s j 5k D (b) PQ u u 10 6 21 2s 2 25 (c) Yes, there are slant asymptotes. Using s Ds 1 21 5 21 2.2361 at s ui j T x, we have x2 2x 1 22 5x2 x 1 10x 2 110 21 → ± 5 21 −11 −4 10 5 x 21 The minimum is D 13. (a) u u cos 0 i 1. (d) y 2.5 ± 105 s 21 1 slant asymptotes. sin 0 j Downward force w T T cos 90 T 0 u u 1 If w sin cos 30 , u ⇒T 2 3 sin i T T T i sin 90 j 0 0 u 60 cos j ui j T sin i cos j (e) Both are increasing functions. (f) → lim 2 T and → lim 2 u . 1 2 T and 1 1.1547 lb and tan ≤ 90 10 20 1.0642 0.3640 30 and T 32T u sec . 1 2 2 3 0.5774 lb (b) From part (a), u Domain: 0 ≤ (c) T u 0 1 0 40 1.3054 0.8391 50 1.5557 1.1918 60 2 1.7321 1.0154 0.1763 1.1547 0.5774 P roblem Solving for Chapter 11 14. (a) The tension T is the same in each tow line. 6000i T cos 20 2T cos 20 i ⇒T 6000 2 cos 20 0 90 0 541 (d) sin 20 j 10,000 cos 20 i T sin 20 3192.5 lbs 2T cos (b) As in part (a), 6000i ⇒T 3000 cos < 90 20 (e) As increases, there is less force applied in the direction of motion. Domain: 0 < (c) T 10 3046.3 30 3464.1 40 3916.2 50 4667.2 60 6000.0 3192.5 15. Let sin For u v u , the angle between u and v. Then uv uv vu . uv cos , sin , 0 , u sin sin cos cos cos cos v sin 1 and k. cos , sin , 0 and v i cos cos j sin sin v u k 0 0 Thus, sin sin . Rio de Janeiro: 4000, 43.23 , 112.90 43.23 43.23 16. (a) Los Angeles: 4000, (b) Los Angeles: x y z 118.24 , 55.95 118.24 118.24 4000 sin 55.95 cos 4000 sin 55.95 sin 4000 cos 55.95 1568.2, 1568.2 2684.7 Rio de Janeiro: x y z 4000 sin 112.90 cos 4000 sin 112.90 sin 4000 cos 112.90 2684.7, x, y, z (c) cos uv uv 2919.7, 2239.7 2919.7 2523.8 4000 4000 2239.7 x, y, z 1556.5 2523.8, 1556.5 0.02047 91.17 or 1.59 radians (d) s r 4000 1.59 6360 miles (e) For Boston and Honolulu: a. Boston: 4000, b. Boston: x y z 71.06 , 47.64 71.06 71.06 Honolulu: 4000, Honolulu: x y z 157.86 , 68.69 157.86 157.86 4000 sin 47.64 cos 4000 sin 47.64 sin 4000 cos 47.64 2795.7, 2695.1 959.4 4000 sin 68.69 cos 4000 sin 68.69 sin 4000 cos 68.69 3451.7, 959.4, c. cos uv uv 1404.4, 1453.7 0.28329 3451.7 2795.7 1404.4 4000 4000 2695.1 1453.7 73.54 or 1.28 radians d. s r 4000 1.28 5120 miles 542 Chapter 11 Vectors and the Geometry of Space 18. Assume one of a, b, c, is not zero, say a. Choose a point d1 a, 0, 0 . The distance in the first plane such as between this point and the second plane is w v . D a d1 a a d1 a 2 2 17. From Theorem 11.13 and Theorem 11.7 (6) we have \ D PQ n n w u u v v u u v v w u u v b0 b c 2 2 c0 c a 2 d2 d2 b 2 d2 b 2 d1 2 c2 . 19. x 2 z y2 2y 1 cylinder plane 2y. 3 2 z Introduce a coordinate system in the plane z The new u-axis is the original x-axis. The new v-axis is the line z 2y, x 0. (0, 1, 2) y 2 −2 2 x Then the intersection of the cylinder and plane satisfies the equation of an ellipse: x2 x2 x2 y2 z 2 z2 4 2 (0, − 1, − 2) 1 1 1 ellipse 20. Essay. ...
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This note was uploaded on 11/13/2010 for the course MATH MAT 231 taught by Professor Thurber during the Spring '08 term at Thomas Edison State.

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