EVEN10 - CHAPTER 10 Vectors and the Geometry of Space...

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Unformatted text preview: CHAPTER 10 Vectors and the Geometry of Space Section 10.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . 474 Section 10.2 Space Coordinates and Vectors in Space . . . . . . . . . . 479 Section 10.3 The Dot Product of Two Vectors . . . . . . . . . . . . . . 483 Section 10.4 The Cross Product of Two Vectors in Space . . . . . . . . 487 Section 10.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . 491 Section 10.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . 496 Section 10.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . 499 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 CHAPTER 10 Vectors and the Geometry of Space Section 10.1 Vectors in the Plane Solutions to Even-Numbered Exercises 2. (a) v (b) −3 −2 −1 −1 −2 −3 −4 −5 −6 3 3, y 2 4 0, 6 4. (a) v (b) 1 2, 3 y 1 3, 2 x 1 2 3 3 (− 3, 2) v v 2 1 (0, − 6) −3 −2 −1 x 6. u v u 1 7 v 3 4 ,8 2, 7 0 1 5, 8 5, 8 8. u v u 11 25 v 5 4, 0, 10 4 13 1 15, 3 15, 3 10. (b) v 2, 6 y 12 10 8 6 4 2 −1 −4 −6 6 (1, 12) v 1, 12 12. (b) v (a) and (c). 0, 1 4 y 5, 3 (a) and (c). (− 5, 3) 4 (3, 6) −6 −4 v 2 x 2 −2 −2 x 1 2 3 4 5 6 7 (− 5, −1) (0, − 4) (2, − 6) 14. (b) v (a) and (c). 3 7, 1 y 3 2 1 10, 0 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). (0.12, 0.60) (0.84, 1.25) (−10, 0) 1 v 2468 −8 −6 −4 −2 x 0.50 0.25 (0.72, 0.65) v x 0.25 0.50 0.75 1.00 1.25 (−3, −1) −2 −3 (7, −1) 18. (a) 4v 4, 20 y (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( 2 474 —CONTINUED— S ection 10.1 18. —CONTINUED— (c) 0v 0, 0 y 6 Vectors in the Plane 475 (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) 20. Twice as long as given vector u. y 22. y u + 2v u 2v 2u u x x 2 24. (a) 3u 2 3 3, 8 2, 3, 3, 8 16 3 26. v 11, 33 34, 109 2 y 2i 3i j j i 3, 1 2j (b) v (c) 2u u 5v 8, 25 2 8 5 8, 25 w 1 v x 1 2 3 u −1 28. v 5u y 2 3w 5 2, 1 3 1, 2 7, 11 30. u1 u2 3 2 u1 u2 Q 4 9 7 7 7, 7 − 4 −2 x 4 6 8 10 −3w −6 −8 −10 −12 5u v 32. v 144 25 13 34. v 100 9 109 36. v 1 1 2 38. u v u u 52 152 5, 15 5 10 250 1 , 10 5 10 3 unit vector 10 40. u v u u 6.2 2 3.4 2 50 52 6.2, 3.4 52 1.24 0.68 , unit vector 2 2 476 42. u Chapter 10 0, 1 , v 0 9 v v u u u u (e) v v (f) u u u u v v v v 3, 1 9 3, 9 0, 1 1 1 32 1 v v Vectors and the Geometry of Space 3 1 32 2 4 13 (d) u u 3, 3 (e) v v 1 3, 13 1 2 (f) u u u u 4 2.236 u u v u u v v 1 2 22 v v 44. u 2, 4 ,v 4 25 v v u u 5, 5 16 25 7, 1 49 1 2, 25 1 1 5, 5 52 1 v v 1 7, 1 52 1 1 4 52 25 52 (a) u (b) v (c) u u (d) (a) u (b) v (c) u u 46. u u v v u u u v v 2 3, 2 13 1, 5 2, 0 2 3.606 48. 1, 1 1, 1 2 2, 2 2 v≤u u u 1 0, 3 3 0, 3 v 0, 3 v 50. 3 52. v 5 cos 120 i 5 i 2 53 j 2 sin 120 j u u 54. v cos 3.5 i 0.9981i sin 3.5 j 0.0610j 0.9981, 0.0610 56. u v u v 4i i 5i 3j 3j 58. u 5 cos 0.5 i 5 sin 0.5 j 60. See page 718: (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 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 u S ection 10.1 62. See Theorem 10.1, page 719. For Exercises 64–68, au bw ai 0, 2a 1, b 2j b 1. 1, 2a 2, b b 3. 72. f x fx (a) m 1 , then w w (b) m w w tan x sec2 x 2 at x 4 7. bi j a bi 2a b j. Vectors in the Plane 477 64. v 3j. Therefore, a b simultaneously, we have a 3. Solving 66. v 3i 3j. Therefore, a simultaneously, we have a b 3, 2a 2, b 1. b 3. Solving 68. v i 7j. Therefore, a b Solving simultaneously, we have a 70. y x3, y 3x2 12. Let w ± 1 12 . 12 at x 2. (a) m w w (b) m w w 1, 12 , then . 1 1, 12 . 145 Let w 1 12, 145 12, 1. 2. Let w ± 1 2. 1, 2 , then 1 1, 2 . 5 Let w 1 5 63.5 8.26 2, 1 . 2, 1 , then ± ± 74. u u v v 2 3i 3i u 2, 4, 3, F1 F1 F2 2j 3 3j 76. magnitude direction 3 2 3i 33 2j v u 10 78. F1 F2 F3 R R F1 F2 F3 140 200 F2 F3 F3 163.0 4.09 80. 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 F3 400 cos 200 3 R R 82. F1 F2 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 sin 135 j 175 2 i 2 200 3 arctan 35 2 200 385.2483 newtons 200 315 2 200 3 35 2 0.6908 478 84. F1 Chapter 10 20, 0 , F2 F2 Vectors and the Geometry of Space 10 cos , sin 20 400 500 10 cos , 10 sin 400 cos 400 cos (c) The range is 10 ≤ F1 F2 ≤ 30. 0 and 2. 100 cos2 100 sin2 (a) F1 (b) 40 The maximum is 30, which occur at The minimum is 10 at 0 0 . 2 (d) The minimum of the resultant is 10. 86. u 1 u 3 P1 P2 7 2, 1 1, 2 1, 2 1, 5 2 6, 3 2, 1 2 2, 1 3, 3 5, 4 y 88. 1 2 24 20 24 arctan 10 arctan u cos v cos 1 2 0.8761 or 50.2 A θ2 v C u B 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. 90. 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 S ection 10.2 92. 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 Space Coordinates and Vectors in Space 479 35.36 ⇒ 364.64 Direction North of East: Speed: 94. u v 336.35 mph cos2 sin 2 sin2 cos 2 1, 1 96. 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 s u r v s v y yv u y x yu x y v. y 1 2. 1 and x 0. Solving we have x 98. The set is a circle of radius 5, centered at the origin. u 100. True x, y x2 y2 5 ⇒ x2 y2 25 104. True 0 102. False a b Section 10.2 2. 8 z Space Coordinates and Vectors in Space 4. 8 z (3, −2, 5) (4, 0, 5) x y x y ( 3 , 4, −2( 2 6. A 2, B 3, 3, 1, 4 1 8. x 7, 7, y 2, 1 2, z 1: (0, 4, − 5) 10. x 0, y 3, z 2: 0, 3, 2 12. The x-coordinate is 0. 16. The point is on the plane z 3. 14. The point is 2 units in front of the xz-plane. 18. The point is behind the yz-plane. 480 Chapter 10 Vectors and the Geometry of Space 4. 22. The point x, y, z is 4 units above the xy-plane, and above either quadrant II or IV. 20. The point is in front of the plane x 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 2 5 96 3 2 2 2 2 28. d 4 4 2 49 2 5 9 2 62 2 6 3 2 16 46 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 32. A 5, 0, 0 , B 0, 2, 0 , C 0, 0, AB AC BC Neither 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. 34. The y-coordinate is changed by 3 units: 5, 6, 4 , 7, 4, 3 , 3, 8, 3 38. Center: 4, Radius: 5 x x 2 2 36. , 6, 4, 7 1, 1 2 2 40. Center: r 3 1 z 2z 1 25 0 3, 2, 4 4 y2 y 8x (tangent to yz-plane) x 10z 10z 2 z2 2y 7 3 2 y 2 2 z 4 2 9 42. x2 9x 81 4 x2 y2 y2 x z2 2y 9 2 2 9x 1 y 2y z2 1 19 25 5 2 0 19 109 4 81 4 1 25 z Center: Radius: 9 , 1, 2 109 2 4x2 5 44. 4y2 x2 y2 y2 x 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 x2 x 1 4 Center: 1 , 4, 2 1 Radius: 3 S ection 10.2 46. x2 4x 4 y2 x 6y 2 2 Space Coordinates and Vectors in Space 13 481 x2 9 y z2 3 2 y2 8z z z2 < 4x 16 < 4 4 2 6y 9 8z 16 13 < 16 Interior of sphere of radius 4 centered at 2, 48. (a) v 4 4i (b) 8 3, 4 . 50. (a) v 2 4k (b) 4 0i 5j z 0 2k 5j 4, 3 5, 2 1k 2i 0, 0, 4 z 3 3j 4 0k 〈 0, 0, 4 〉 〈4, − 5, 2〉 x y x y 52. 1 4, 7 5 5, 3 25 2 144 25 5, 12, 5 54. 2 1, 4 6 2, 1 1 , 73 2 36 6 , 73 4 36 6 73 1, 6, 73 6 5, 12, Unit vector: 194 12 , 194 2k 12 1, 6, 5 194 z 5, 12, 5 194 4 2i 4j 3 9k 1j 5 , 194 7 Unit vector: 56. (b) v 6i (a) and (c). 6, 4, 9 (− 4, 3, 7) (− 6, 4, 9) x y (2, −1, − 2) 58. q1, q2, q3 Q 60. (a) 1, v 8 3, 0, 2, 5 2 3 2, 2, z 4 1, 21 3, 2 1 (b) 2v 4, 4, 2 z 8 〈4, − 4, 2〉 〈− 2, 2, −1〉 x y x y (c) 1 2 v 1, 1, 1 2 z 2 (d) 5 2v 5, 5, 5 2 z 8 〈1, −1, 1 〈 2 x y 〈5, −5, 〈 5 2 x y 482 62. z 64. z 66. 2u Chapter 10 u 5u v v 3v w 2w 1 2 Vectors and the Geometry of Space 1, 2, 3 w 5, 10, 15 2 1, 2, 3 0, 0, 0 2, 2, 1 6, 6, 2, 2, 1 3 8, 0, 8 2, 0, 4, 0, 4 2 7, 0, 4 3, 4, 20 3 z1, z2, z3 0, 0, 0 3z 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 68. (b) and (d) are parallel since 70. z 7, 8, 3 zz i 4 3 j 3 2 k 2 1i 2 2 3 j 3 4k and 3 i 4 j 9 8 k 31 22 i 2 3 j 3 4 k. 72. P 4, \ 2, 7 , Q 6, 2, 3, 1, 2 1, 2 1 2 2, 0, 3 , R 7, 4 3, 9 (b) is parallel since 14, 16, 6. PQ \ PR 3, 6, 2, 4 → → Therefore, PQ and PR are parallel. The points are collinear. 74. P 0, 0, 0 , Q 1, 3, \ 2 , R 2, 6, 4 76. A 1, 1, 3 , B 9, \ 1, 5 5 7 7 \ 2 , C 11, 2, 9 , D 3, 4, 4 PQ \ 1, 3, 2, \ 2 6, 4 \ AB \ 8, 8, 2, 3, 2, 3, \ 2, 2, PR DC \ Since PQ and PR are not parallel, the points are not collinear. AD \ BC \ \ Since AB DC and AD BC , the given points form the vertices of a parallelogram. 78. v 1 0 9 10 80. v v 84. u u (a) (b) u u u u v v 6, 0, 8 36 0 64 10 (a) (b) 4, 3, 7 16 9 49 86. 74 u u u u u u 8, 0, 0 8 1, 0, 0 1, 0, 0 82. v v 1, 3, 1 2 9 4 14 1 6, 0, 8 10 1 6, 0, 8 10 4, 7.5, 8.732 2 90. 88. (a) u (b) u (c) u (d) v cu cu 14c 2 c c, 2c, 3c c2 9 ± 4c 2 9c 2 3 5.099 9.014 3 14 14 S ection 10.3 u u 1 1 1 , , 3 3 3 3 3 3 , , 3 3 3 u u The Dot Product of Two Vectors 2 , 14 7 3 , 14 1 14 483 92. v 3 3 94. v 5 5 70 3 70 70 , , 14 14 96. v v 5 cos 45 i 5 cos 135 i z 8 sin 45 k sin 135 k 52 i 2 52 2 k or i k 98. v 2 3v 5, 6, 10 3, 13 3, 3 2 4, 6, 3 1, 2, 5 10 3, 4, 2 52 (i + k) 2 x y 100. x0 is directed distance to yz-plane. y0 is directed distance to xz-plane. z0 is directed distance to xy-plane. 104. A sphere of radius 4 centered at x1, y1, z1 . v x x x x1 550 302,500 c2 c F 2 102. x x0 2 y y0 2 z z0 2 r2 x2, y x1 y 2 y1, z y y1 2 z1 y1 z 2 106. As in Exercise 105(c), x a will be a vertical asymptote. Hence, lim T . r0 → a z z1 2 z1 2 4 16 sphere 110. 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, 0, 6 and 10, 6 . → 10, 6 , AC 10, → 10 2, AC 2 59 → AB 420 → , F2 AB F2 237.6, 423.1, 185.5, F 860.0 lb 108. c 75i 18,125c 2 50j 100k 16.689655 4.085 4.085 75i 306i 204j 50j 100k 409k Thus, F1 F F1 → AC 650 → AC 297.0, 178.2 423.1, 253.9 720.1, 432.1 Section 10.3 2. u (a) u (b) u (c) u (d) u (e) u 2 The Dot Product of Two Vectors 2, 3 2 10 3 10 10 22 116 4. u i, v v u 2 4, 10 , v v u 4 44 116 vv 2v 22 2u i 1 1 1 vv 2v i 2u v 2 (a) u (b) u (c) u 2, 3 v 2 22 44, 66 44 (d) u (e) u 484 6. u Chapter 10 2i v u 2 Vectors and the Geometry of Space i 1 11 3 2 3j 2k 22 2 9 5 8. u v 3240, 1450, 2235 2.22, 1.85, 3.25 j 2k, v 21 22 9 (a) u (b) u (c) u (d) u (e) u uv uv u v Increase prices by 4%: 1.04 2.22, 1.85, 3.25 . New total amount: 1.04 u v 1.04 17,139.05 $17,824.61 vv 2v 5i 2u v 3j 2 2k 5 5i 10 15j 10k 10. cos 40 25 cos 5 6 500 3 12. u cos 3, 1 , v uv uv 4 2, 1 5 10 5 1 2 14. u v cos cos cos 6 i sin sin 6 j 3 i 2 1 j 2 2 i 2 2 j 2 16. u cos 3i 2j uv uv 2 k, v 32 2i 3j 23 uv 0 0 3 i 4 3 j 4 uv uv 3 2 arccos 2 2 2 1 4 1 2 3 2 2 2 1 4 105 3 18. u cos 2i 3j uv uv k, v i 2j k 20. u u u 2, 18 , v 3 , 2 1 6 c v ⇒ not parallel v 0 ⇒ orthogonal 9 14 6 arccos 9 2 21 3 21 14 10.9 3 21 14 2i 22. u u 1 3 1 6 i 2j , v 4j 24. u u u v 2i 3j k, v 2i j k v ⇒ parallel c v ⇒ not parallel 0 ⇒ orthogonal 26. u v u u v cos , sin , sin , 1, 28. u cos cos cos cos2 5, 3, 1 5 35 3 35 1 35 u 35 cos , 0 c v ⇒ not parallel 0 ⇒ orthogonal cos2 cos2 25 35 9 35 1 35 1 S ection 10.3 30. u cos cos cos cos2 a, b, c , u a2 a2 a2 cos2 a b2 b b2 c b2 a2 c2 c2 c2 cos2 a2 50 ⇒ ⇒ 1 ⇒ 2 a2 b2 52 2.1721 or 124.4 1.1326 or 64.9 or 45 c2 a2 b2 b2 c2 a2 c2 b2 c2 1 b2 c2 The Dot Product of Two Vectors 485 32. u cos cos cos 4, 3, 5 4 52 3 52 5 52 300 F1 100 F2 F2 u 34. u cos cos cos 2, 6, 1 u 41 1.8885 or 108.2 0.3567 or 20.4 1.4140 or 81.0 2 ⇒ 41 6 ⇒ 41 1 ⇒ 41 4 36. F1: C1 F2: C2 F F1 13.0931 6.3246 38. v1 v1 v2 v2 s, s, s s3 s, s, 0 s2 s2 s3 arccos 6 3 6 3 35.26 z v1 (s, s, s) y 13.0931 20, 10, 5 6.3246 5, 15, 0 v2 x cos (s, s, 0) 230.239, F cos cos cos 242.067 lb 36.062, 65.4655 230.239 ⇒ F 36.062 ⇒ F 65.4655 ⇒ F C1 0, 10, 10 . F1 162.02 98.57 74.31 40. F1 200 C1 10 2 ⇒ C1 10 2 42. w2 u w1 9, 7 3, 9 6, 2 and F1 F2 F2 F F C2 C3 4, 0, 100 2, 100 2 4, 6, 10 6, 10 0, 0, w F1 4C2 F2 F3 4C3 6C2 0 0 ⇒ C2 6C3 C3 C3 25 2 N 3 2, 1, 3 100 2 0 ⇒ C2 44. w2 u w1 8, 2, 0 6, 3, 3 486 46. u Chapter 10 2, 3 ,v u v u v 2 Vectors and the Geometry of Space 3, 2 v 0v 2, 3 0, 0 48. 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 22 33 , 0, 13 13 (a) w1 (b) w2 (a) w1 (b) w2 w1 w1 50. The vectors u and v are orthogonal if u The angle cos between u and v is given by uv . uv v 0. 52. (a) and (b) are defined. 54. See figure 10.29, page 739. 56. Yes, u u v v 2 v v v2 1 v u v 1 u v v u u u 2 u u u2 v 58. (a) u (b) u 5, v 8.602, 5.745, 91.33 90 60. (a) (b) 64 16 , 17 17 21 63 42 ,, 26 26 13 9.165, v 62. 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 64. u v \ 8i 3i 3j. Want u 8j and v v 3i 0. 8j are orthogonal to u. 66. u v 0, 3, 6 . Want u v v 0, 0. 6, 3 are orthogonal to u. 0, 6, 3 and 68. OA 10, 5, 20 , v \ 0, 0, 1 0, 0, 20 70. F v W 25 cos 20 i 50 i F v sin 20 j projvOA \ 20 0, 0, 1 12 20 4, 2, 10 1250 cos 20 1174.6 ft lb projvOA \ 72. PQ \ 74. True w u v w 0 and u u 0 w v V W 2, 3, 6 \ \ PQ V 74 0⇒w v are orthogonal. Section 10.4 \ The Cross Product of Two Vectors in Space kkk ,, 222 kkk ,, 222 k2 4 k 2 2 487 76. (a) (k, 0, k) k z (d) r1 (0, k, k) \ k, k, 0 0, 0, 0 kk ,, 22 k , 2 k 2 k , 2 k 2 r2 k x k y cos (k, k, 0) 3 1 3 (b) Length of each edge: k2 (c) cos k2 02 k2 1 2 60 109.5 k2 k 2k 2 arccos 1 2 78. The curves y1 x2 and y2 x1 3 intersect at 0, 0 and at 1, 1 . 2 y y1 At 0, 0 : 1, 0 is tangent to y1 and 0, 1 is tangent to y2. The angle between these vectors is 90 . At 1, 1 : 1 5 1, 2 is tangent to y1 and 3 10 1, 1 3 1 is tangent to y2. To find the angle between these vectors, cos 1 5 1 3 10 2 1 ⇒ 2 45 . 10 3, 1 y2 1 (1, 1) x 1 2 (0, 0) 80. u u v v u v cos u v cos u v cos ≤ u v since cos ≤ 1. 82. 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 w2. w2 u Taking the dot product of both sides with v produces u w2 v cv v w2 v c v 2, since w2 and v are orthogonol. Thus, u v cv 2 θv w1 ⇒c u v v 2 and w1 projv u cv u v v 2 v. Section 10.4 ijk 100 010 z The Cross Product of Two Vectors in Space ijk 001 010 z 2. i j k 4. k j i 6. k i ijk 001 100 z j 1 1 1 k j 1 x k −i j k j 1 y x i −1 1 y x 1 −1 1 i −1 1 y 488 Chapter 10 i 3 2 u 0 j 0 3 Vectors and the Geometry of Space k 5 2 v 15, i 3 1 u 0 7, 0, 0 j 0 0 10 0 0⇒u 0 1 v u v 70 0⇒v u 0 42 u v k 6 0 42j 0 42 v 00 0, 42, 0 60 j 2 5 v k 2 1 8. (a) u (b) v (c) v 12. u u u v u v u v 15, 16, 9 16, 9 10. (a) u (b) v (c) v v u v 8, 8, 5, 5, 17 17 1, 1, 2 , v i 1 0 v j 1 1 1 0, 1, 0 k 2 0 2 u 2i 10 v k 2 2, 0, 1 1 14. u u u v u 10, 0, 6 , v i 10 7 v 0⇒u v u v 0 2 0⇒v 10 u v 16. u u v 18. v u u v v z 6 5 4 3 2 1 1 4 3 2 ijk 160 211 16 26 6 6i 1 1 1 j 13k 0⇒u 1 13 u v u v 20. 6 5 4 3 2 1 1 2 z 0⇒v v v u 4 6 y x 4 3 u 4 6 y x 22. u v u u u v v v 8, 10, 6, 4 12, 2 24. u v u u u v v v i j v u j k 2 k 3 1 i 2 6k 60, 24, 156 1 60, 24, 156 36 22 5 2 13 , , 3 22 3 22 3 22 1 0, , 0 3 0, 1, 0 26. (a) u u (b) u u v v v v 18, 52.650 12, 48 28. u v k 50, 40, 72.498 34 u A ijk 111 011 v j k j k 2 S ection 10.4 The Cross Product of Two Vectors in Space 489 30. u v 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 3, 1 , B 6, 5, \ 1 , C 3, 1, \ AB 4, 8, \ 2 , AC \ 3, 3 , CD \ 4, 8, 2 , BD 1, 3, 3 Since AB \ \ CD and AC BD , the figure is a parallelogram. AB and AC are adjacent sides and \ \ AB Area 34. A 2, \ AC \ i 4 1 \ j 8 3 k 2 3 920 18, 2 230 14, 20 . AB AC 3, 4 , B 0, 1, 2 , C \ 1, 2, 0 3, 5, k 2 4 44 6i 4 2j 2k 36. A 1, 2, 0 , B \ 2, 1, 0 , C 0, 0, 0 \ AB \ \ 2, 4, AC 1 AB 2 2 , AC i 2 3 j 4 5 1 2 AB \ \ 3, AC 1 AB 2 z 1, 0 , AC i 3 1 j 1 2 5 2 k 0 0 1, 5k 2, 0 AB AB \ \ \ \ A 38. F \ AC 11 1000 3 j 1000k A AC 2000 cos 30 j 0.16 k F \ sin 30 k PQ \ PQ PQ PQ i 0 0 j 0 1000 3 lb k 0.16 1000 0.16 ft 60° F 160 3 i y F 160 3 ft x 40. (a) B is 15 12 \ 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 −300 490 Chapter 10 1 2 0 1 0 4 w Vectors and the Geometry of Space 1 1 0 1 0 1 3 6 0 72 50. See Theorem 10.8, page 746. 1 6 4 2 1 0 0 1 2 0 1 2 42. u v w 1 44. u v w 0 46. u V 48. u v w u V v u w v 72 1, 1, 0 1, 0, 2 0, 1, 1 v u w v 1 1 0 w 1 0 1 3 0 2 1 3 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 1, 0, 0 . z1 54. False, let u Then, u 56. u u v u v u 1, 0, 0 , v 1, 0, 0 , w w 0, but v w. w1, w2, w3 u1, u2, u3 , v u1i w v u2 j v2w3 w u3 k v1, v2, v3 , w 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 58. u cu u1, u2, u3 , v v i j cu1 cu2 v1 v2 cu2v3 c u2v3 v1, v2, v3 , c is a scalar. k cu3 v3 cu3v2 i u3v2 i u3 v3 w3 v w1 w2 u1 u2 v1 v2 v1u 3 w1v3 w3 u3 v3 w3 u 1v2 u 3 v1w2 v1u 2 w1v2 cu1v3 u1v3 cu3v1 j u3v1 j cu1v2 u1v2 cu2v1 k cu v u2v1 k 60. u v w u1 u2 v1 v2 w1 w2 w v2 u 3 w2v3 w u u v w w1 u 2v3 u 1 v2w3 u v w2 u 1v3 u 2 v1w3 S ection 10.5 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, v 0. Thus, sin 0, c v for some scalar c. Lines and Planes in Space 491 0, then u v sin c v for some scalar c. 0. Assume u 0, and u and v are parallel. Therefore, 64. u a1, b1, c1 , v v w a2, b2, c2 , w k c2 c3 b2c3 j b1 b3c2 a3b2 a2c3 b1b3 b1b3 b1b3 a3c2 a3, b3, c3 b3c2 i a2c3 k c1 a2c3 a2b3 a2c3 i b3c2 k b1b2 b1b2 b1b2 c1c2 i c1c2 j c1c2 k c1c2 a3, b3, c3 a3b2 a1 a2b3 a3b2 c1 b2c3 b3c2 j a3c2 j a2b3 a3b2 k ij a2 b2 a3 b3 i a1 b2c3 b1 a2b3 a1 a3c2 a2 a1a3 b2 a1a3 c2 a1a3 a1a3 u u u v v w w c1 a3c2 b1 b2c3 c1c3 c1c3 c1c3 a3 a1a2 b3 a1a2 c3 a1a2 b1b3 u c1c3 a2, b2, c2 vw a1a2 b1b2 wv Section 10.5 2. x (a) 2 3t, y z Lines and Planes in Space 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 4. Point: 0, 0, 0 Direction vector: v Direction numbers: (a) Parametric: x x (b) Symmetric: 4 5 2, , 1 2 4, 5, 2 4t, y y 5 z 2 5t, z 2t 6. Point: 3, 0, 2 0, 6, 3 Direction vector: v Direction numbers: 0, 2, 1 (a) Parametric: x (b) Symmetric: y 2 z 3, y 2, x 2t, z 3 2 t 492 Chapter 10 3, 5, 4 Vectors and the Geometry of Space 10. Points: 2, 0, 2 , 1, 4, 2, 1 3 3 3 y 3t, y 5 2 5 z 2t, z 4 4 t 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 8. Point: Directions numbers: 3, (a) Parametric: x (b) Symmetric: x 12. Points: 0, 0, 25 , 10, 10, 0 Direction vector: 10, 10, Direction numbers: 2, 2, (a) Parametric: x (b) Symmetric: x 2 2t, y y 2 z 25 5 2t, z 25 5 25 5t 14. Point: 2, 3, 4 Direction vector: v Parametric: x 2 3i 3t, y 2j 1 3 2t, z 4 t k Direction numbers: 3, 2, 16. Points: 2, 0, 3 , 4, 2, 2i 2t, y y 2 1 2 2 2j k 3 t 18. L1: v L 2: v L 3: v L 4: v 4, 2, 3 8, 5, 9 on line Direction vector: v Parametric: x Symmetric: x 2 2 2 2, 1, 5 8, 4, 6 8, 5, 9 on line Direction numbers: 2, 2, 1 2t, z z 1 1 3 2, 1, 1.5 L1 and L 2 are identical. (a) Not on line 1 (b) On line (c) Not on line 3 2 3 2 1 20. 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 22. 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 4s. Thus, t 1, s 1 and the 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) (Second line) 4 46 21 x 5s 3s 2s 12 11 4 3 x 2 2 4 966 2 966 483 24. x y z 2t 4t t 1 10 y z x = 2t − 1 y = − 4t + 10 z=t 3 2 z x = − 5s − 12 y = 3s + 11 z = − 2s − 4 −2 −3 Point of intersection: 3, 2, 2 (3, 2, 2) 3 y S ection 10.5 26. 2x P 3y \ Lines and Planes in Space 493 4z 4 2, 0, 0 , R \ 28. Point: 1, 0, 3, 2, 3, 2, k 1 3 3 2, 3, 4 2 n 0x z 3 k 1 0 3 0 1z 3 0 0, 0, 1 , Q 2, 0, \ 0, 0, 1 0y (a) PQ \ 1 , PR i 2 3 j 0 2 (b) PQ PR 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. 30. Point: (0, 0, 0 Normal vector: n 3x 3x 0 2z 0y 0 2 to 3, 4, 2 : 1, 1, 4 . 2 to 1, i 1 1 1z 6x 2y j 1 4 2 z k 4 2 0 8 40. 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 : k 3k v 2i 2j k v 0 i 2 2 j 1 2 k 1 1 i 2k 1, 0 : 18, 3 6x 2 2y 3 1, 6, 4, 2 . 3 0 3i 2k 2z 0 0 32. Point: 3, 2, 2 Normal vector: v 4x 4x 3 y 3z y 8 i, 1 x 1 0, x 1 2 4i 3z j 3k 2 0 34. Let u be vector from 2, 3, Let v be vector from 2, 3, Normal vector: u v 36. 1, 2, 3 , Normal vector: v 6, 2, 1 38. 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 Normal vector: u x x 2 2z 2z 0 1 42. 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 5: 6i 7j 2k 44. Let u k and let v be the vector from 4, 2, 1 to 7i 3j 6k 3, 5, 7 : v Since u and v both lie in the plane P, the normal vector to P is: u v 4 7y 7y 26 i 0 7 2 j 0 3 k 1 6 0 3i 7j 3i 7j 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 3 18y 18 y 3z 2 27 3z 1 36j 18j 0 6k 3x 3k 3x 494 Chapter 10 Vectors and the Geometry of Space 3, 1, 4 , n2 9, 3, 12 . Since n2 3n1, the planes are parallel, but not equal 46. The normal vectors to the planes are n1 48. 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, 50. The normal vectors to the planes are n1 cos Thus, 2, 0, 1 , n2 0. 4, 1, 8 , 14 21 65.9 . 1 . 6 n1 n2 n1 n2 2 and the planes are orthogonal. 52. 3x 6y z 3 2z 6 54. 2x y z z 4 4 56. x 2y 4 z 4 2 −4 3 x x 2 2 3 y 3 x 1 1 y 4 3 y 58. z 8 z 8 60. x 3z 3 z 2 1 62. 2.1x z 4.7y z 3 0 3 2 3 x 2 1 1 2 y 1 1 Generated by Mathematica 5 x 5 y x 2 y Generated by Mathematica 64. P1: n P2: n P3: n P4: n 6, 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. 66. If c If c 0, z 0, cy 0 is xy-plane. z 0⇒y 68. 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 16, 31, 3 . 6, 3, 1 . 1 z is a plane parallel to c x-axis and passing through the points 0, 0, 0 and 0, 1, c . 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 2⇒x 16t, y S ection 10.5 Lines and Planes in Space 495 70. 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 72. 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 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. 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. 74. Point: Q 0, 0, 0 Plane: 8x 4y z 8 8, 4, 1 76. Point: Q 3, 2, 1 Plane: x y 2z 4 1, 1, 2 Normal to plane: n Normal to plane: n Point in plane: P 1, 0, 0 \ Point in plane: P 4, 0, 0 \ Vector: PQ \ 1, 0, 0 8 81 8 9 Vector: PQ \ 1, 2, 1 1 6 1 6 6 6 D PQ n n D PQ n n 4, 4, 9 and 78. 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 \ 2, 0, 4 and 80. The normal vectors to the planes are n1 n2 2, 0, 4 . Since n1 n2, the planes are parallel. Choose a point in each plane. P 2, 0, 0 is a point in 2x is a point in 2x 4z 10. \ \ 5,0, 3 is a point in 4x 0, 0, 2 is a point in 4x 5, 0, \ 4y 4y 9z 9z 7. 18. 4z 4. Q 6 20 5, 0, 0 35 5 PQ D 1 11 113 11 113 113 PQ 3, 0, 0 , D PQ n1 n1 PQ n1 n1 82. 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 \ 0. 84. 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. PQ \ PQ ijk 112 212 5 9 You need n and P to find the equation. 0, 2, 5 3 1 D PQ u u 86. 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 88. (a) t v represents a line parallel to v. (b) u t v represents a line through the terminal point of u parallel to v. (c) su t v represent the plane containing u and v. 496 Chapter 10 Vectors and the Geometry of Space 1, 1, 8 . 90. 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 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 1, 1, 8 . (− 1, − 1, 8) 6 4 2 z cos n1 n2 n1 n2 arccos 1 65 6 x (0, 0, 0) 4 (6, 0, 0) (0, 6, 0) 6 y 92. False. They may be skew lines. (See Section Project) Section 10.6 Surfaces in Space 4. Elliptic cone Matches graph (b) 10. x2 z2 25 6. Hyperbolic paraboloid Matches graph (a) 2. Hyperboloid of two sheets Matches graph (e) 8. x 4 4 z Plane parallel to the yz-coordinate plane 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 2 4 y 6 4 4 x 8 x 8 y 12. z 4 y2 14. y 2 y2 4 z2 z2 4 4 1 16. z ey 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 The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a hyperbola. z 5 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. z 20 15 10 5 3 1 8 x 4 8 12 y 5 x 5 y x 2 3 4 y Section 10.6 18. y2 z2 4 (b) From 0, 10, 0 : z Surfaces in Space 497 (a) From 10, 0, 0 : z (c) From 10, 10, 10 : z 3 y y 3 x 3 y 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 22. z 2 Ellipsoid xy-trace: xz-trace: 1 ellipse 5 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 12 34 5 y 1 ellipse 25 circle x yz-trace: z2 z ± yz-trace: y2 10 : 24. z x2 4y2 26. 3z y2 x2 ±x 12 3x 12 3y 28. x2 2y 2 2z 2 ± ± Elliptic paraboloid xy-trace: point 0, 0, 0 xz-trace: z yz-trace: z z Hyperbolic paraboloid xy-trace: y xz-trace: z yz-trace: z z 28 24 20 Elliptic Cone xy-trace: x xz-trace: x 2y 2z x2 parabola 4y2 parabola yz-trace: point: 0, 0, 0 z 5 4 5 x 10 x 10 y 5 y 3 x 2 1 1 2 y 30. 9 x2 6x 9x2 9 y2 y2 9x x 9z2 4y 3 2 2 54x 4 y y 4 4y 9 z2 2 2 2 54z 6z 9z z 4 9 3 2 2 0 4 z 81 4 4 81 2 1 23 45 y 3 49 2 3 49 3. x 1 Hyperboloid of one sheet with center 3, 2, 32. z x2 0.5y 2 z 4 34. z 2 z 4y ± x2 4y z 36. x2 x2 ln y2 x2 e y2 z z z 8 4 −2 4 1 2 x 1 2 y x 8 −4 −8 3 y 4 x −8 −4 4 −3 −3 −4 4 y 498 Chapter 10 x x2 z Vectors and the Geometry of Space 38. z 8 y2 40. 9x 2 z ± 4y 2 8z 2 92 x 8 72 12 y 2 9 42. z y x 4 4 0, y 5 x2 x2 0, z z 0 4 2 20 2 4 y 10 z 2 x 4 10 10 x 20 20 y x 43 4 y 44. z y z 2z 0 4 x2 y2 3 z 46. x2 x2 z2 z2 ry 9y 2. 2 and z ry 3y; therefore, −3 3 3 x y 48. y2 z2 y2 z2 rx 2 and z rx 1 2 4y 2 4 4z 2 x2 ; therefore, 4. 50. x2 x2 y2 y2 rz e2z. 2 and y rz ez; therefore, 1 4 4 x2 , x 2 52. x 2 z2 cos2 y Equation of generating curve: x cos y or z cos y 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. x2 2 y2 4 56. About x-axis: y2 About y-axis: x2 About z-axis: x2 z2 z2 y2 rx ry rz 2 2 2 58. V 2 0 y sin y dy sin y y cos y 0 60. z 2 2 2 z (a) When y Focus: (b) When x z 2 4 4 we have z 0, 4, 9 2 x2 2 4, 4 1 z 2 4 x2. 1.0 2 we have y2 ,4 z 2 y 2. 0.5 π 2 π y Focus: 2, 0, 3 62. If x, y, z is on the surface, then z2 z2 8z x2 x2 x2 y2 y2 y2 z z2 4 2 8z 16 x2 8 y2 8 2 16 ⇒ z Elliptic paraboloid shifted up 2 units. Traces parallel to xy-plane are circles. S ection 10.7 64. z (a) 0.775x2 Year z Model 0.007y2 1980 37.5 37.8 1985 72.2 72.0 22.15x 1990 111.5 112.2 0.54y 1995 185.2 185.8 45.4 1996 200.1 204.5 1997 214.6 214.7 (b) Cylindrical and Spherical Coordinates 499 z 250 200 150 100 10 20 x 100 200 y (c) For y constant, the traces parallel to the xz-plane are concave downward. That is, for fixed y (public assistance), the rate of increase of z (Medicare) is decreasing with respect to x (worker’s compensation). (d) The traces parallel to the yz-plane (x constant) are concave upward. That is, for fixed x (worker’s compensation), the rate of increase of z (Medicare) is increasing with respect to y (public assistance). 66. Equating twice the first equation with the second equation, 2x2 6y2 4z2 4y 4y 3x 8 8 4y 2x2 3x 6y2 2 4z2 3x 2 6, a plane Section 10.7 2. x y z 0, 4, 4, , 2 Cylindrical and Spherical Coordinates 4. x y z 6, , 2 , cylindrical 32 32 6. x y z 1, 3 , 1 , cylindrical 2 cos sin 1 1, 1 , rectangular 3 2 3 2 0 1 2 , cylindrical 0 4 4 4 cos 4 sin 2 2 2 6 cos 6 sin 2 4 4 2 , rectangular 0, 3 2, 2 3 2, 8. 2 2, r 2 2, 4 , rectangular 22 2 10. 2 3, 4 r 2, 6 , rectangular 12 4 4 1 3 5 6 12. r 3, 2, 1 , rectangular 3 2 22 4 2 22 13 arctan 2 3 arctan z 4, 4 4 1 arctan z 1 2 arctan 3 z 1 13, 2 arctan , 3 , 4 , cylindrical 4, 6 , 1 , cylindrical 1 , cylindrical 14. z z x2 r2 y2 2 2 rectangular equation cylindrical equation 16. x2 y2 r2 r 8x 8r cos 8 cos rectangular equation cylindrical equation 500 Chapter 10 Vectors and the Geometry of Space z 2 x2 x2 y2 z 4 18. z 2 20. r 22. r y2 z 4 2 2 cos 2r cos y2 y2 1 2 z Same z 3 z 2 0 r2 x2 x2 x 2x 2x y 2 0 1 2 3 x 1 3 2 y −2 2 x x −2 2 y 2 3 y −2 24. z z r 2 cos2 x2 z 26. 1, 1, 1 , rectangular 12 arctan 1 1 arccos 3 3, , arccos 4 1 , spherical 3 12 4 12 3 9 1 32 x 12 34 56 y 28. 2, 2, 4 2 , rectangular 22 arctan 1 2 arccos 5 2 10, , arccos 4 2 , spherical 5 22 4 42 2 30. 2 10 4, 0, 0 , rectangular 4 2 02 02 4 arccos 0 4, , 2 2 , spherical 32. x y z 12, 3 , , spherical 49 3 12 sin cos 9 4 12 sin 12 cos 9 9 sin 3 4 11.276 34. 2.902 2.902 x y z 9, , 4 9 sin 9 sin 9 cos , spherical cos sin 0 0 9 4 4 0, 0, 9 , rectangular 2.902, 2.902, 11.276 , rectangular 38. (a) Programs will vary. (b) 2 2 cos sin 0 6 ,, x, y, z 5, 1, 0.5 1.295, 2.017, 4.388 36. x y z 6, , 2 , spherical 6 sin 6 sin 6 cos 0 2 6, 0, 0 , rectangular S ection 10.7 40. x2 x2 y2 y2 3z2 z2 2 Cylindrical and Spherical Coordinates 10 rectangular equation cos 10 csc 10 sec spherical equation 501 0 4z2 4 4 1 2 3 2 rectangular equation 42. x sin cos 2 1 cos cos2 (cone) spherical equation 44. tan 1 x 3 4 y x y x y 3 46. cos 0 0 z 48. 2 x2 z y2 z y2 z2 z2 z 2 sec cos 2 z 3 2 x2 0 z z xy-plane 2 3 3 2 x −3 1 3 y −3 −3 3 x 3 −3 x −2 −3 y 3 y 50. 4 csc sec 52. 3, 4 , 0 , cylindrical 02 3 54. 2, 2 , 3 22 2 3 2 , cylindrical 2 2 4 sin cos sin x 4 z 6 4 32 4 4 arccos 3, 22 cos 0 9 arccos 2 2 2, 1 2 3 4 , , spherical 42 23 , spherical , 34 4 6 x 6 y 56. 4, , 4 , cylindrical 3 4 3 arccos 1 2 4 2 58. 4, , 3 , cylindrical 2 42 2 arccos 3 5 32 5 60. r 4, , , spherical 18 2 4 sin 2 4 42 42 18 z 4, 4 cos 2 0 4 2, , , spherical 34 3 5, , arccos , spherical 2 5 18 , 0 , cylindrical 502 Chapter 10 Vectors and the Geometry of Space 5 , 6 5 sin 5 6 z 5 cos 0, 5 , 6 5 5 , cylindrical z 3 7, , , spherical 44 r 7 sin 3 4 72 2 62. r 18, , , spherical 33 sin 18 sin 3 9 64. r 5, , spherical 0 66. 3 z cos 18 cos 3 93 4 7 cos 72 ,, 24 3 4 72 2 72 , cylindrical 2 9, , 9 3 , cylindrical 3 Rectangular 68. 6, 2, 3 6.816, 6 Cylindrical 6.325, 10, 0.322, 3 Spherical 7.000, 11.662, 0.322, 2.014 70. 7.317, 0.75, 6 0.750, 1.030 72. 6.115, 1.561, 4.052 74. 3 2, 3 2, 76. 0, 78. 5, 4 1.732, 1, 3 3 6.311, 0.25, 4.052 6, 0.785, 5, 2, 3 7.5, 0.25, 1 6.708, 0.785, 2.034 6.403, 1.571, 0.896 1.571, 4 11 ,3 6 3.606, 2.618, 0.588 Note: Use the cylindrical coordinate 2, 80. 2.207, 7.949, 4 5 ,3 6 4 9.169, 1.3, 2.022 8.25, 1.3, 82. 4 Plane Matches graph (e) 84. Cone 4 Matches graph (a) 90. a Sphere 86. 4 sec , z Plane Matches graph (b) cos 4 88. r a Cylinder with z-axis symmetry b Plane perpendicular to xy-plane b Vertical half-plane c Half-cone 94. x2 y2 r2 2 z 92. 4 x2 (a) c Plane parallel to xy-plane y2 z 2, 2 z2 2r cos z 2 2 z z 4r 2 (a) sin 2 (b) 4 sin 2 sin 1 , 4 2 2 cos 2 , (b) sin2 cos , sin2 cos , csc sin2 cot cos , 4 sin2 tan 1 , 2 16 16, r sin2 sin cos 2 , tan2 arctan 1 2 96. x2 y2 98. y 4 16, 4 2 4 4, r sin 4 csc 4, 4 csc csc (a) r 2 (b) 2 (a) r sin sin2 4 16 0, 0, 4 csc (b) sin sin R eview Exercises for Chapter 10 102. 0 ≤ z2 ≤ ≤2 r2 z 4 3 −5 5 4 4 y x 5 y 2 503 100. 2 ≤ ≤ 2 104. 0 ≤ 6r 8 4 ≤ ≤2 ≤ 2 0≤r≤3 0 ≤ z ≤ r cos z 4 3 2≤r≤4 0≤ ≤1 z −4 −2 −2 x 2 x 2 y 106. Cylindrical: 0.75 ≤ r ≤ 1.25, z 108. Cylindrical 1 2 8 z 110. 2 sec 4 sphere ⇒ cos 2⇒z 2 plane ≤r≤3 ≤2 9 r2 −4 4 0≤ 9 The intersection of the plane and the sphere is a circle. 4 x −4 y r2 ≤ z ≤ Review Exercises for Chapter 10 2. P (a) u (b) v (c) 2u v 2, \ 1, Q 7, 0 42 14i 52 5, 1R 7i, v 41 PR 2, 4 \ 4. v 4, 5 4i 5j v cos i v sin j 1 cos 225 i 2 2 i 4 2 j 4 1 sin 225 j 2 PQ 4i 5j 18i 5j y 6. (a) The length of cable POQ is L. \ OQ L 9i yj y2 ⇒ \ O 2 92 L2 4 y2 −9 x 9 81 y θ P 500 lb Q Tension: T Also, cy c OQ c 81 250 y 250 ⇒ T 81 y2 ⇒ T 250 L2 4 81 L 2 250L L2 324 18 in. Domain: L > 18 inches (b) L T (c) 1000 19 780.9 20 573.54 21 485.36 22 434.81 23 401.60 24 377.96 25 360.24 (e) lim T L→ (d) The line T curve at L 400 intersects the 250 23.06 inches. The maximum tension is 250 pounds in each side of the cable since the total weight is 500 pounds. 18 0 25 504 8. x Chapter 10 z 0, y Vectors and the Geometry of Space 7: 0, 7, 0 10. 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. 12. Center: Radius: x 14. x2 x 2 2 0 2 2 y 40 , 0 3 2 2 2 64 , 3 z y2 0 2 0 2 6y 2 2 2, 3, 2 2 17 9 z2 4 4z 4 34 25 9 4 6 4 2 z 4 2 4 9 4 17 10x 5 2 25 y 3 2 z 2 2 Center: 5, Radius: 2 3, 2 2 4 6 8 x y 16. v 3 6, z 8 7 6 5 4 3 2 1 1 2 1 3 2, 8 0 3, 5, 8 18. v w 8 11 5, 5 4, 5 4, 3 7 7 3, 1, 4 2 (3, −3, 8) 5, 6 6, 10, Since v and w are not parallel, the points do not lie in a straight line. v 3 y 6 x 5 4 (6, 2, 0) 20. 8 6, 3, 2 49 8 6, 7 3, 2 48 , 7 24 16 , 77 22. P (a) u v (b) u (c) v 2, 1, 3 , Q \ 0, 5, 1 , R 2, 6, 2 3 66 9 3, 2, 54 2 3i 2i 6j 2 5, 5, 0 6j 3k 3 36 2k, PQ \ PR v v 4, 9 3, 6, 23 36 24. u Since v 4, 3, 6,v 16, 12, 24 26. u u v 1, 5 , v 4u, the vectors are parallel. 0 ⇒ is orthogonal to v. 2 \ \ 28. u v u u v cos v 1, 0, 2, 3 2, 1 1 30. W F PQ F PQ cos 75 8 cos 30 300 3 ft lb 10 3 uv uv 83.9 1 3 10 R eview Exercises for Chapter 10 505 In Exercises 32–40, u < 3, 2, 1 , v > < 2, 4, 3,w > < 1, 2, 2 . > 32. cos uv uv arccos 11 14 29 11 14 29 56.9 34. Work u w 3 4 2 5 36. 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 v u 10i 11j 8k Thus, u 38. 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 1 v 2 4i 6i 4j 1 2 4k 7j u 4k v w 5 (See Exercise 35) 2 40. Area triangle w 2 2 1 2 42. V u v w 2 0 0 1 2 1 0 1 2 25 10 44. Direction numbers: 1, 1, 1 (a) x (b) x 1 1 t, y y 2 2 z t, z 3 3 t 46. u v i 2 3 j 5 1 k 1 4 21 i 11j 13k 48. P \ 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 PQ 4 13t n 1 , PR \ Direction numbers: 21, 11, 13 (a) x (b) x 21 21t, y y 11 1 1 z 13 11t, z 4 27 x 3 PQ PR i 0 4 4y 4j 32k 0 33 506 Chapter 10 Vectors and the Geometry of Space 50. The normal vectors to the planes are the same, n 5, 3, 1 . 52. Q u P \ 5, 1, 3 point 1, 2, 1 direction vector Choose a point in the first plane, P 0, 0, 2 . Choose a 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 54. y z2 56. y cos z 58. 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 a parabola in the yz-coordinate plane. z 2 1 2 x 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 4 xy-trace: point 0,0, 0 xz-trace: z yz-trace: z z 4, x 2 z 4 ± 4x 3 4y 3 9 ± y2 3 4 y −2 x 2 2 y −3 3 23 y −3 x 60. x2 25 y2 4 z2 100 1 12 z 62. Let y r x the x-axis. −5 2 x and revolve the curve about Hyperboloid of one sheet xy-trace: xz-trace: yz-trace: x2 25 x2 25 y2 4 y2 4 z2 100 z2 100 1 x 5 y 1 1 64. 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 3 33 ,, , cylindrical 222 3 , 10 30 , , arccos 23 3 , spherical 10 2 2 (b) 3 , ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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