# U chapter 10 2 1 2 v u v v 2 vectors and the geometry

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Unformatted text preview: hogonal. 57. Programs will vary. 59. Programs will vary. 61. 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. 63. u v 1 i 2 8i 2 j. Want u 3 6j and v v 8i 0. 6j are orthogonal to u. 65. u v 3, 1, 2 . Want u v v 0, 0. 2, 1 are orthogonal to u. 0, 2, 1 and 67. (a) Gravitational Force F v w1 cos 10 i F v v 2 48,000 j (b) w2 F w1 48,000 j 46,552.6 j 8335.1 cos 10 i sin 10 j sin 10 j F vv 48,000 sin 10 v 8335.1 cos 10 i sin 10 j w2 8208.5 i v 47,270.8 lb w1 1 i 2 v 8335.1 lb 3 j 2 425 ft lb 1, 7 and w 5, 5 . Then u v and u v u 0 v v v v v v. u u+v v \ 69. F v W 85 10 i F 71. PQ v W 4, 7, 5 1, 4, 8 \ PQ v 72 73. False. Let u 2, 4 , v 2 28 30 and u w 10 20 30. 75. In a rhombus, u u v u v v . The diagonals are u u u u 2 u−v v u v v u u 2 u Therefore, the diagonals are orthogonal. S ection 10.4 The Cross Product of Two Vectors in Space 241 77. u cos , sin , 0 , v cos , sin , 0 . Assuming that cos cos 11 sin sin &gt; . Also, cos cos sin sin . The angle between u and v is cos uv uv 79. u v 2 u u u u u 2 2 v v u u v 2 u u v v u v u u u 2u v v v v v v v 2 81. u v 2 u u v v u 2 2 u u v 2u u v v u u v 2 v v v v v v u u ≤u ≤ Therefore, u u 2u v v 2 v 2 from Exercise 66 v≤u v. Section 10.4 ijk 010 100 z The Cross Product of Two Vectors in Space ijk 010 001 z 1. j i k 3. j k i 5. i k ijk 100 001 z j 1 1 1 k j 1 x −1 −j k j 1 i −1 −k 1 y x 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 9. (a) u (b) v (c) v v u v i 7 1 u 0 j 3 1 v k 2 5 17, 33, 10 17, 33, 10 ijk 372 372 11. u u u v v 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 j 1 1 k 1 1 1, 1 1 1, 1 u u v v u u 0⇒u 0⇒v 242 13. u u u Chapter 10 12, v u v 3, 0 , v i 12 2 j 3 5 12 0 Vectors and the Geometry of Space 2, 5, 0 k 0 0 54k 30 u 50 u v 19. 6 5 4 3 2 1 1 2 15. u 0, 0, 54 0 54 u u i v u j i 1 2 v k, v j 1 1 1 2i k 1 1 2 j 2i 13 u 13 u k 3j 1 v 1 v 1 1 k 2, 3, 1 0⇒u v u v 20 0⇒v 17. 6 5 4 3 2 1 1 4 x 3 2 v 0 54...
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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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