AN72 Answers: Chapter 9 a b c a b c aa ab ac ba bb bc ac bc cc a a b b c c aa bb cc aa bb cc aa aabb aacc bbcc bbaa bb bbcc aacc cc a a b b c c a a b b b b c c a a c c u v u v 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 1 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 ( )( ) ( ) ( )( ) ( ) =++ ++=++++++++ ⋅ = + + = + + + + = + + + + + + + + = + + + + + ab ac ba ac bc bc a a b b b b c c a a c c u v u v u v 2 2 2 , which equals . 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 2 ( ) − ⋅ = + + + + + − − − × 59. By Problem 58, since u and v are orthogonal, × = u v u v . If, in addition, u and v are unit vectors, × = ⋅ = u v 1 1 1. 61. Assume that a b c d e f u i j k v i j k , , = + + = + + and l m n w i j k. = + + Then bf ec af dc ae db u v i j k, ( ) ( ) ( ) × = − − − + − bn mc an lc am lb u w i j k, ( ) ( ) ( ) × = − − − + − and d l e m f n v w i j k. ( ) ( ) ( ) + = + + + + + Therefore, bf ec bn mc af dc an lc ae db am lb u u w i j k v ( ) ( ) ( ) ( ) ( ) ×+×= −+− − −+− + −+− and b f n e m c a f n d l c a e m d l b u v w i j k ( ) [ ] ( ) [ ] ( ) ( ) ( ) ( ) ( ) [ ] ×+= +−+ − +−+ + +−+ bf ec bn mc af dc an lc ae db am lb i j k, ( ) ( ) ( ) = −+− −−+− +−+− which equals u v u w . ( ) ( ) × + × 63. Let v v v v i j k 1 2 3 = + + and w w w w i j k. 1 2 3 = + + Then vw vw vw vw vw vw v w v w i j k 2 3 6 6 . 2 3 3 2 1 3 3 1 1 2 2 1 ( ) ( ) [ ] ( ) ( ) × = × = − − − + − v w v w v w v w v w v w vw vwv vw vwv vw vwv v v w v i v j v k 2 3 6 6 0 2 3 3 2 1 3 3 1 1 2 2 1 2 3 3 2 1 1 3 3 1 2 1 2 2 1 3 ( ) ( ) [ ] ( ) ( ) [ ] ( ) ( ) ( ) ⋅ × = − ⋅ − − ⋅ + − ⋅ = − − − + − = Since v v w v 2 3 0, ( ) ⋅ × = is orthogonal to v w 2 3 . × Similarly, w v w 2 3 0, ( ) ⋅ × = so w is also orthogonal to v w 2 3 . × 66. 4 π 67. 17, 4.22 , 17, 1.08 ( ) ( ) − 68. f x x log 5 1 1 7 ( ) ( ) = − + − 69. x z 1 2 log 3 log 4 4 − 70. 2 71. { } ≠ − ≠ x x x 4, 4 72. 11 4 73. 18 3 square units 74. x x x 16 4 ( ) − + 75. 3 π − Review Exercises (page 676) 1. 3 3 2 , 3 2 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ P 6 3, O P 6 2. 1, 3 ( ) 4P 3 22, 4P 3 O 3. 0, 3 ( ) P 22 P 2 23, 2 O 4. 3 2, 3 4 , 32, 4 π π ( ) ( ) − − 5. 2, 2 , 2, 2 π π ( ) ( ) − − 6. 5, 0.93 , ( ) 5, 4.07 ( ) − 7. (a) x y 1 1 2 2 ( ) + − = (b) circle, radius 1, center 0, 1 ( ) in rectangular coordinates 3P 4 U 5 5P 4 U 5 3P 2 U 5 7P 4 U 5 P 2 U 5 P 4 U 5 y x 1 2 U 5 P U 5 0 8. (a) x y 25 2 2 + = (b) circle, radius 5, center at pole 3P 4 U 5 5P 4 U 5 3P 2 U 5 7P 4 U 5 P 2 U 5 P 4 U 5 y x 1 2 3 4 U 5 P U 5 0 5 1 2 3 4 5 9. (a) x y 0 − = (b) line though pole, making an angle of 4 π with polar axis 3P 4 U 5 5P 4 U 5 3P 2 U 5 7P 4 U 5 P 2 U 5 P 4 U 5 y x 1 2 3 4 5 U 5 P U 5 0 10. (a) x y 4 2 25 2 2 ( ) ( ) − + + = (b) circle, radius 5, center 4, 2 ( ) − in rectangular coordinates 3P 4 U 5 5P 4 U 5 3P 2 U 5 7P 4 U 5 P 2 U 5 P 4 U 5 y x 4 U 5 P U 5 0 4 8 11. Circle; radius 2, center 2, 0 ( ) in rectangular coordinates; symmetric with respect to the polar axis 3P 4 U 5 5P 4 U 5 3P 2 U 5 7P 4 U 5 P 2 U 5 P 4 U 5 y x 1 2 3 5 U 5 P U 5 0 (4, 0) P 3 2, P 2 0, 12. Cardioid; symmetric with respect to the line 2 θ π = 3P 4 U 5 5P 4 U 5 3P 2 U 5 7P 4 U 5 P 2 U 5 P 4 U 5 y x 2 4 6 8 10 U 5 P U 5 0 P 2 0, (3, 0) P 2 6, 2
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