Unit vectors directed from the origin to the points a

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Unformatted text preview: e; Vectors EXERCISE SET 13.1 1. (a) (0, 0, 0), (3, 0, 0), (3, 5, 0), (0, 5, 0), (0, 0, 4), (3, 0, 4), (3, 5, 4), (0, 5, 4) (b) (0, 1, 0), (4, 1, 0), (4, 6, 0), (0, 6, 0), (0, 1, −2), (4, 1, −2), (4, 6, −2), (0, 6, −2) 3. corners: (4, 2, −2), (4,2,1), (4,1,1), (4, 1, −2), (−6, 1, 1), (−6, 2, 1), (−6, 2, −2), (−6, 1, −2) 2. corners: (2, 2, ±2), (2, −2, ±2), (−2, 2, ±2), (−2, −2, ±2) z (–2, –2, 2) z (–6, 2, 1) (–2, 2, 2) (–6, 1, –2) (2, –2, 2) (2, 2, 2) (–2, –2, –2) y (–2, 2, –2) (4, 1, 1) (2, –2, –2) (–6, 2, –2) y (4, 2, 1) (2, 2, –2) x (4, 1, –2) x 4. (a) (x2 , y1 , z1 ), (x2 , y2 , z1 ), (x1 , y2 , z1 )(x1 , y1 , z2 ), (x2 , y1 , z2 ), (x1 , y2 , z2 ) (b) The midpoint of the diagonal has coordinates which are the coordinates of the midpoints 1 (x1 + x2 ), y1 , z1 ; of the edges. The midpoint of the edge (x1 , y1 , z1 ) and (x2 , y1 , z1 ) is 2 1 the midpoint of the edge (x2 , y1 , z1 ) and (x2 , y2 , z1 ) is x2 , (y1 + y2 ), z1 ; the midpoint 2 1 of the edge (x2 , y2 , z1 ) and (x2 , y2 , z2 )) is x2 , y2 , (z1 + z2 ) . Thus the coordinates of the 2 1 1 1 midpoint of the diagonal are (x1 + x2 ), (y1 + y2 ), (z1 + z2 ). 2 2 2 √ √ √ 5. The diameter is d = (1 − 3)2 + (−2 − 4)2 + (4 + 12)2 = 296, so the radius is 296/2 = 74. The midpoint (2, 1, −4) of the endpoints of the diameter is the center of the sphere. 6. Each side has length √ 14 so the triangle is equilateral. √ 7. (a) The sides have lengths 7, 14, and 7 5; it is a right triangle because the sides satisfy the √2 Pythagorean theorem, (7 5) = 72 + 142 . (b) (2,1,6) is the vertex of the 90◦ angle because it is opposite the longest side (the hypotenuse). (c) area = (1/2)(altitude)(base) = (1/2)(7)(14) = 49 8. (a) 3 (d) (2)2 + (−3)2 √ = 13 (b) 2 (e) (−5)2 9. (a) (x − 1)2 + y 2 + (z + 1)2 = 16 (b) r = (−1 − 0)2 + (3 − 0)2 + (2 − 0)2 = + (−3)2 √ = 34 (c) 5 (f ) (−5)2 + (2)2 = √ 14, (x + 1)2 + (y − 3)2 + (z − 2)2 = 14 448 √ 29 449 Chapter 13 1 1√ (−1 − 0)2 + (2 − 2)2 + (1 − 3)2 = 5, center (−1/2, 2, 2), 2 2 (x + 1/2)2 + (y − 2)2 + (z − 2)2 = 5/4 (c) r = 10. r = |[distance between (0,0,0) and (3, −2, 4)] ± 1| = √ √ 2 x2 + y 2 + z 2 = r 2 = 29 ± 1 = 30 ± 2 29 √ 29 ± 1, 11. (x − 2)2 + (y + 1)2 + (z + 3)2 = r2 , (a) r2 = 32 = 9 (b) r2 = 12 = 1 12. (a) The sides have length 1, so the radius is (b) The diagonal has length √ 1+1+1= (c) r2 = 22 = 4 1 1 ; hence (x + 2)2 + (y − 1)2 + (z − 3)2 = 2 4 √ 3 3 and is a diameter, so (x+2)2 +(y −1)2 +(z −3)2 = . 4 13. (x + 5)2 + (y + 2)2 + (z + 1)2 = 49; sphere, C (−5, −2, −1), r = 7 14. x2 + (y − 1/2)2 + z 2 = 1/4; sphere, C (0, 1/2, 0), r = 1/2 √ 15. (x − 1/2)2 + (y − 3/4)2 + (z + 5/4)2 = 54/16; sphere, C (1/2, 3/4, −5/4), r = 3 6/4 16. (x + 1)2 + (y − 1)2 + (z + 1)2 = 0; the point (−1, 1, −1) 17. (x − 3/2)2 + (y + 2)2 + (z − 4)2 = −11/4; no graph 18. (x − 1)2 + (y − 3)2 + (z − 4)2 = 25; sphere, C (1, 3, 4), r = 5 z 19. (a) z (b) y y y x x x z 20. (a) z (c) (b) z (c) z y=1 x=1 z=1 y y y x x x z 21. (a) z (b) z (c) y y 5 y 5 5 x x x Exercise Set 13.1 450 z 22. (a) z (b) z (c) y y y x x x 23. (a) −2y + z = 0 (b) −2x + z = 0 (c) (x − 1) + (y − 1) = 1 2 (d) (x − 1)2 + (z − 1)2 = 1 2 24. (a) (x − a)2 + (z − a)2 = a2 (b) (x − a)2 + (y − a)2 = a2 (c) (y − a)2 + (z − a)2 = a2 z 25. y x z 26. z 27. 1 1 y x y x z 28. z 29. 3 y 3 2 y x x z 30. z 31. 2 3 y 2 z 32. √3 -3 y y x 3 3 x x 451 Chapter 13 z 33. -2 z 34. 2 y x y x 1.4 35. (a) z (b) -1.4 1.4 x y -1.4 36. (a) -2 z (b) 1 2 x y -1 37. Complete the square to get (x + 1)2 + (y − √ 2 + (z − 2)2 = 9; center (−1, 1, 2), radius 3. The 1) distance between the origin and the center is 6 < 3 so the origin is inside the sphere. The largest √ √ distance is 3 + 6, the smallest is 3 − 6. 38. (x − 1)2 + y 2 + (z + 4)2 ≤ 25; all points on and inside the sphere of radius 5 with center at (1, 0, −4). 39. (y + 3)2 + (z − 2)2 > 16; all points outside the circular cylinder (y + 3)2 + (z − 2)2 = 16. 40. (x − 1)2 + (y + 2)2 + z 2 = 2 x2 + (y − 1)2 + (z − 1)2 , square and simplify to get 3x2 + 3y 2 + 3z 2 + 2x − 12y − 8z + 3 = 0, then complete the square to get √ (x + 1/3)2 + (y − 2)2 + (z − 4/3)2 = 44/9; center (−1/3, 2, 4/3), radius 2 11/3. 41. Let r be the radius of a styrofoam sphere. The distance from the origin to the center of the bowling ball is equal to the sum of the distance from the origin to the center of the styrofoam sphere nearest the origin and the distance between the center of this sphere and the center of the bowling ball so √ √ √ √ √ √ 3−1 3R = 3r + r + R, ( 3 + 1)r = ( 3 − 1)R, r = √ R = (2 − 3)R. 3+1 42. (a) Complete the square to get (x + G/2)2 + (y + H/2)2 + (z + I/2)2 = K/4, so the equation represents a sphere when K > 0, a point when K = 0, and no graph when K < 0. √ (b) C (−G/2, −H/2, −I/2), r = K/2 Exercise Set 13.2 452 43. (a sin φ c...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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