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Unformatted text preview: 2 2 2 + 2 2 + 3 2 = v1 + v2 + v3 / v v2 v v 2 = v 2/ v 2 =1 18. Let v = x, y, z , then x = x2 + y 2 cos θ, y = x2 + y 2 sin θ, x2 + y 2 = v cos λ, and z = v sin λ, so x/ v = cos θ cos λ, y/ v = sin θ cos λ, and z/ v = sin λ. √ √√ √ 3 1 31 3 33 = , cos β = = , cos γ = ; α ≈ 64◦ , β ≈ 41◦ , γ = 60◦ 19. cos α = 22 4 22 4 2 20. Let u1 = u1 cos α1 , cos β1 , cos γ1 , u2 = u2 cos α2 , cos β2 , cos γ2 , u1 and u2 are perpendicular if and only if u1 · u2 = 0 so u1 u2 (cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 ) = 0, cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 = 0. 21. (a) b = 3/5, 4/5 , so projb v = 6/25, 8/25 b and v − projb v = 44/25, −33/25 y 2 projbv x 2 v -2 -2 (b) √ √ b = 1/ 5, −2/ 5 , so projb v = −6/5, 12/5 b and v − projb v = 26/5, 13/5 5 v – projbv y v – projbv v projbv -5 x 5 -5 Exercise Set 13.3 (c) 460 √ √ b = 2/ 5, 1/ 5 , so projb v = −16/5, −8/5 b and v − projb v = 1/5, −2/5 y x -4 v – projbv projbv v -4 22. (a) (b) b = 1/3, 2/3, 2/3 , so projb v = 2/3, 4/3, 4/3 and v − projb v = 4/3, −7/3, 5/3 b b = 2/7, 3/7, −6/7 , so projb v = −74/49, −111/49, 222/49 b and v − projb v = 270/49, 62/49, 121/49 23. (a) projb v = −1, −1 , so v = −1, −1 + 3, −3 (b) projb v = 16/5, 0, −8/5 , so v = 16/5, 0, −8/5 + −1/5, 1, −2/5 24. (a) projb v = 1, 1 , so v = 1, 1 + −4, 4 (b) projb v = 0, −8/5, 4/5 , so v = 0, −8/5, 4/5 + −2, 13/5, 26/5 −→ −→ −→ 25. AP = −i + 3j, AB = 3i + 4j, proj −→ AP AB −→ √ AP = 10, 10 − 81/25 = 13/5 −→ −→ −→ −→ −→ = | AP · AB |/ AB −→ 26. AP = −4i + 2k, AB = −3i + 2j − 4k, proj −→ AP AB −→ √ AP = 20, 20 − 16/29 = 564/29 = 9/5 −→ −→ −→ √ = | AP · AB |/ AB = 4/ 29. 27. Let F be the downward force of gravity on the block, then F = 10(9.8) = 98 N, and if F1 and √ F2 are the forces parallel to and perpendicular to the ramp, then F1 = F2 = 49 2 N. Thus √ √ the block exerts a force of 49 2 N against the ramp and it requires a force of 49 2 N to prevent the block from sliding down the ramp. 28. Let x denote the magnitude of the force in the direction of Q. Then the force F acting on the 1 1 block is F = xi − 10j. Let u = − √ (i + j) and v = √ (i − j) be the unit vectors in the directions 2 2 x + 10 x − 10 along and against the ramp. Then F decomposes as F = − √ u + √ v, and thus the block 2 2 will not slide down the ramp provided x ≥ 10 N. 29. Three forces act on the block: its weight −300j; the tension √ cable A, which has the form in j a(−i + j); and the tension in cable B, which has the form b( 3i −√), where a, b are positive √ constants. The sum of these forces is zero, which yields a = 450 + 150 3, b = 150 + 150 3. Thus the forces√ along cables A and B are, √ respectively, √ √ √ √ 150(3 + 3)(i − j) = 450 2 + 150 6 lb, and 150( 3 + 1)( 3i − j) = 300 + 300 3 lb. 461 Chapter 13 30. (a) Let T and T be the forces exerted on the block by cables A and B . Then A B T = a(−10i + dj) and T = b(20i + dj) for some positive a, b. Since T + T − 100j = 0, we A B A B 100 200 100 2000 2000 200 ,b = ,T = − i+ j, and T = i+ j. ﬁnd a = A B 3d 3d 3d 3 3d 3 500 -20 100 -100 (b) An increase in d will decrease both forces. 500 -20 100 -100 40 ≤ 150 is equivalent to d ≥ √ , and T B 65 40 40 . d ≥ √ . Hence we must have d ≥ 65 77 (c) The inequality T A −→ ≤ 150 is equivalent to −→ 31. Let P and Q be the points (1,3) and (4,7) then P Q = 3i + 4j so W = F · P Q = −12 ft · lb. −→ 32. W = F · P Q= F −→ P Q cos 45◦ = (500)(100) √ √ 2/2 = 25,000 2 N · m 33. W = F ·15i = 15 · 50 cos 60◦ = 375 ft · lb. √ √ 34. W = F ·(15/ 3)(i + j + k) = −15/ 3 N · m 35. With the cube as shown in the diagram, and a the length of each edge, d1 = ai + aj + ak, d2 = ai + aj − ak, cos θ = (d1 · d2 ) / ( d1 d2 ) = 1/3, θ ≈ 71◦ z d2 y d1 x 36. Take i, j, and k along adjacent edges of the box, then 10i + 15j + 25k is along a diagonal, and a √ 2 3 5 unit vector in this direction is √ i + √ j + √ k. The direction cosines are cos α = 2/ 38, 38 38 38 √ √ ◦ cos β = 3/ 38, and cos γ = 5/ 38 so α ≈ 71 , β ≈ 61◦ , and γ ≈ 36◦ . Exercise Set 13.3 462 37. u + v and u − v are vectors along the diagonals, (u + v) · (u − v) = u · u − u · v + v · u − v · v = u 2 −v 2 so (u + v) · (u − v) = 0 if and only if u = v . 38. The diagonals have lengths u + v and u − v but u+v 2 = (u + v) · (u + v) = u 2 + 2u · v + v 2 , and u − v 2 = (u − v) · (u − v) = u 2 − 2u · v + v 2 . If the parallelogram is a rectangle then u · v = 0 so u + v 2 = u − v 2 ; the diagonals are equal. If the diagonals are equal, then 4u · v = 0, u · v = 0 so u is perpendicular to v and hence the parallelogram is a rectangle. u+v 2 = (u + v) · (u + v) = u 2 + 2u · v + v u−v 2 = (u − v) · (u − v) = u 2 − 2u · v + v , add to get u+v 39. 2 + u−v 2 =2 u 2 +2 v 2 and 2 2 The sum of th...
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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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