1 32 26 r r32 r0 3i 3j v 3 cos ti 3 sin tj

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Unformatted text preview: y = y0 + (y1 − y0 ) t, z = z0 + (z1 − z0 ) t (b) The line is parallel to the vector a, b, c so x = x1 + at, y = y1 + bt, z = z1 + ct 42. Solve each of the given parametric equations (2) for t to get t = (x − x0 ) /a, t = (y − y0 ) /b, t = (z − z0 ) /c, so (x, y, z ) is on the line if and only if (x − x0 ) /a = (y − y0 ) /b = (z − z0 ) /c. 43. (a) It passes through the point (1, −3, 5) and is parallel to v = 2i + 4j + k (b) x, y, z = 1 + 2t, −3 + 4t, 5 + t −→ −→ 44. Let the desired point be P (x0 , y0 , z0 ), then P1 P = (2/3) P1 P2 , x0 − 1, y0 − 4, z0 + 3 = (2/3) 0, 1, 2 = 0, 2/3, 4/3 ; equate corresponding components to get x0 = 1, y0 = 14/3, z0 = −5/3. 469 Chapter 13 45. (a) Let t = 3 and t = −2, respectively, in the equations for L1 and L2 . (b) u = 2i − j − 2k and v = i + 3j √ k are parallel to L1 and L2 , − cos θ = u · v/( u v ) = 1/(3 11), θ ≈ 84◦ . (c) u × v = 7i + 7k is perpendicular to both L1 and L2 , and hence so is i + k, thus x = 7 + t, y = −1, z = −2 + t. 46. (a) Let t = 1/2 and t = 1, respectively, in the equations for L1 and L2 . (b) u = 4i − 2j + 2k and v = i − j + 4k are parallel to L1 and L2 , √ cos θ = u · v/( u v ) = 14/ 432, θ ≈ 48◦ . (c) u × v = −6i − 14j − 2k is perpendicular to both L1 and L2 , and hence so is 3i + 7j + k, thus x = 2 + 3t, y = 7t, z = 3 + t. 47. (0,1,2) is on the given line (t = 0) so u = j − k is a vector from this point to the point (0,2,1), v = 2i − j + k is parallel to the given line. u × v = −2j − 2k, and hence w = j + k, is perpendicular to both lines so v × w = −2i − 2j + 2k, and hence i + j − k, is parallel to the line we seek. Thus x = t, y = 2 + t, z = 1 − t are parametric equations of the line. 48. (−2, 4, 2) is on the given line (t = 0) so u = 5i − 3j − 4k is a vector from this point to the point (3, 1, −2), v = 2i + 2j + k is parallel to the given line. u × v = 5i − 13j + 16k is perpendicular to both lines so v × (u × v) = 45i − 27j − 36k, and hence 5i − 3j − 4k is parallel to the line we seek. Thus x = 3 + 5t, y = 1 − 3t, z = −2 − 4t are parametric equations of the line. 49. (a) When t = 0 the bugs are at (4, 1, 2) and (0,1,1) so the distance between them is √ √ 42 + 02 + 12 = 17 cm. 10 (b) 0 (c) The distance has a minimum value. 5 0 (d) Minimize D2 instead of D (the distance between the bugs). D2 = [t − (4 − t)]2 + [(1 + t) − (1 + 2t)]2 + [(1 + 2t) − (2 + t)]2 = 6t2 − 18t + 17, d(D2 )/dt = 12t − 18 = 0 when t = 3/2; the minimum √ distance is 6(3/2)2 − 18(3/2) + 17 = 14/2 cm. 50. The line intersects the xz -plane when t = −1, the xy -plane when t = 3/2. Along the line, T = 25t2 (1 + t)(3 − 2t) for −1 ≤ t ≤ 3/2. Solve dT /dt = 0 for t to find that the maximum value of T is about 50.96 when t ≈ 1.073590. EXERCISE SET 13.6 1. x = 3, y = 4, z = 5 2. x = x0 , y = y0 , z = z0 3. (x − 2) + 4(y − 6) + 2(z − 1) = 0, x + 4y + 2z = 28 4. −(x + 1) + 7(y + 1) + 6(z − 2) = 0, −x + 7y + 6z = 6 Exercise Set 13.6 470 6. 2x − 3y − 4z = 0 5. z = 0 7. n = i − j, x − y = 0 8. n = i + j, P (1, 0, 0), (x − 1) + y = 0, x + y = 1 9. n = j + k, P (0, 1, 0), (y − 1) + z = 0, y + z = 1 −→ −→ −→ 10. n = j − k, y − z = 0 −→ 11. P1 P2 × P1 P3 = 2, 1, 2 × 3, −1, −2 = 0, 10, −5 ,for convenience choose 0, 2, −1 which is also normal to the plane. Use any of the given points to get 2y − z = 1 12. P1 P2 × P1 P3 = −1, −1, −2 × −4, 1, 1 = 1, 9, −5 , x + 9y − 5z = 16 13. (a) parallel, because 2, −8, −6 and −1, 4, 3 are parallel (b) perpendicular, because 3, −2, 1 and 4, 5, −2 are orthogonal (c) neither, because 1, −1, 3 and 2, 0, 1 are neither parallel nor orthogonal 14. (a) neither, because 3, −2, 1 and 6, −4, 3 are neither parallel nor orthogonal (b) parallel, because 4, −1, −2 and 1, −1/4, −1/2 are parallel (c) perpendicular, because 1, 4, 7 and 5, −3, 1 are orthogonal 15. (a) parallel, because 2, −1, −4 and 3, 2, 1 are orthogonal (b) neither, because 1, 2, 3 and 1, −1, 2 are neither parallel nor orthogonal (c) perpendicular, because 2, 1, −1 and 4, 2, −2 are parallel 16. (a) parallel, because −1, 1, −3 and 2, 2, 0 are orthogonal (b) perpendicular, because −2, 1, −1 and 6, −3, 3 are parallel (c) neither, because 1, −1, 1 and 1, 1, 1 are neither parallel nor orthogonal 17. (a) 3t − 2t + t − 5 = 0, t = 5/2 so x = y = z = 5/2, the point of intersection is (5/2, 5/2, 5/2) (b) 2(2 − t) + (3 + t) + t = 1 has no solution so the line and plane do not intersect 18. (a) 2(3t) − 5t + (−t) + 1 = 0, 1 = 0 has no solution so the line and the plane do not intersect. (b) (1 + t) − (−1 + 3t) + 4(2 + 4t) = 7, t = −3/14 so x = 1 − 3/14 = 11/14, y = −1 − 9/14 = −23/14, z = 2 − 12/14 = 8/7, the point is (11/14, −23/14, 8/7) 19. n 1 = 1, 0, 0 , n2 = 2, −1, 1 , n1 · n2 = 2 so √ √ 2 n 1 · n2 = √ √ = 2/ 6, θ = cos−1 (2/ 6) ≈ 35◦ cos θ = n1 n2 16 20. n1 = 1, 2, −2 , n2 = 6, −3, 2 , n1 · n2 = −4 so 4 (−n1 ) ·...
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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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