R 2i 2tj t2 k r 2j 2tk r 2k r r 2t2 i 4tj

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Unformatted text preview: 4, v = i − 4j; x = −2 + t, y = 4 − 4t 15. x = −1 + 3t, y = 2 − 4t, z = 4 + t 16. x = 2 − t, y = −1 + 2t, z = 5 + 7t 17. The line is parallel to the vector 2, −1, 2 so x = −2 + 2t, y = −t, z = 5 + 2t. 18. The line is parallel to the vector 1, 1, 0 so x = t, y = t, z = 0. 19. (a) y = 0, 2 − t = 0, t = 2, x = 7 (b) x = 0, 1 + 3t = 0, t = −1/3, y = 7/3 √ √ √ −1 ± 85 43 −7 ± 85 85 2 2 2 ,x = ,y = (c) y = x , 2 − t = (1 + 3t) , 9t + 7t − 1 = 0, t = 18 6 18 20. (4t)2 + (3t)2 = 25, 25t2 = 25, t = ±1, the line intersects the circle at ± 4, 3 21. (a) z = 0 when t = 3 so the point is (−2, 10, 0) (b) y = 0 when t = −2 so the point is (−2, 0, −5) (c) x is always −2 so the line does not intersect the yz -plane 22. (a) z = 0 when t = 4 so the point is (7,7,0) (b) y = 0 when t = −3 so the point is (−7, 0, 7) (c) x = 0 when t = 1/2 so the point is (0, 7/2, 7/2) 23. (1 + t)2 + (3 − t)2 = 16, t2 − 2t − 3 = 0, (t + 1)(t − 3) = 0; t = −1, 3. The points of intersection are (0, 4, −2) and (4,0,6). 24. 2(3t) + 3(−1 + 2t) = 6, 12t = 9; t = 3/4. The point of intersection is (5/4, 9/4, 1/2). 25. The lines intersect if we can find values of t1 and t2 that satisfy the equations 2 + t1 = 2 + t2 , 2 + 3t1 = 3 + 4t2 , and 3 + t1 = 4 + 2t2 . Solutions of the first two of these equations are t1 = −1, t2 = −1 which also satisfy the third equation so the lines intersect at (1, −1, 2). 26. Solve the equations −1 + 4t1 = −13 + 12t2 , 3 + t1 = 1 + 6t2 , and 1 = 2 + 3t2 . The third equation yields t2 = −1/3 which when substituted into the first and second equations gives t1 = −4 in both cases; the lines intersect at (−17, −1, 1). 27. The lines are parallel, respectively, to the vectors 7, 1, −3 and −1, 0, 2 . These vectors are not parallel so the lines are not parallel. The system of equations 1 + 7t1 = 4 − t2 , 3 + t1 = 6, and 5 − 3t1 = 7 + 2t2 has no solution so the lines do not intersect. 28. The vectors 8, −8, 10 and 8, −3, 1 are not parallel so the lines are not parallel. The lines do not intersect because the system of equations 2 + 8t1 = 3 + 8t2 , 6 − 8t1 = 5 − 3t2 , 10t1 = 6 + t2 has no solution. 29. The lines are parallel, respectively, to the vectors v1 = −2, 1, −1 and v2 = −4, 2, −2 ; v2 = 2v1 , v1 and v2 are parallel so the lines are parallel. 30. The lines are not parallel because the vectors 3, −2, 3 and 9, −6, 8 are not parallel. −→ −→ 31. P1 P2 = 3, −7, −7 , P2 P3 = −9, −7, −3 ; these vectors are not parallel so the points do not lie on the same line. Exercise Set 13.5 468 −→ −→ −→ −→ 32. P1 P2 = 2, −4, −4 , P2 P3 = 1, −2, −2 ; P1 P2 = 2 P2 P3 so the vectors are parallel and the points lie on the same line. 33. If t2 gives the point −1 + 3t2 , 9 − 6t2 on the second line, then t1 = 4 3t2 yields the point 3 − (4 − 3t2 ), 1 + 2(4 − 3t2 ) = −1 + 3t2 , 9 − 6t2 on the first line, so each point of L2 is a point of L1 ; the converse is shown with t2 = (4 − t1 )/3. 34. If t1 gives the point 1 + 3t1 , −2 + t1 , 2t1 on L1 , then t2 = (1 − t1 )/2 gives the point 4 − 6(1 − t1 )/2, −1 − 2(1 − t1 )/2, 2 − 4(1 − t1 )/2 = 1 + 3t1 , −2 + t1 , 2t1 on L2 , so each point of L1 is a point of L2 ; the converse is shown with t1 = 1 − 2t2 . 35. The line segment joining the points (1,0) and (−3, 6). 36. The line segment joining the points (−2, 1, 4) and (7,1,1). 37. A(3, 0, 1) and B (2, 1, 3) are on the line, and (method of Exercise 25) −→ −→ −→ AP = −5i + j, AB = −i + j + 2k, proj −→ AP AB so distance = √ −→ −→ −→ = | AP · AB |/ AB = √ −→ 6 and AP −→ −→ √ 26 − 6 = 4 5. Using the method of Exercise 26, distance = AP × AB −→ = √ 26, √ = 2 5. AB 38. A(2, −1, 0) and B (3, −2, 3) are on the line, and (method of Exercise 25) −→ −→ −→ AP = −i + 5j − 3k, AB = i − j + 3k, proj −→ AP AB −→ AP = √ 35, so distance = −→ distance = −→ = | AP · −→ AB |/ −→ AB 15 =√ and 11 35 − 225/11 = 4 10/11. Using the method of Exercise 26, −→ AP × AB −→ = 4 10/11. AB 39. The vectors v1 = −i + 2j + k and v2 = 2i − 4j − 2k are parallel to the lines, v2 = −2v1 so v1 and v2 are parallel. Let t = 0 to get the points P (2, 0, 1) and Q(1, 3, 5) on the first and second lines, −→ respectively. Let u = P Q = −i + 3j + 4k, v = 1 v2 = i − 2j − k; u × v = 5i + 3j − k; by the method 2 of Exercise 26 of Section 13.4, distance = u × v / v = 35/6. 40. The vectors v1 = 2i + 4j − 6k and v2 = 3i + 6j − 9k are parallel to the lines, v2 = (3/2)v1 so v1 and v2 are parallel. Let t = 0 to get the points P (0, 3, 2) and Q(1, 0, 0) on the first and second −→ lines, respectively. Let u = P Q = i − 3j − 2k, v = 1 v1 = i + 2j − 3k; u × v = 13i + j + 5k, 2 distance = u × v / v = 195/14 (Exer. 26, Section 13.4). 41. (a) The line is parallel to the vector x1 − x0 , y1 − y0 , z1 − z0 so x = x0 + (x1 − x0 ) t,...
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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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