59 30 s0 0 so x 3v0 t2 y v0 t2 16t2 dydt

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Unformatted text preview: n2 = = 4/21, θ = cos−1 (4/21) ≈ 79◦ cos θ = − n1 n2 (3)(7) (Note: −n1 is used instead of n1 to get a value of θ in the range [0, π/2]) 21. 4, −2, 7 is normal to the desired plane and (0,0,0) is a point on it; 4x − 2y + 7z = 0 22. v = 3, 2, −1 is parallel to the line and n = 1, −2, 1 is normal to the given plane so v × n = 0, −4, −8 is normal to the desired plane. Let t = 0 in the line to get (−2, 4, 3) which is also a point on the desired plane, use this point and (for convenience) the normal 0, 1, 2 to find that y + 2z = 10. 471 Chapter 13 23. Find two points P1 and P2 on the line of intersection of the given planes and then find an equation of the plane that contains P1 , P2 , and the given point P0 (−1, 4, 2). Let (x0 , y0 , z0 ) be on the line of intersection of the given planes; then 4x0 − y0 + z0 − 2 = 0 and 2x0 + y0 − 2z0 − 3 = 0, eliminate y0 by addition of the equations to get 6x0 − z0 − 5 = 0; if x0 = 0 then z0 = −5, if x0 = 1 then z0 = 1. Substitution of these values of x0 and z0 into either of the equations of the planes gives the corresponding values y0 = −7 and y0 = 3 so P1 (0, −7, −5) and P2 (1, 3, 1) are on the −→ −→ line of intersection of the planes. P0 P1 × P0 P2 = 4, −13, 21 is normal to the desired plane whose equation is 4x − 13y + 21z = −14. 24. 1, 2, −1 is parallel to the line and hence normal to the plane x + 2y − z = 10 25. n1 = 2, 1, 1 and n2 = 1, 2, 1 are normals to the given planes, n1 × n2 = −1, −1, 3 so 1, 1, −3 is normal to the desired plane whose equation is x + y − 3z = 6. −→ 26. n = 4, −1, 3 is normal to the given plane, P1 P2 = 3, −1, −1 is parallel to the line, −→ n × P1 P2 = 4, 13, −1 is normal to the desired plane whose equation is 4x + 13y − z = 1. 27. n1 = 2, −1, 1 and n2 = 1, 1, −2 are normals to the given planes, n1 × n2 = 1, 5, 3 is normal to the desired plane whose equation is x + 5y + 3z = −6. 28. Let t = 0 and t = 1 to get the points P1 (−1, 0, −4) and P2 (0, 1, −2) that lie on the line. Denote the −→ −→ given point by P0 , then P0 P1 × P0 P2 = 7, −1, −3 is normal to the desired plane whose equation is 7x − y − 3z = 5. 29. The plane is the perpendicular bisector of the line segment that joins P1 (2, −1, 1) and P2 (3, 1, 5). −→ The midpoint of the line segment is (5/2, 0, 3) and P1 P2 = 1, 2, 4 is normal to the plane so an equation is x + 2y + 4z = 29/2. 30. n1 = 2, −1, 1 and n2 = 0, 1, 1 are normals to the given planes, n1 × n2 = −2, −2, 2 so n = 1, 1, −1 is parallel to the line of intersection of the planes. v = 3, 1, 2 is parallel to the given line, v × n = −3, 5, 2 so 3, −5, −2 is normal to the desired plane. Let t = 0 to find the point (0,1,0) that lies on the given line and hence on the desired plane. An equation of the plane is 3x − 5y − 2z = −5. 31. The line is parallel to the line of intersection of the planes if it is parallel to both planes. Normals to the given planes are n1 = 1, −4, 2 and n2 = 2, 3, −1 so n1 × n2 = −2, 5, 11 is parallel to the line of intersection of the planes and hence parallel to the desired line whose equations are x = 5 − 2t, y = 5t, z = −2 + 11t. −→ −→ 32. Denote the points by A, B , C , and D, respectively. The points lie in the same plane if AB × AC −→ −→ −→ −→ −→ −→ and AB × AD are parallel (method 1). AB × AC = 0, −10, 5 , AB × AD = 0, 16, −8 , these vectors are parallel because 0, −10, 5 = (−10/16) 0, 16, −8 . The points lie in the same plane −→ −→ if D lies in the plane determined by A, B, C (method 2), and since AB × AC = 0, −10, 5 , an equation of the plane is −2y + z + 1 = 0, 2y − z = 1 which is satisfied by the coordinates of D. 33. v = 0, 1, 1 is parallel to the line. (a) For any t, 6 · 0 + 4t − 4t = 0, so (0, t, t) is in the plane. (b) n = 5, −3, 3 is normal to the plane, v · n = 0 so the line is parallel to the plane. (0,0,0) is on the line, (0, 0, 1/3) is on the plane. The line is below the plane because (0,0,0) is below (0, 0, 1/3). Exercise Set 13.6 472 (c) n = 6, 2, −2 , v · n = 0 so the line is parallel to the plane. (0,0,0) is on the line, (0, 0, −3/2) is on the plane. The line is above the plane because (0,0,0) is above (0, 0, −3/2). −→ −→ 34. The intercepts correspond to the points A(a, 0, 0), B (0, b, 0), and C (0, 0, c). AB × AC = bc, ac, ab is normal to the plane so bcx + acy + abz = abc or x/a + y/b + z/c = 1. 35. v1 = 1, 2, −1 and v2 = −1, −2, 1 are parallel, respectively, to the given lines and to each other so the lines are parallel. Let t = 0 to find the points P1 (−2, 3, 4) and P2 (3, 4, 0) that lie, −→ respectively, on the given lines. v1 × P1 P2 = −7, −1, −9 so 7, 1, 9 is normal to the desired plane whose equation is 7x + y + 9z = 25. 36. The system 4t1 − 1 = 12t2 − 13, t1 + 3 = 6t2 + 1, 1 = 3t2 + 2 has the solution (Exercise 26, Section 13.5) t1 = ...
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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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