From example 1 the orbit is 22352 mi above the earth

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Unformatted text preview: 4, t2 = −1/3 so (−17, −1, 1) is the point of intersection. v1 = 4, 1, 0 and v2 = 12, 6, 3 are (respectively) parallel to the lines, v1 × v2 = 3, −12, 12 so 1, −4, 4 is normal to the desired plane whose equation is x − 4y + 4z = −9. 37. n1 = −2, 3, 7 and n2 = 1, 2, −3 are normals to the planes, n1 × n2 = −23, 1, −7 is parallel to the line of intersection. Let z = 0 in both equations and solve for x and y to get x = −11/7, y = −12/7 so (−11/7, −12/7, 0) is on the line, a parametrization of which is x = −11/7 − 23t, y = −12/7 + t, z = −7t. 38. Similar to Exercise 37 with n1 = 3, −5, 2 , n2 = 0, 0, 1 , n1 × n2 = −5, −3, 0 . z = 0 so 3x − 5y = 0, let x = 0 then y = 0 and (0,0,0) is on the line, a parametrization of which is x = −5t, y = −3t, z = 0. √ 39. D = |2(1) − 2(−2) + (3) − 4|/ 4 + 4 + 1 = 5/3 √ 40. D = |3(0) + 6(1) − 2(5) − 5|/ 9 + 36 + 4 = 9/7 √ √ 41. (0,0,0) is on the first plane so D = |6(0) − 3(0) − 3(0) − 5|/ 36 + 9 + 9 = 5/ 54. √ √ 42. (0,0,1) is on the first plane so D = |(0) + (0) + (1) + 1|/ 1 + 1 + 1 = 2/ 3. 43. (1,3,5) and (4,6,7) are on L1 and L2 , respectively. v1 = 7, 1, −3 and v2 = −1, 0, 2 are, respectively, parallel to L1 and L2 , v1 × v2 = 2, −11, 1 so the plane 2x − 11y + z + 51 = 0 √ √ contains L2 and is parallel to L1 , D = |2(1) − 11(3) + (5) + 51|/ 4 + 121 + 1 = 25/ 126. 44. (3,4,1) and (0,3,0) are on L1 and L2 , respectively. v1 = −1, 4, 2 and v2 = 1, 0, 2 are parallel to L1 and L2 , v1 × v2 = 8, 4, −4 = 4 2, 1, −1 so 2x + y − z − 3 = 0 contains L2 and is parallel to √ √ L1 , D = |2(3) + (4) − (1) − 3|/ 4 + 1 + 1 = 6. √ √ 45. The distance between (2, 1, −3) and the plane is |2 − 3(1) + 2(−3) − 4|/ 1 + 9 + 4 = 11/ 14 which is the radius of the sphere; an equation is (x − 2)2 + (y − 1)2 + (z + 3)2 = 121/14. 46. The vector 2i + j − k is normal to the plane and hence parallel to the line so parametric equations of the line are x = 3 + 2t, y = 1 + t, z = −t. Substitution into the equation of the plane yields 2(3 + 2t) + (1 + t) − (−t) = 0, t = −7/6; the point of intersection is (2/3, −1/6, 7/6). 47. v = 1, 2, −1 is parallel to the line, n = 2, −2, −2 is normal to the plane, v · n = 0 so v is parallel to the plane because v and n are perpendicular. (−1, 3, 0) is on the line so √ √ D = |2(−1) − 2(3) − 2(0) + 3|/ 4 + 4 + 4 = 5/ 12 473 Chapter 13 48. (a) (b) n r – r0 P(x0, y0) n · (r − r0 ) = a(x − x0 ) + b(y − y0 ) = 0 P(x, y) r r0 O (c) See the proof of Theorem 13.6.1. Since a and b are not both zero, there is at least one point (x0 , y0 ) that satisfies ax + by + d = 0, so ax0 + bx0 + d = 0. If (x, y ) also satisfies ax + by + d = 0 then, subtracting, a(x − x0 ) + b(y − y0 ) = 0, which is the equation of a line with n = a, b as normal. √ (d) Let Q(x1 , y1 ) be a point on the line, and position the normal n = a, b , with length a2 + b2 , so that its initial point is at Q. The distance is the orthogonal projection of −→ QP0 = x0 − x1 , y0 − y1 onto n. Then −→ D = projn QP 0 −→ |ax0 + by0 + d| QP0 · n √ = n= . 2 n a2 + b2 √ √ 49. D = |2(−3) + (5) − 1|/ 4 + 1 = 2/ 5 50. (a) If x0 , y0 , z0 lies on the second plane, so that ax0 + by0 + cz0 + d2 = 0, then by Theorem | − d2 + d1 | |ax0 + by0 + cz0 + d1 | √ =√ 13.6.2, the distance between the planes is D = 2 + b2 + c2 a a2 + b2 + c2 5 (b) The distance between the planes −2x + y + z = 0 and −2x + y + z + = 0 is 3 5 |0 − 5/3| = √. D= √ 4+1+1 36 EXERCISE SET 13.7 1. (a) elliptic paraboloid, a = 2, b = 3 (b) hyperbolic paraboloid, a = 1, b = 5 (c) hyperboloid of one sheet, a = b = c = 4 (d) circular cone, a = b = 1 (e) elliptic paraboloid, a = 2, b = 1 (f ) hyperboloid of two sheets, a = b = c = 1 √ √ 2. (a) ellipsoid, a = 2/2, b = 1/2, e = 3/3 (b) hyperbolic paraboloid, a = b = 1 (c) hyperboloid of one sheet, a = 1, b = 3, c = 1 (d) hyperboloid of two sheets, a = 1, b = 2, c = 1 √ √ (e) elliptic paraboloid, a = 2, b = 2/2 √ (f ) elliptic cone, a = 2, b = 3 Exercise Set 13.7 474 3. (a) −z = x2 + y 2 , circular paraboloid opening down the negative z -axis z x y (b) z = x2 + y 2 , circular paraboloid, no change (c) z = x2 + y 2 , circular paraboloid, no change (d) z = x2 + y 2 , circular paraboloid, no change z x (e) x = y 2 + z 2 , circular paraboloid opening along the positive x-axis y (f ) y = x2 + z 2 , circular paraboloid opening along the positive y -axis z z y x x y z 4. (a) (b) (c) (d) x +y −z x2 + y 2 − z 2 x2 + y 2 − z 2 x2 + y 2 − z 2 2 2 2 = 1, = 1, = 1, = 1, no no no no change change change change x y 475 Chapter 13 (e) −x2 + y 2 + z 2 = 1, hyperboloid of one sheet with x-axis as axis (f ) x2 − y 2 + z 2 = 1, hyperboloid of one sheet with y -axis as axis z z x x 5. (a) (b) (c) (d) (e) (f ) y y hyperboloid of one sheet, axis is y -axis hyperboloid of two sheets separated by yz -plane elliptic paraboloid opening along the positive x-axis elliptic cone with x-ax...
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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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