Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

132 D CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
49. Setting x = 0, we see that (0, 1, 0) satisfies the equatio~s of both planes, so that they do in fact have a line of intersection.
v = o 1 x 0 2 = (1, 1, 1} x (1, 0, 1) = (1, 0, -1} is the direction of this line. Therefore, direction numbers of the intersecting
line are 1, 0, - 1.
51. Normal vectors for the planes are 01 = (1, 4, -3} and 0 2 = ( -3, 6, 7), so the normals (and thus the planes) aren't parallel.
. .
But 01 · 0 2 ,;, -3 + 24 - 21 = 0, so the normals (and thus the planes) are perpendicular.
53. Normal vectors for the planes are 01 = (1, 1, 1) and 02 = (1, -1, 1). The normals are not parallel, so neither are tl1e planes.
Furthe'rmore, o 1 · o 2 = 1 - 1 + 1 = 1 =F 0, so the planes aren't perpendicular. The angle between them is given by·
= __ 1_ = ~ 1(1)
lolllo21 .j3.j3 3 => O=cos- 3 ::::::70.5o.
cos8 = O t . 0 2
55. The normals are 0 1 = (1, -4, 2) and 0 2 = (2, -8, 4). Since 02 = 2o 1 , the normals (and tlms the planes) are parallel.
57. (a) To find a point on the line of intersection, set one of the variables equal to a constant, say z = 0. (This will fail if the line .of
intersection does not cross the xy-plane; in that case, try setting x or y equal to 0.) The equations of the two planes reduce
to x + y = 1 and x + 2y = 1. Solving these two equations gives x = 1, y = 0. Thus a point on ilie line is (1, 0,·0).
A vector v in the direction of this intersecting line is perpendicular to the normal vectors of both planes, so we can take
v = o 1 x o 2 = (1, 1, 1) x (1, 2, 2) = (2- 2, 1 - 2, 2- 1) = (0, - 1, 1}. By Equations 2, parametric equations for tlle
line are .'1: = 1, y = - t, z = t.
. 0] . 0 2 1 + 2 + 2 5 h fi 0 -1 ( 5 ) 15 80
(b) The angle between the planes satisfies cos 0 = I II I = r;; If\ = r;;. T ere ore = cos r;; :::::: . .
0] 02 v 3 v 9. 3 v 3 3 v 3
59. Setting z = 0, tlle equations of the two planes become 5x- 2y = 1 and 4x + y = 6. Solving tllese two equations gives
x = 1, y = 2 so a point on the line of intersection is (1, 2, 0). A vector v in the direction of this intersecting line is
perpendicular to the normal vectors ofbotll planes. So we can use v = o 1 x 0 2 = (5, - 2, -2) x (4, 1, 1} = (0, - 13, 13} or
equivalently we can take v = (0, -1, 1}, and symmetric equations for the line are x = 1, y _=- 1
2 = I or x = 1, y-2 = -z.
61. The distance from a point (x, y , z) to (1, 0, - 2) is d 1 = ..j(x- 1)2 + y 2 + (z + 2)2 and the distance from (x, y, z) to
(3,4, 0) is d 2 = ..j(x- 3) 2 + (y- 4) 2 + z 2 . The plane consists of all points (x, y, z) where d1 = d2 => d{ = di