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

SECTION 12.5 EQUATIONS OF LINES AND PLANES 0 133
therefore bcx + aetJ + abz = abc + 0 + 0 or bcx + acy + abz = abc. Notice that if a ol 0, b ol 0 and c ol 0 then we can
' . X '1j Z 1 Th' . d . be I
rewnte the equat1on as - + - + - = . ts 1s a goo equat1on to remem r.
a b c
65. Two vectors which are perpendicular to the·required line are the normal of the given plane, (1, 1, 1), and a direction vector for
the given line, (1, - 1, 2). So a direction vector for the required line is (1, 1, 1} x (1, -1, 2} = (3, - 1, -2}. Thus L is given
by (x, y, z} = (0, 1, 2} + t(3, -1, - 2}, or in parametric form, x = 3t, y = 1- t, z = 2- 2t.
67. Let Pi have normal vector D i. Then n1 = (3, 6, -3}, n 2 = (4, -12, 8), n a = (3, -9, 6), ll4 = (1, 2, - 1}. Now n 1 = 3D4,
so n1 and n4 are parallel, and hence P1 and P4 are parallel; similarly P 2 and P3 are parallel because n2 = ! n 3. However, n 1
and n 2 are not parallel (so not all four planes are parallel). Notice that the point (2, 0, 0) lies on both P 1 and P 4 , so these two
planes are identical. The point ( ~ , 0, 0) lies on P2 but not on P3, so these are different planes.
69. Let Q = (1, 3, 4) and R = (2, 1, 1), points on the line corresponding to t = 0 and t = 1. Let
. .
---> --+
P = ( 4, 1, - 2). Then a = QR = (1, - 2, -3), b = QP = (3, - 2, - 6) . The distance is
d = Ia x bl = 1(1, -2,-3} x (3,-2,-6}/ = 1(6, - 3,4)1 = J6 2 + (-3)2 + 42 = J6I = {61
/al 1(1, -2, -3}/ 1(1, - 2, - 3}1 JP+(- 2)2 + (-3)2 v'I4 V14'
1 BE
. tl
9
d'ta . D iax1+by1+cz1 +d/ /3(1) +2(- 2)+6(4)-5/ /181 18
7 . y quat1on , 1e IS nee IS = = = r.n = - .
Ja2+b2+ c2 . J 32 + 22 + 62 v49 7
73. Put y = z = 0 in the equation of the first plane to get the point (2, 0, 0) on the plane. Because the planes are parallel, the
distance D between them is the distance from (2, 0, 0) to the second plane. By Equation 9,
D = /4(2) - 6(0) + 2(0) - 31 = _ 5_ = _5_ or 5 VT4.
J42 + ( - 6)2 + (2)2 V56 2 VI4 28
75. The distance between two parallel planes is the same as the distance between a point on on~ ofilie planes and the oilier plane.
Let Po = (xo, yo, zo) be a point on the plane given by ax+ by + cz + dt = 0. Then axo + byo + czo + d 1 = 0 and the
distance between Po and ilie plane given by ax + by + cz + d2 = 0 is, from Equation 9,
D = /ax o + byo + czo + d2/ = /-dl + d2 / , /d1 - d2 /
Ja2 +b2+c2 Ja2 + b2+ c2 Ja2+ b2+c2
77. £ 1: x = y = z =? x = y (1). £2: x + 1 = y/2 = z/3 =? x + 1 = y/2 (2). The solution of (l ) and (2) is
x = y = - 2. However, when x = -2, x = z =? z = - 2, but x + 1 = z/3 =? z = - 3, a contradiction. Hence the
lines do not intersect. For L1. v1 = (1, 1, 1), and for L2, v2 = (1, 2, 3}, so the lines are not parallel. Thus ilie lines are skew
lines. If two lines are skew, they can be viewed as lying in two parallel planes and so ilie distance between the skew lines
would be the same as the distance between these parallel planes. The common normal vector to the planes must be
perpendicular to boili (1, 1, 1) and (1, 2, 3), the direction vectors of the two lines. So set
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