Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 13.5 EQUATIONS OF LINES AND PLANES ET SECTION 12.5 ¤ 123
61. The plane contains the points (a, 0, 0), (0,b,0) and (0, 0,c). Thus the vectors a = h−a, b, 0i and b = h−a, 0,ci lie in the
plane, and n = a × b = hbc − 0, 0+ac, 0+abi = hbc, ac, abi is a normal vector to the plane. The equation of the plane is
therefore bcx + acy + abz = abc +0+0or bcx + acy + abz = abc. Notice that if a 6= 0, b 6= 0and c 6= 0then we can
rewrite the equation as x a + y b + z c
=1. This is a good equation to remember!
63. Two vectors which are perpendicular to the required line are the normal of the given plane, h1, 1, 1i, and a direction vector for
the given line, h1, −1, 2i. So a direction vector for the required line is h1, 1, 1i×h1, −1, 2i = h3, −1, −2i. Thus L is given
by hx, y, zi = h0, 1, 2i + th3, −1, −2i, or in parametric form, x =3t, y =1− t, z =2− 2t.
65. Let P i have normal vector n i .Thenn 1 = h4, −2, 6i, n 2 = h4, −2, −2i, n 3 = h−6, 3, −9i, n 4 = h2, −1, −1i. Now
n 1 = − 2 3 n 3,son 1 and n 3 are parallel, and hence P 1 and P 3 are parallel; similarly P 2 and P 4 are parallel because n 2 =2n 4 .
However, n 1 and n 2 are not parallel. 0, 0, 1 2
lies on P1,butnotonP 3, so they are not the same plane, but both P 2 and P 4
contain the point (0, 0, −3),sothesetwoplanesareidentical.
67. Let Q =(1, 3, 4) and R =(2, 1, 1), points on the line corresponding to t =0and t =1.Let
P =(4, 1, −2). Thena = −→
QR = h1, −2, −3i, b = −→
QP = h3, −2, −6i. The distance is
d =
|a × b|
|a|
=
|h1, −2, −3i×h3, −2, −6i|
|h1, −2, −3i|
=
|h6, −3, 4i|
|h1, −2, −3i| =
62 +(−3) 2 +4 2
12 +(−2) 2 +(−3) 2 = √
61
√
14
=
69. By Equation 9, the distance is D = |ax 1 + by 1 + cz 1 + d| |3(1) + 2(−2) + 6(4) − 5|
√ = √ = √ |18| = 18
a2 + b 2 + c 2 32 +2 2 +6 2 49 7 .
64
14 .
71. Put y = z =0in 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) − 3|
42 +(−6) 2 +(2) 2 = 5 √
56
= 5
2 √ 14 or 5 √ 14
28 .
73. The distance between two parallel planes is the same as the distance between a point on one of the planes and the other plane.
Let P 0 =(x 0,y 0,z 0) be a point on the plane given by ax + by + cz + d 1 =0.Thenax 0 + by 0 + cz 0 + d 1 =0and the
distance between P 0 and the plane given by ax + by + cz + d 2 =0is, from Equation 9,
D = |ax 0 + by 0 + cz 0 + d 2 |
√
a2 + b 2 + c 2 = |−d 1 + d 2 |
√
a2 + b 2 + c 2 = |d 1 − d 2 |
√
a2 + b 2 + c 2 .
75. L 1 : x = y = z ⇒ x = y (1). L 2 : x +1=y/2 =z/3 ⇒ x +1=y/2 (2). The solution of (1)and(2)is
x = y = −2. However,whenx = −2, x = z ⇒ z = −2,butx +1=z/3 ⇒ z = −3, a contradiction. Hence the
lines do not intersect. For L 1, v 1 = h1, 1, 1i,andforL 2, v 2 = h1, 2, 3i, so the lines are not parallel. Thus the lines are skew
lines. If two lines are skew, they can be viewed as lying in two parallel planes and so the 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 both h1, 1, 1i and h1, 2, 3i, the direction vectors of the two lines. So set