James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

828 Chapter 12 Vectors and the Geometry of Space
Figure 8 shows the portion of the
plane in Example 5 that is enclosed by
triangle PQR.
x
Q(3, _1, 6)
FIGURE 8
z
R(5, 2, 0)
P(1, 3, 2)
y
ExamplE 5 Find an equation of the plane that passes through the points Ps1, 3, 2d,
Qs3, 21, 6d, and Rs5, 2, 0d.
SOLUtion The vectors a and b corresponding to PQ l and PR l are
a − k 2, 24, 4 l
b − k4, 21, 22 l
Since both a and b lie in the plane, their cross product a 3 b is orthogonal to the plane
and can be taken as the normal vector. Thus
Z
i j k
Z
n − a 3 b − 2 24 4 − 12 i 1 20 j 1 14 k
4 21 22
With the point Ps1, 3, 2d and the normal vector n, an equation of the plane is
12sx 2 1d 1 20sy 2 3d 1 14sz 2 2d − 0
or 6x 1 10y 1 7z − 50 ■
ExamplE 6 Find the point at which the line with parametric equations x − 2 1 3t,
y − 24t, z − 5 1 t intersects the plane 4x 1 5y 2 2z − 18.
SOLUtion We substitute the expressions for x, y, and z from the parametric equations
into the equation of the plane:
4s2 1 3td 1 5s24td 2 2s5 1 td − 18
This simplifies to 210t − 20, so t − 22. Therefore the point of intersection occurs
when the parameter value is t − 22. Then x − 2 1 3s22d − 24, y − 24s22d − 8,
z − 5 2 2 − 3 and so the point of intersection is s24, 8, 3d.
■
FIGURE 9
n ¨ n¡
¨
Two planes are parallel if their normal vectors are parallel. For instance, the planes
x 1 2y 2 3z − 4 and 2x 1 4y 2 6z − 3 are parallel because their normal vectors are
n 1 − k1, 2, 23 l and n 2 − k2, 4, 26 l and n 2 − 2n 1 . If two planes are not parallel, then
they intersect in a straight line and the angle between the two planes is defined as the
acute angle between their normal vectors (see angle in Figure 9).
Figure 10 shows the planes in Example
7 and their line of intersection L.
x+y+z=1
x-2y+3z=1
6
L
4
2
z 0
_2
_4
_2 0
y 2
0
_2
2 x
FIGURE 10
ExamplE 7
(a) Find the angle between the planes x 1 y 1 z − 1 and x 2 2y 1 3z − 1.
(b) Find symmetric equations for the line of intersection L of these two planes.
SOLUTION
(a) The normal vectors of these planes are
n 1 − k1, 1, 1 l
n 2 − k1, 22, 3 l
and so, if is the angle between the planes, Corollary 12.3.6 gives
cos − n 1 ? n 2 1s1d 1 1s22d 1 1s3d
| n 1 || n − − 2
2 | s1 1 1 1 1 s1 1 4 1 9 s42
− cos 21S 2
s42
D < 72°
(b) We first need to find a point on L. For instance, we can find the point where the line
intersects the xy-plane by setting z − 0 in the equations of both planes. This gives the
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