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

Section 12.5 Equations of Lines and Planes 829
Another way to find the line of intersection
is to solve the equations of the
planes for two of the variables in terms
of the third, which can be taken as the
parameter.
z
2
1
0
_1
_2
y z
=
2 3
_1
0
y
FIGURE 11
x-1
5
=
y
_2
L
0
1 2 1 _1 _2
x
Figure 11 shows how the line L in
Example 7 can also be regarded as the
line of intersection of planes derived
from its symmetric equations.
equations x 1 y − 1 and x 2 2y − 1, whose solution is x − 1, y − 0. So the point
s1, 0, 0d lies on L.
Now we observe that, since L lies in both planes, it is perpendicular to both of the
normal vectors. Thus a vector v parallel to L is given by the cross product
v − n 1 3 n 2 − Z
i
1
1
j
1
22
and so the symmetric equations of L can be written as
x 2 1
5
−
k
1
3
Z − 5i 2 2 j 2 3 k
y
22 − z
23
Note Since a linear equation in x, y, and z represents a plane and two nonparallel
planes intersect in a line, it follows that two linear equations can represent a line. The points
sx, y, zd that satisfy both a 1 x 1 b 1 y 1 c 1 z 1 d 1 − 0 and a 2 x 1 b 2 y 1 c 2 z 1 d 2 − 0 lie
on both of these planes, and so the pair of linear equations represents the line of intersection
of the planes (if they are not parallel). For instance, in Example 7 the line L was
given as the line of intersection of the planes x 1 y 1 z − 1 and x 2 2y 1 3z − 1. The
symmetric equations that we found for L could be written as
x 2 1
5
−
y
22
and
y
22 − z
23
which is again a pair of linear equations. They exhibit L as the line of intersection of the
planes sx 2 1dy5 − yys22d and yys22d − zys23d. (See Figure 11.)
In general, when we write the equations of a line in the symmetric form
x 2 x 0
a
− y 2 y 0
b
− z 2 z 0
c
we can regard the line as the line of intersection of the two planes
■
x 2 x 0
a
− y 2 y 0
b
and
y 2 y 0
b
− z 2 z 0
c
Distances
ExamplE 8 Find a formula for the distance D from a point P 1 sx 1 , y 1 , z 1 d to the
plane ax 1 by 1 cz 1 d − 0.
SOLUtion Let P 0 sx 0 , y 0 , z 0 d be any point in the given plane and let b be the vector
corresponding to P¸P¡ A. Then
b − k x 1 2 x 0 , y 1 2 y 0 , z 1 2 z 0 l
P¸
FIGURE 12
b
¨
n
P¡
D
From Figure 12 you can see that the distance D from P 1 to the plane is equal to the
absolute value of the scalar projection of b onto the normal vector n − ka, b, c l . (See
Section 12.3.) Thus
D − | comp n b | − | n ? b |
| n |
− | asx 1 2 x 0 d 1 bsy 1 2 y 0 d 1 csz 1 2 z 0 d |
sa 2 1 b 2 1 c 2
− | sax 1 1 by 1 1 cz 1 d 2 sax 0 1 by 0 1 cz 0 d |
sa 2 1 b 2 1 c 2
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