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

830 Chapter 12 Vectors and the Geometry of Space
Since P 0 lies in the plane, its coordinates satisfy the equation of the plane and so we
have ax 0 1 by 0 1 cz 0 1 d − 0. Thus the formula for D can be written as
9 D − | ax 1 1 by 1 1 cz 1 1 d |
sa 2 1 b 2 1 c 2
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ExamplE 9 Find the distance between the parallel planes 10x 1 2y 2 2z − 5 and
5x 1 y 2 z − 1.
SOLUtion First we note that the planes are parallel because their normal vectors
k10, 2, 22 l and k5, 1, 21 l are parallel. To find the distance D between the planes, we
choose any point on one plane and calculate its distance to the other plane. In particular,
if we put y − z − 0 in the equation of the first plane, we get 10x − 5 and so
( 1 2 , 0, 0) is a point in this plane. By Formula 9, the distance between ( 1 2 , 0, 0) and the
plane 5x 1 y 2 z 2 1 − 0 is
D − | 5( 1 2 ) 1 1s0d 2 1s0d 2 1 |
s5 2 1 1 2 1 s21d 2
−
3
2
− s3
3s3 6
So the distance between the planes is s3 y6.
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ExamplE 10 In Example 3 we showed that the lines
L 1 : x − 1 1 t y − 22 1 3t z − 4 2 t
L 2 : x − 2s y − 3 1 s z − 23 1 4s
FIGURE 13
Skew lines, like those in Example 10,
always lie on (nonidentical) parallel
planes.
are skew. Find the distance between them.
SOLUtion Since the two lines L 1 and L 2 are skew, they can be viewed as lying on two
parallel planes P 1 and P 2 . The distance between L 1 and L 2 is the same as the distance
between P 1 and P 2 , which can be computed as in Example 9. The common normal vector
to both planes must be orthogonal to both v 1 − k1, 3, 21 l (the direction of L 1 ) and
v 2 − k2, 1, 4 l (the direction of L 2 ). So a normal vector is
Z
i
n − v 1 3 v 2 − 1
2
j
3
1
k
Z
21 − 13i 2 6 j 2 5k
4
If we put s − 0 in the equations of L 2 , we get the point s0, 3, 23d on L 2 and so an
equation for P 2 is
13sx 2 0d 2 6sy 2 3d 2 5sz 1 3d − 0 or 13x 2 6y 2 5z 1 3 − 0
If we now set t − 0 in the equations for L 1 , we get the point s1, 22, 4d on P 1 . So
the distance between L 1 and L 2 is the same as the distance from s1, 22, 4d to
13x 2 6y 2 5z 1 3 − 0. By Formula 9, this distance is
| D − 13s1d 2 6s22d 2 5s4d 1 3 |
− 8 < 0.53
s13 2 1 s26d 2 1 s25d 2 s230
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