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

Section 12.1 Three-Dimensional Coordinate Systems 795
Distance Formula in Three Dimensions The distance | P 1P 2 | between the points
P 1 sx 1 , y 1 , z 1 d and P 2 sx 2 , y 2 , z 2 d is
| P 1P 2 | − ssx 2 2 x 1 d 2 1 sy 2 2 y 1 d 2 1 sz 2 2 z 1 d 2
z
P¡(⁄, ›, z¡)
P(¤, fi, z)
To see why this formula is true, we construct a rectangular box as in Figure 11, where
P 1 and P 2 are opposite vertices and the faces of the box are parallel to the coordinate
planes. If Asx 2 , y 1 , z 1 d and Bsx 2 , y 2 , z 1 d are the vertices of the box indicated in the figure,
then
| P 1 A | − | x 2 2 x 1 | | AB | − | y 2 2 y 1 | | BP 2 | − | z 2 2 z 1 |
x
0
FIGURE 11
FIGURE 11
B(¤, fi, z¡)
A(¤, ›, z¡)
7et120111
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MasterID: 01380
y
Because triangles P 1 BP 2 and P 1 AB are both right-angled, two applications of the Pythagorean
Theorem give
| P 1P 2 | 2 − | P 1B | 2 1 | BP 2 | 2
and | P 1B | 2 − | P 1A | 2 1 | AB | 2
Combining these equations, we get
| P 1P 2 | 2 − | P 1A | 2 1 | AB | 2 1 | BP 2 | 2
− | x 2 2 x 1 | 2 1 | y 2 2 y 1 | 2 1 | z 2 2 z 1 | 2
− sx 2 2 x 1 d 2 1 sy 2 2 y 1 d 2 1 sz 2 2 z 1 d 2
Therefore | P 1P 2 | − ssx 2 2 x 1 d 2 1 sy 2 2 y 1 d 2 1 sz 2 2 z 1 d 2
Example 4 The distance from the point Ps2, 21, 7d to the point Qs1, 23, 5d is
| PQ | − ss1 2 2d2 1 s23 1 1d 2 1 s5 2 7d 2 − s1 1 4 1 4 − 3 ■
z
Example 5 Find an equation of a sphere with radius r and center Csh, k, ld.
r
P (x, y , z)
SOLUtion By definition, a sphere is the set of all points Psx, y, zd whose distance from
− r. Squaring both
C is r. (See Figure 12.) Thus P is on the sphere if and only if | PC |
sides, we have | PC | 2 − r 2 or
C (h, k, l)
sx 2 hd 2 1 sy 2 kd 2 1 sz 2 ld 2 − r 2
■
x
0
FIGURE 12
y
The result of Example 5 is worth remembering.
Equation of a Sphere An equation of a sphere with center Csh, k, ld and radius r
is
sx 2 hd 2 1 sy 2 kd 2 1 sz 2 ld 2 − r 2
In particular, if the center is the origin O, then an equation of the sphere is
x 2 1 y 2 1 z 2 − r 2
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