Michael Corral: Vector Calculus

1.6 Surfaces 41
Example 1.27. Find the intersection of the sphere x 2 + y 2 + z 2 = 169 with the plane
z=12.
Solution: The sphere is centered at the origin and has
radius 13= √ 169, so it does intersect the plane z=12.
Putting z=12 into the equation of the sphere gives
x 2 +y 2 +12 2 = 169
x 2 +y 2 = 169−144=25=5 2
whichisacircleofradius5centeredat(0,0,12), parallel
to the xy-plane (see Figure 1.6.2).
x
0
z
Figure 1.6.2
z=12
If the equation in formula (1.29) is multiplied out, we get an equation of the form:
x 2 +y 2 +z 2 +ax+by+cz+d=0 (1.31)
for some constants a, b, c and d. Conversely, an equation of this form may describe a
sphere, which can be determined by completing the square for the x, y and z variables.
Example 1.28. Is 2x 2 +2y 2 +2z 2 −8x+4y−16z+10=0the equation of a sphere?
Solution: Dividing both sides of the equation by 2 gives
x 2 +y 2 +z 2 −4x+2y−8z+5=0
(x 2 −4x+4)+(y 2 +2y+1)+(z 2 −8z+16)+5−4−1−16=0
(x−2) 2 +(y+1) 2 +(z−4) 2 = 16
which is a sphere of radius 4 centered at (2,−1,4).
Example 1.29. Find the points(s) of intersection (if any) of the sphere from Example
1.28 and the line x=3+t, y=1+2t, z=3−t.
Solution: Put the equations of the line into the equation of the sphere, which was
(x−2) 2 +(y+1) 2 +(z−4) 2 = 16, and solve for t:
(3+t−2) 2 +(1+2t+1) 2 +(3−t−4) 2 = 16
(t+1) 2 +(2t+2) 2 +(−t−1) 2 = 16
6t 2 +12t−10=0
The quadratic formula gives the solutions t=−1± 4 √
6
. Putting those two values into
the equations of the line gives the following two points of intersection:
(2+ √ 4 ,−1+ √ 8 ,4− 4 )
√ and
(2−√ 4 ,−1−√ 8 ,4+ 4 )
√
6 6 6 6 6 6
y