Calculus: Analytic Geometry and Calculus, with Vectors

2.4 Planes and lines in E, 79

A, B, C are not all 0, then the equation

14x+By+Cz+D=0

is the equation of a plane. In case .4 s 0, we can accomplish the result
by putting the equation in the form

_4 (x- Dl+B(y-0)+C(z-0)=0

and noticing that it is the equation of the plane which passes through the
point (-D/A, 0, 0) and is normal to the vector having scalar components
11, B, C. In case B 0 or C 54 0, the proof is similar.
The information which we have obtained can be useful in various ways.
Suppose, for example, we are required to find the equation of the plane 7r
which contains three given noncollinear (not on a line) points Pi(xl,yj,zj),

P2(x2,Y2,z2), P3(x3,y3,z3) The obvious way to solve this problem is to use
the fact that the equation of 9r must have the form of the first equation in
the system
14x+By+Cz+D=0

44 ) Ax1 + By, + Cz1 + D = 0

(2 14x2+By2+Cz2+D=0

Ax3 + By3 + Cz3 + D =O,

where 4, B, C, D are constants for which 14, B, C are not all 0. Since it

contains P1, P2, P3, the remaining three equations must be satisfied. In

case A 0 0, the equation of a can be obtained by solving the last three

equations for B, C, D in terms of J and substituting the results in the

first equation. In case A = 0, we must have either B 34 0 or C 0 0, and

a similar procedure will work. Except for cases in which some of the

coordinates of the given points are zero, solving the problem in this way

can be tedious. The next section will show how answers to this and

other problems can be expressed in terms of determinants.

We now look at an interesting procedure which often provides a good

way to find the equation of a plane Irl which contains two given points

P.1(x1,y1,z1) andP2(x2,y2,z2) and satisfies another condition. We simplify

matters by supposing that x2 F6 x1, y2 0 yl, and z2 0 z1. We know that

P1 and P2 determine a line L and that the family F of planes 7r that contain

P1 and P2 is identical with the family of planes in that contain L. We

capitalize this fact. If a point P(x,y,z) lies on L, then

x-x1 = y - yl Z -%I

X2 - x1 y2 - yi z2 z1

and hence

(2.45) x-x1- z-zl

+Y-yl - z -zl0

(x2- x1 z2 -zl) \Y- Yl z2 - zl

or

(2.451) (x - x1) + (Y - yl) -

A

+ fM (z- Si) = 0,

x2 x1 Ys - Yl z2 - zl