*a*(

*x*-

*x*

_{0}) +

*b*(

*y*-

*y*

_{0}) +

*c*(

*z*-

*z*

_{0}) = 0,

*a x*+

*b y*+

*c z*=

*d*and

*z*=

*d*+

*m x*+

*n y*

*y*-

*y*

_{0}=

*m*(

*x*-

*x*

_{0}),

*a x*+

*b y*=

*c*and

*y*=

*m x*+

*b*

Geometrically, a plane is just a linear object in more than two
dimensions. A line is, of course, a linear object in two dimensions.
And linear means that it has constant slope (in each direction). Thus
all of the equations
*a*(*x* - *x*_{0}) +
*b*(*y* - *y*_{0}) +
*c*(*z* - *z*_{0}) = 0,

*a x* + *b y* + *c z* = *d* and

*z* = *d* + *m x* + *n y*
are equations of planes, in the same way that each of the equations
*y* - *y*_{0} =
*m* ( *x* - *x*_{0} ),

*a x* + *b y* = *c* and

*y* = *m x* + *b*
is an equation of a line.

We find that different forms of the equation for a plane are useful in
different situations, in the same way that the different forms of the
equation for a line are useful in different situations. If we're
given a slope and the *y*-intercept, we use the slope-intercept
form of a line (the last one given above); if we have the slope and a
point on the line, we use the point-slope form (the first, above).
This same idea holds for planes. All three of the forms written above
really are the same thing, just rendered in a different way. In the
following we look at the same plane in each of these ways to see how
they are equivalent.

Consider the plane with normal vector **n** = <2,4,1>
that goes through the point *P*(1/2,1/2,1). The normal vector
and point are shown without adding the plane and then adding the plane
in figure 1 to the right. (Notice how the normal vector and the
point do exactly determine the plane!) We can easily write the
equation of the plane in all three ways:

2(*x*-1/2) + 4(*y*-1/2) + (*z*-1) = 0,
or, expanding,

2*x* + 4*y* + *z* = 4, or, solving for
*z*,

*z* = 4 - 2*x* - 4*y*.

2

Now suppose we were trying to find the plane through the points
(0,0,4), (2,0,0) and (0,1,0). We can easily find the slope in the
*x* and *y* direction as (change in *z*)/(change in
*x*) and (change in *z*)/(change in
*y*), respectively, using the first two and first and third
points:

which leads us directly to the third of the forms of the equation for the plane above,

Now, how are the *x*-slope and *y*-slopes related to the
coefficients of the normal vector? Going between the two forms

we see that the *x*-slope is *m* = -*a*/*c*
and that the *y*-slope is *n* = -*b*/*c*. Is
this the case for our specific example? Yup! *a* =2,
*b* =4, and *c* =1.

Ok, two more thoughts. Thought 1: *This second example is
pretty special: all of the points we were given were on the
axes*. When one of the points that we're given isn't the
*z*-intercept, or when the points that we're given aren't in
lines in the *x* and *y* directions, it's harder to just
find the intercept and slopes. For example, if we're given
three arbitrary points, say (1,3,-2), (-1,5,2) and (3,3,-8), we're
sort of stuck and find the plane by getting two vectors,
calculating a normal with the cross-product, etc.

Thought 2: is it obvious that the slopes in the *x* and *y*
directions are constant for a plane? By the slope in the
*x*-direction we mean the slope with the *y* coordinate fixed: we
can illustrate this by drawing a plane with a fixed *y*
coordinate and seeing what the slope of the line of intersection is.
This is shown for two values of *y* in figure 2 to the
left. (I've made the plane at *y*=0 red and that at *y*=0.5
grey and wire-frame, so that you can see through it.) Notice that in
either case the slope of the line of intersection with the plane we're
interested in is the same! So the *x*-slopes are in fact the same!

*(How were the figures here generated? In Maple, with this
maple worksheet.)*

Last modified: Wed Jan 21 17:25:37 EST 2004

Comments to:glarose@umich.edu

©2004 Gavin LaRose, UM Math Dept.