equations of planes

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 - x₀) + b(y - y₀) + c(z - z₀) = 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 - y₀ = m ( x - x₀ ),
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,
2x + 4y + z = 4, or, solving for z,
z = 4 - 2x - 4y.

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:

x-slope = (delta z)/(delta x) = (0 - 4)/(2 - 0) = -2
y-slope = (delta z)/(delta y) = (0 - 4)/(1 - 0) = -4

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!