Equations of planes

A plane is a flat, two-dimensional surface that extends to infinity. You can have a plane in `RR^2`, but we don’t talk about that often because it’s not very interesting. There is only one plane in two-space, the xy plane. However, in three-space, there are `oo` planes.

Consider a point, defined by a tuple of coordinates. The point has zero dimensions because there is only one point. Now consider a line—a line is just a set of `oo` points that happen to lie in a straight line (yes, this is a circular definition). The line is one-dimensional and has one degree of freedom; it is controlled by the free parameter t in

`vec r = vec r_0 + tvec m, qquad t in RR`.

Now consider a plane. A plane another set of `oo` points, and these ones all lie on a flat surface. You could also think of it as `oo` parallel lines. The plane is two-dimensional and has two degrees of freedom; it is controlled by the free parameters s and t in

`vec r = vec r_0 + svec u + tvec v, qquad s in RR and t in RR`.

That’s the vector equation of a plane. As long as `vec u` and `vec v` are non-collinear, this allows `vec r` to move around anywhere on the plane by choosing different values for the free parameters.

We are trying to generalize ideas about sets of points in different dimensions. It is easier to see the natural progression of adding degrees of freedom with a drawing—but don’t forget that these are really two-dimensional representations:

pointlineplanesolid
Geometric figures that can be represented in zero, one, two, and three dimensions

The arrows on the line indicates that it extends to infinity in both directions. You may be tempted to think of the plane as a quadrilateral, but it is most definitely not. The plane doesn’t really have edges—it’s just hard to draw it without them. The same thing goes for the solid: it’s not a cube. It extends to infinity, so there is only one solid in three-space: the xyz solid.

As I mentioned before, there is a single plane in `RR^2` but many planes in `RR^3`. We can generalize this idea to all the geometrical figures (sets of points) that we’ve seen so far:

Dimensions Space Many One
0 `RR^0` point
1 `RR^1` point line
2 `RR^2` line plane
3 `RR^3` plane solid
4 `RR^4` solid hypersolid

Let’s look at the equation of the plane again:

`vec r = vec r_0 + svec u + tvec v, qquad s in RR and t in RR`.

We can substitute components like we did for the equation of the line:

`[x,y,z] = [x_0,y_0,z_0] + s[u_1,u_2,u_3] + t[v_1,v_2,v_3]`.

From this, we can get the parametric equations of the plane:

`x = x_0 + su_1 + tv_1, quad y = y_0 + su_2 + tv_2, and z = z_0 + su_3 + tv_3`.

Another way to describe a plan is with a point and a normal vector. The normal vector is perpendicular to the plane, and we get it by crossing the plane’s two direction vectors:

`vec n = vec u xx vec v`.

Given this normal vector and a position vector `vec r_0`, every point `vec r` on the plane satisfies the equation

`vec n * (vec r - vec r_0) = 0`.

This works because the dot product of perpendicular vectors is always zero, and the vector `vec r - vec r_0` is parallel to the plane. You can use points instead of position vectors if you wish: `vec n * vec(AP) = 0`, where A is the initial point and P represents all other points on the plane. Suppose we replace the vectors in that equation with components:

`[a,b,c] * ([x,y,z] - [x_0,y_0,z_0]) = 0`.

We can simplify this to give us

`a(x-x_0) + b(y-y_0) + c(z-z_0) = 0`.

When we actually have numbers for `vec r_0`, we usually write this as

`ax + by + cz + d = 0`,

where `d = -(ax_0 + by_0 + cz_0)`. This is called the standard form of the equation of a plane. It is just like the standard form of a line in `RR^2`, but with an added term for the third dimension. The standard form of a plane is also useful because the normal vector is in plain sight. Given a normal vector `vec n = [a,b,c]` and a point P, we can jump straight to the standard form equation by substituting the components of `vec n` into the equation above. We still don’t know d, so we need to substitute in the point and solve for it.

You’ll notice that the value of d and the normal vector are independent. Since d doesn’t affect the direction of the plane, it must affect something else. If we set it to zero, the plane passes through the origin. If we set it to one, the shortest distance between the origin and the plane is one. The value of d effectively moves the plane away from the origin in the direction of the normal vector.

The normal vector is useful for another reason: we can use it to find the angle between two planes. If we have planes with normal vectors `vec n_1` and `vec n_2`, the acute angle separating them is given by

`theta_"acute" = cos^-1((|vec n_1 * vec n_2|)/(|vec n_1||vec n_2|))`.

The vertical bars in the numerator of this equation mean absolute value. In the bottom, they mean vector magnitude. Don’t get confused and think that the dot product should produce a vector instead of a scalar. (For this reason, it is more common to use the notation `||vec v||` for the norm and reserve `|a|` for the absolute value.)