Intersecting planes example

Example

Find intersection of planes given by x + y + z + 1 = 0 and x + 2y + 3z + 4 = 0.

Solution: In three dimensions (which we are implicitly working with here), what is the intersection of two planes? As long as the planes are not parallel, they should intersect in a line. So our result should be a line.

By now, you should know that you need to think a bit to find the equation for a line in three dimensions. In particular, we need to find a parametrization for a line.

How can we get a parametrization for the line formed by the intersection of these two planes? We need an equation for all (x,y,z) that satisfy both x + y + z + 1 = 0 and x + 2y + 3z + 4 = 0. This is a system of equations, and we attempt to solve this system. (This is a system of two equations and three unknowns, so we know we can’t solve it for a unique (x,y,z). But that makes sense, because we want a whole bunch of (x,y,z), i.e., a whole line of points.)

We could use either the method of “substitution” or the method of “elimination” to solve these equations. We’ll use the method of elimination. Since in both equations, x has a coefficient of 1, we’ll subtract the two equations to eliminate x. If we subtract x + y + z + 1 = 0 from x + 2y + 3z + 4 = 0, we are left with

y + 2z + 3 = 0
which we can rewrite as y = -2z - 3. Now, we don’t have enough equations to uniquely solve for either y or z. So, we’ll simply set z = t, where t can be any real number. Then y = -2t - 3. To find x in terms of t, we simply plug these values of y and z into one of the original equations. We’ll use the first one so that
x = -y - z - 1 = 2t + 3 - t - 1 = t + 2.
We can write our parametrization of the line as
(x,y,z) = (t + 2,-2t - 3,t).