Systems of three planes

There are five ways that three distinct planes can relate to one another:

type	intersection	condition
I	none	all normals collinear
II	2 lines	two normals collinear
III	3 lines	all normals coplanar
IV	1 line	all normals coplanar
V	1 point	normals not coplanar

Strictly speaking, systems of type II and III really have no intersection because the three planes do not share any points at all. It might seem inconsistent to talk about these lines of intersection between two of the three planes, since we ignore them in type V (which has many such lines in addition to the point), but it is just for the purpose of identifying the type.

When we say two normals ${\vec{n}}_{1}$ and ${\vec{n}}_{2}$ are collinear, we mean ${\vec{n}}_{1} = k {\vec{n}}_{2} .$ If all three are collinear, as in type I, then we must have ${\vec{n}}_{1} = k {\vec{n}}_{2} = c {\vec{n}}_{3},$ where k and c are constants. Recall that three vectors are coplanar if and only if ${\vec{n}}_{1} \cdot {\vec{n}}_{2} \times {\vec{n}}_{3} = 0,$ and the order of the vectors does not matter.

Here is how we should go about determining the type of a system:

Are all three normals collinear? Type I.
Are two normals collinear? Type II.
Attempt to solve the system using Gauss-Jordan elimination.
Can you get to the reduced row-echelon form? Type V.
Do you end up with a row of three zeros followed by a nonzero value in the right column (an impossible equation)? Type III.
You must eliminate an entire row (four zeros). Type IV.

In the case of type IV, you can find the equation of the line of intersection by abandoning the matrix once you get a row zeroed out and rewrite the other two rows as equations. One variable will be in both equations, and the other two will not. Suppose this variable is x. Write “let $x = t,$ ” then replace x with t in the equations and solve for y and z. You now have the parametric equations of a line. Its direction should be collinear with the result of crossing any two of the normal vectors.