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 = kvec n_2`. If all three are collinear, as in type I, then we must have `vec n_1 = kvec n_2 = cvec n_3`, where *k* and *c* are constants. Recall that three vectors are coplanar if and only if `vec n_1 * vec n_2 xx 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.