Relative position of two planes

Let's see now the relative positions that can have two planes, $\pi (P;\overrightarrow{u},\overrightarrow{v})$ and $\pi (Q;\overrightarrow{u^{\prime }},\overrightarrow{v^{\prime }})$, both expressed by means of their general equations: $$\begin{array}{rrcl}\pi :& Ax+By+Cz+D& =& 0\\ {\pi }^{\prime }:& A^{\prime }x+B^{\prime }y+C^{\prime }z+D^{\prime }& =& 0\end{array}$$

To find the relative positions, let's consider the system formed by two equations, with its matrix $M$ and its extended matrix $M^{\prime }$: $$M=\left(\begin{array}{ccc}A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right)$$ $$M^{\prime }=\left(\begin{array}{cccc}A& B& C& -D\\ A^{\prime }& B^{\prime }& C^{\prime }& -D^{\prime }\end{array}\right)$$

Equal planes

$$rank(M)=rank(M^{\prime })=1$$

It is equivalent to: $${\displaystyle \frac{A}{A^{\prime }}=\frac{B}{B^{\prime }}=\frac{C}{C^{\prime }}=\frac{D}{D^{\prime }}}$$ Indeterminate compatible system.

The solution to the system depends on two parameters. The planes are equal.

Parallel planes

$$rank(M)=1,rank(M^{\prime })=2$$

It is equivalent to: $${\displaystyle \frac{A}{A^{\prime }}=\frac{B}{B^{\prime }}=\frac{C}{C^{\prime }}\ne \frac{D}{D^{\prime }}}$$ Incompatible system.

The system has no solution. There are no common points. The planes are parallel.

Secant planes

$$rank(M)=rank(M^{\prime })=2$$

It is equivalent to: $${\displaystyle \frac{A}{A^{\prime }}\ne \frac{B}{B^{\prime }}\text{ o }\frac{A}{A^{\prime }}\ne \frac{C}{C^{\prime }}\text{ o }\frac{B}{B^{\prime }}\ne \frac{C}{C^{\prime }}}$$ Indeterminate compatible system.

The solution of the system depends on a parameter. The planes are secant, that is, they cut in a straight line.