Relative position of two planes

Let's see now the relative positions that can have two planes, $\pi(P;\overrightarrow{u}, \overrightarrow{v})$ and $\pi(Q;\overrightarrow{u'}, \overrightarrow{v'})$, both expressed by means of their general equations:

$$\begin{array}{rrcl} \pi:&Ax+By+Cz+D &=&0\\ \pi':&A'x+B'y+C'z+D'&=&0\end{array}$$

To find the relative positions, let's consider the system formed by two equations, with its matrix $M$ and its extended matrix $M'$:

$$M=\begin{pmatrix} A & B & C \\ A' & B' & C' \end{pmatrix}$$

$$M'=\begin{pmatrix}A & B & C & -D \\ A' & B' & C' & -D' \end{pmatrix}$$

Equal planes

$$rank (M) = rank (M') = 1$$

It is equivalent to:

$$\displaystyle \frac{A}{A'}=\frac{B}{B'}=\frac{C}{C'}=\frac{D}{D'}$$

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

Consider the planes $\pi$ and $\pi'$'

$$\begin{array}{rrcl}\pi:& 2x - 3y + z - 1 &=& 0\\ \pi':&-4x + 6y - 2z +2& =& 0\end {array}$$

They are the same plane since:

$$\displaystyle \frac{2}{-4}=\frac{-3}{6}=\frac{1}{-2}=\frac{-1}{2}$$

Parallel planes

$$rank(M) = 1, rank (M') = 2$$

It is equivalent to:

$$\displaystyle \frac{A}{A'}=\frac{B}{B'}=\frac{C}{C'}\neq \frac{D}{D'}$$

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

Consider the planes $\pi$ and $\pi'$'

$$\begin{array}{rrcl}\pi:& 2x - 3y + z - 1 &=& 0\\ \pi':&-4x + 6y - 2z +7& =& 0\end {array}$$

They are parallel planes since:

$$\displaystyle \frac{2}{-4}=\frac{-3}{6}=\frac{1}{-2}\neq\frac{-1}{7}$$

Secant planes

$$rank(M) = rank (M') = 2$$

It is equivalent to:

$$\displaystyle \frac{A}{A'} \neq \frac{B}{B'} \mbox{ o } \frac{A}{A'} \neq \frac{C}{C'} \mbox{ o } \frac{B}{B'} \neq \frac{C}{C'}$$

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

Consider the planes $\pi$ and $\pi'$

$$\begin{array}{rrcl}\pi:& 2x - 3y + z - 1 &=& 0\\ \pi':&-x + y - 2z +2& =& 0\end {array}$$

It is a question of secant planes since:

$$\displaystyle \frac{2}{-1}\neq\frac{-3}{1}$$