# Relative position of two planes

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