Prove that an affine transformation of space maps planes to…

Prove that an affine transformation of space maps planes to planes.

Step 1: Setup

Let $\vec r\,'=M\vec r+\vec b$ with $\det M\ne0$, and let $\pi:\ \vec n\cdot\vec r+D=0$ with $\vec n=(A,B,C)\ne\vec0$. Since $M$ is invertible, $\vec r=M^{-1}(\vec r\,'-\vec b)$.

Step 2: Image equation

A point $\vec r$ lies on $\pi$ iff its image $\vec r\,'$ satisfies

\[ \vec n\cdot M^{-1}(\vec r\,'-\vec b)+D=0\ \Longleftrightarrow\ \vec n\,'\cdot\vec r\,'+D'=0, \]

\[ \vec n\,'=(M^{-1})^T\vec n,\qquad D'=D-\vec n\,'\cdot\vec b. \]

This is a linear equation in $x',y',z'$.

Step 3: It is a plane

If $\vec n\,'=\vec 0$, then $\vec n=M^T\vec n\,'=\vec0$, a contradiction. Hence $\vec n\,'\ne\vec0$ and the image is the plane $\pi'$. Every point of $\pi'$ is the image of a point of $\pi$, because the map is a bijection. $\blacksquare$

\[\mathcal A(\pi)=\pi':\quad \big((M^{-1})^T\vec n\big)\cdot\vec r\,'+D'=0\]

Step 4: Corollaries

Parallel planes. Take $\pi_{1,2}:\ \vec n\cdot\vec r+D_{1,2}=0$ with $D_1\ne D_2$. The images have the same normal $\vec n\,'$, and $D_1'-D_2'=D_1-D_2\ne0$, so $\pi_1'\parallel\pi_2'$.

Lines and ratios. A line is the intersection of two planes, so it maps to a line. A point dividing $M_1M_2$ in the ratio $\lambda$ is $\vec r=\tfrac{\vec r_1+\lambda\vec r_2}{1+\lambda}$. Its image is $\tfrac{\vec r_1\,'+\lambda\vec r_2\,'}{1+\lambda}$, so the simple ratio is preserved.