Transformation of the points on a plane

Question

How do I transform a point $(x,y,z)$ on plane $\Pi (ax + by + cz = 0)$ to a point $(x',y',z')$ on plane $\Phi(ax+by+cz+d=0)$?

What matrix should I use?

Here is a 2-D representation of what I'm trying to do:

Answer 1

Any element of the first plane is the form:

$$\begin{bmatrix}x \\ y \\ \frac{-ax-by}{c} \end{bmatrix}=\begin{bmatrix}1 \\ 0 \\ -\frac{a}{c} \end{bmatrix}x+\begin{bmatrix}0 \\ -1 \\ -\frac{b}{c} \end{bmatrix}y$$

$$\Pi=\text{span} \left< \begin{bmatrix}1 \\ 0 \\ -\frac{a}{c} \end{bmatrix}, \begin{bmatrix}0 \\ -1 \\ -\frac{b}{c} \end{bmatrix} \right>$$

Any element of the second plane is:

$$\begin{bmatrix}x \\ y \\ \frac{-ax-by-d}{c} \end{bmatrix}=\begin{bmatrix}1 \\ 0 \\ -\frac{a}{c} \end{bmatrix}x+\begin{bmatrix}0 \\ -1 \\ -\frac{b}{c} \end{bmatrix}y+\begin{bmatrix}0 \\ 0 \\ -\frac{d}{c} \end{bmatrix}$$

$$\Phi=\text{span} \left< \begin{bmatrix}1 \\ 0 \\ -\frac{a}{c} \end{bmatrix}, \begin{bmatrix}0 \\ -1 \\ -\frac{b}{c} \end{bmatrix} \right> + \begin{bmatrix}0 \\ 0 \\ -\frac{d}{c} \end{bmatrix} $$

Call $v_1,v_2,v_3$ the vectors involved in the last equation (which are lineary independent). Any element of the planes $\Pi$ and $\Phi$ is writed in the form:

$$v_\Pi=\alpha v_1 + \beta v_2$$ $$v_\Phi=\delta v_1 + \epsilon v_2 + v_3$$

If such transformation exists a matrix $A$ would exist with the property:

$$Av_\Pi=\alpha A v_1 + \beta A v_2$$

$$\text{span}<Av_1,Av_2>=\text{span}<v_1,v_2>+v_3$$

Answer 2

Lets use vector notation.

Denote plane $\Pi$ as $\Pi(\textbf x) = \textbf n \cdot \textbf x = 0$, where $\textbf n=(a,b,c)$, $\textbf x = (x,y,z)$, $\Phi(\textbf x) = \textbf n \cdot \textbf x + d = 0$.

Now we have point $\textbf x$ on plane $\Pi$, $\Pi(\textbf x) = 0 $. We want to translate it by vector parallel to $\textbf n$ and get point on $\Phi$. Therefore we are looking for $s$ for which $\Phi(\textbf x+s\textbf n) = 0$

$$\Phi(\textbf x+s\textbf n) = \textbf n \cdot (\textbf x + s\textbf n ) + d =\textbf n \cdot \textbf x + s \textbf n \cdot \textbf n + d = \Pi(\textbf x) + s \|\textbf n\|^2 + d = s\|\textbf n\|^2 + d = 0$$

so $$ s = - \frac{d}{\|\textbf n\|^2} $$ Thus we need to translate point $\textbf x$ by vector $- \textbf n \frac{d}{\|\textbf n\|^2}$.