Projection matrix How do I find a square $3\times 3$ matrix so that the appropriate function is an orthogonal projection onto a plane $a - 2b + 3c = 0$
I know how to do it with lines. I just rotate it so that my line merges with one of the axes, then use the matrix of orthogonal projection onto an axis $x_1$ or $x_2$ and then finally rotate my line back to its original position. With the plane however, I cannot figure it out.
Thanks for your help.
 A: you can use the householder transformation to get the reflection on the plane as $R = I - \frac 2{u^\top u} uu^\top$ with $u = (1, -2, 3)^\top$ the normal vector to the plane. 
the projection is given $P = \frac 1{u^\top u}uu^\top$
$P$ and $R$ are connected by the relation $2P - I = R.$
A: The projection matrix onto the plane is given by $\bf{P} = I - \bf{v}(\bf{v}^T\bf{v})^{-1}\bf{v}^T$ where $\bf{v}$ is the vector normal to the plane ${\bf{v}} = (1, -2, 3)^T$ in your case. 
A: The following will calculate the part $v_p$ of a vector $v$ which lies in the given plane. I hope this is the projection you have in mind.
Assuming the plane consists of points whose coordinates $(x,y,z)$ fulfill
$$
x - 2 y + 3 z = 0
$$
a normal vector to the plane is
$$
(1, -2, 3)^t
$$
a unit normal vector then is
$$
n = (1/\sqrt{14}) \, (1, -2, 3)^t
$$
A projection of a vector $v$ into the plane has to remove the parts of $v$ which are in normal direction:
$$
v_p = v - (n \cdot v) \, n = (I - n \, n^t) \, v
$$
where I used the idea from Nir's answer for the last equation.
Then 
$$
A =
\left(
\begin{matrix}
1 - n_1^2 & -n_1 n_2 & - n_1 n_3 \\
-n_2 n_1 & 1 - n_2^2 & - n_2 n_3 \\
-n_3 n_1 & -n_3 n_2 & 1 - n_3^2
\end{matrix}
\right)
$$
allows to write this as
$$
v_p = A v
$$
A: How about saying that our orthogonal projection equals to the identity minus the orthogonal projection onto $W^\perp?$
$W^\perp$ is the span of the normal vector $v=(1,−2,3)$, and the orthogonal projection onto which is $x\mapsto \frac{(v\mid x)}{(v\mid v)}v$, and whose matrix is:  $14^{-1}\begin{pmatrix}1\\-2\\3\end{pmatrix}
\begin{pmatrix}1&-2&3\end{pmatrix}
= 14^{-1}\begin{pmatrix}1&-2&3\\-2&4&-6\\3&-6&9\end{pmatrix}.$
Then we substract that from the identity and we should get the orthogonal projection matrix
$ P = \begin{pmatrix}13/14&1/7&-3/14\\1/7&5/7&3/7\\-3/14&3/7&5/14\end{pmatrix}. $
