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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.

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  • $\begingroup$ Calling your plane $x-2y+3z=0$ seems to be more standard. $\endgroup$ – Element118 Nov 9 '15 at 11:22
  • $\begingroup$ Hint: $Tx=xTe_1+yTe_2+zTe_3$ where $e_1, e_2, e_3$ is the standard basis. You have lots of choices as to where $Te_1, Te_2$ and $Te_3$ end up on the plane. $\endgroup$ – Paul Nov 9 '15 at 11:26
  • $\begingroup$ ah, orthogonal projection. Was orthogonal in there originally?? $\endgroup$ – Paul Nov 9 '15 at 12:49
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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.

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  • $\begingroup$ Looks nice, but $(v^t v)^{-1} = v^{-1} (v^t)^{-1}$ and would cancel out to $P=0$, or? $\endgroup$ – mvw Nov 9 '15 at 12:06
  • $\begingroup$ what you wrote is true for square matrices only. $\bf{v}$ is a vector. It won't cancel out. Try it. $\endgroup$ – Nir Regev Nov 9 '15 at 12:08
  • $\begingroup$ Ah, I see. Thanks. $\endgroup$ – mvw Nov 9 '15 at 12:08
  • $\begingroup$ @abel what you wrote is a reflection, not a projection. $\endgroup$ – Nir Regev Nov 9 '15 at 12:16
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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 $2P - I = R.$

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  • $\begingroup$ what you wrote is a reflection, not a projection. $\endgroup$ – Nir Regev Nov 9 '15 at 21:07
  • $\begingroup$ @NirRegev, you are right. i have edited my answer. $\endgroup$ – abel Nov 10 '15 at 0:22
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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 $$ v_p = A v $$

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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}. $

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