Calculating matrix for linear transformation of orthogonal projection onto plane. The question goes like this:
"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane  
$$ 2x_1+2x_2+x_3^{}= 0" $$
So I am thinking that projection is the way to go.  What I basically will do is use the normal of the plane. Which is: 
$$ \left[
\begin{array}{cc|c}
  2\\
  2\\
  1
\end{array}
\right] $$
That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$
 |e_1| = \left[
\begin{array}{cc|c}
  1\\
  0\\
  0
\end{array}
\right], |e_2| =\left[
\begin{array}{cc|c}
  0\\
  1\\
  0
\end{array}
\right], |e_3| =\left[
\begin{array}{cc|c}
  0\\
  0\\
  1
\end{array}
\right]$$
Basically, what I will do is set up an equation
$$ Proj V_n + |n| = \left[
\begin{array}{cc|c}
  1\\
  0\\
  0
\end{array}
\right]$$ 
So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?
Thanks in advance.
 A: Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $\mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $\mathbf e_1$ is equal to $\mathbf e_1-\mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.  
The problem is that you’ve set the rejection of $\mathbf e_1$ from the plane to be equal to $\mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $P\mathbf e_1$ of $\mathbf e_1$ onto the plane is $\mathbf e_1-k\mathbf n$ for some as-yet-undetermined scalar $k$. However, $k\mathbf n$ here is simply the orthogonal projection of $\mathbf e_1$ onto $\mathbf n$, which I suspect that you know how to compute.
A: Since $(2,2,1)$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $(1,0,-2)$ and $(0,1,-2)$. You want the each one is projected into itself.
So, take the only linear map $P\colon\mathbb{R}^3\longrightarrow\mathbb{R}^3$ such that


*

*$P(2,2,1)=(0,0,0)$;

*$P(1,0,-2)=(1,0,-2)$;

*$P(0,1,-2)=(0,1,-2)$.


A simple computation shows that the matrix of $P$ with respect to the canonical basis is$$\frac19\begin{bmatrix}5 & -4 & -2 \\ -4 & 5 & -2 \\ -2 & -2 & 8\end{bmatrix}.$$
