Finding Matrix of Projection 
Let $P:R4→R4$ be the orthogonal projection onto the plane
$ W={−(y+2z+t)=0}$
(That is, the projection parallel to the normal vector $(0,−1,−2,−1)$.)
Find the matrix $M^{E}_{E} (P)$ of the projection $P$ relative to the standard
  basis of $R4$.
$M^{E}_{E} (P)$ = ...

So, I'm kinda new to the topic and confused about this and don't know how to do it. Can someone help me with this.
 A: The simplest way to compute this matrix is to consider how you would normally compute this projection. Let $v$ denote the normal vector $v = (0,-1,-2,-1)$.  For a vector $x$, the projection onto the orthogonal complement $W^\perp$ of the hyperplane is given by 
$$
\operatorname{proj}_{v}(x) = \frac{v^Tx}{v^Tv}\cdot v.
$$
So, the projection of $x$ onto $W$ is given by $x - \operatorname{proj}_{v}(x)$. Our goal is to produce a matrix $P$ for which 
$$
Px = x - \operatorname{proj}_{v}(x) = x - \frac{v^Tx}{v^Tv}\cdot v.
$$
A nice way to find this matrix $P$ is to factor the right-hand side into the form $Mx$ for some matrix $M$. In particular, note that
$$
x - \frac{v^Tx}{v^Tv}\cdot v = x - \frac 1{v^Tv}v\cdot v^Tx = x - \frac 1{v^Tv}vv^T x = \left(I - \frac{vv^T}{v^Tv} \right)x,
$$
where $I$ dentoes the identity matrix. So, our matrix $P$ should be given by $P = I - \frac{vv^T}{v^Tv}.$ If you compute this correctly, you should end up with the answer
$$
P = \pmatrix{1 & 0 & 0 & 0\\ 0 & \frac{5}{6} & -\frac{1}{3} & -\frac{1}{6}\\ 0 & -\frac{1}{3} & \frac{1}{3} & -\frac{1}{3}\\ 0 & -\frac{1}{6} & -\frac{1}{3} & \frac{5}{6}}.
$$
