<s>[INST] Orthogonal Projection of $v$ on sub-space $U$ 
What is the orthogonal projection of $v = (1,5,-10)$ on the sub-space $U = Sp\{(5,-2,1),(1,2,-1)\}$?

Well, I managed to compute it on the way I've been taught by the book, but it doesn't seem to work. Any chance you guys can solve it? If needed, I can provide the way I tried to solve it.
 [/INST] Hint: 
1) Compute an orthonormal base $u_1,u_2$ of $U$ using Gramm-SChmidt
2) Consider the projector $p_U(x) = \sum_{i=1}^2 <u_i,x>u_i$
3) Compute $p_U(v)$
So let's do it:
Note that $$\langle \begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix}5\\ -2 \\ 1 \end{pmatrix}\rangle = 0 \quad \text{ and }\quad \|\begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}\|_2 = \sqrt{6}, \ \|\begin{pmatrix} 5\\ -2 \\ 1 \end{pmatrix}\|_2 = \sqrt{30}.$$
So we may choose 
$$ u_1 = \frac{1}{\sqrt{30}}\begin{pmatrix} 5\\ -2 \\ 1 \end{pmatrix}\quad \text{ and }\quad u_2=\frac{1}{\sqrt{6}}\begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}$$
and thus $p_U(v)$, the projection of $v$ on $U$ is given by
$$p_U(v) = \langle u_1,v\rangle u_1 +\langle u_2,v\rangle u_2= \frac{-15}{\sqrt{30}}u_1+\frac{1}{\sqrt{6}}u_2 = 
\begin{pmatrix}\frac{-75}{30}+\frac{1}{6} \\ \frac{30}{30}+\frac{2}{6}  \\ \frac{-15}{30}-\frac{1}{6}\end{pmatrix}=\frac{1}{3}\begin{pmatrix}-7 \\ 4\\ -2\end{pmatrix}.$$
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