Find the closest point p in S to the point w, given Let $S$ be the subspace of $\mathbb R^3$ spanned by vectors $u$ and $v$. Find the closest point $p$ in $S$ to the point $w$, given:
$$u^T = [1,−2,2]$$
$$v^T = [−4,−7,−5]$$
$$w^T = [3,3,1].$$
How would I solve this? Step by step would be helpful.
Thanks a lot!
 A: Here is a step by step stretch:


*

*First you want to find the normal vector $\mathbf{n}$ of the plane $S$ spanned by $\mathbf{u}$ and $\mathbf{v}$. The normal vector $\mathbf{n}$ is perpendicular to every vectors on $S$ so $$\mathbf{n} = \mathbf{u} \times \mathbf{v}$$

*The points on the that line passes through $\mathbf{w}$ and is parallel to $\mathbf{n}$ have the form
$$l = \{\mathbf{x}~:~\mathbf{x}=t\cdot \mathbf{n} + \mathbf{w}\}$$
where $t$ is a scalar in the underlying field of the vector space

*Our goal is the find the intersection between line $l$ and plane $S$ containing points of the form $a\cdot \mathbf{u} + b \cdot \mathbf{v}$. To do so, simply equate the equation of the line to the equation of the plane and solve the equation:
$$t\cdot \mathbf{n} + \mathbf{w} = a\cdot \mathbf{u} + b \cdot \mathbf{v}$$
Observe that this is essentially a system of 3 equations with 3 unknowns scalars $(t,a,$and $ b)$.

*The closest point $\mathbf{p}$ in $S$ to $\mathbf{w}$ is $$\mathbf{p}=t\cdot \mathbf{n} + \mathbf{w}$$ 
A: recall that the projection matrix is given by $P = A(A^TA)^{-1}A^T$
Where the matrix $A$ has the vectors $u$ and $v$ in the columns. 
$$A = \begin{bmatrix}1&-4\\-2&-7\\2&-5\end{bmatrix}$$ 
We need to compute $P$ but, let's start with $A^TA$.
$$(A^TA) = \begin{bmatrix}1&-2&2\\-4&-7&-5\end{bmatrix}\begin{bmatrix}1&-4\\-2&-7\\2&-5\end{bmatrix} = \begin{bmatrix}9&0\\0&90\end{bmatrix}$$
$$(A^TA)^{-1} = \frac{1}{90}\begin{bmatrix}10&0\\0&1\end{bmatrix}$$
$$P = A(A^TA)^{-1}A^T = \frac{1}{90}\begin{bmatrix}1&-4\\-2&-7\\2&-5\end{bmatrix} \begin{bmatrix}10&0\\0&1\end{bmatrix}\begin{bmatrix}1&-2&2\\-4&-7&-5\end{bmatrix}$$
$$P = A(A^TA)^{-1}A^T = \frac{1}{90}\begin{bmatrix}1&-4\\-2&-7\\2&-5\end{bmatrix} \begin{bmatrix}10&-20&20\\-4&-7&-5\end{bmatrix}$$
$$P = A(A^TA)^{-1}A^T = \frac{1}{90}\begin{bmatrix}26&8&40\\8&89&-5\\40&-5&65\end{bmatrix}$$
Ok, Great, now we have the projection matrix. All we need to do is multiply $Pw$ to find the projection of $w$ onto the space spanned by the columns of $A$.
$$Pw = \frac{1}{90}\begin{bmatrix}26&8&40\\8&89&-5\\40&-5&65\end{bmatrix}\begin{bmatrix}3\\3\\1\end{bmatrix} = \frac{1}{90}\begin{bmatrix}142\\286\\170\end{bmatrix}$$
Please check my arithmetic. I did this in a hurry. 
