Vector Project onto Subspace So the question is:
Let S be the subspace of $\mathbb{R}^3$ spanned by the vectors $ u_2 = \begin{pmatrix} \frac{2}{3}\\\frac{2}{3}\\\frac{1}{3}\end{pmatrix} u_3 = \begin{pmatrix} \frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\\0\end{pmatrix}$. Let $x=(1,2,2)^T$. Find the projection p of x onto S. Show that $(p-x)\perp u_2$ and $(p-x)\perp u_2$.
I understand how to show that they are perpendicular and I actually found the answer for the projection. It's:
$\begin{pmatrix} \frac{23}{18}\\\frac{41}{18}\\\frac{8}{9} \end{pmatrix}$
The problem is, i have no idea why i am doing what I am doing, I just followed my notes. Can someone explain why I was supposed to do:
$(xu_2)u_2 + (xu_3)u_3$
To find p.
 A: Let $\{v_1,\dotsc,v_m\}$ be a basis for a subspace $V$ of $\Bbb R^n$. The matrix of the orthogonal projection onto $V$ is
$$
P=A\left(A^\top A\right)^{-1}A^\top
$$
where $A$ is the matrix whose columns are $\{v_1,\dotsc,v_m\}$. A nice discussion of this can be found in these notes.
In our case we have
$$
A=
\left[\begin{array}{rr}
\frac{2}{3} & \frac{1}{2} \, \sqrt{2} \\
\frac{2}{3} & -\frac{1}{2} \, \sqrt{2} \\
\frac{1}{3} & 0
\end{array}\right]
$$
Thus
$$
P = 
\left[\begin{array}{rrr}
\frac{17}{18} & -\frac{1}{18} & \frac{2}{9} \\
-\frac{1}{18} & \frac{17}{18} & \frac{2}{9} \\
\frac{2}{9} & \frac{2}{9} & \frac{1}{9}
\end{array}\right]
$$
Now, the projection of $v=(1,2,2)$ is
$$
Pv=
\left(\frac{23}{18},\,\frac{41}{18},\,\frac{8}{9}\right)
$$
A: First, the projection has to be on the span of u2 and u3 so it will look like a linear combination of the two.
Second, what you write as (xu2) can be thought of as the length of x in the u2 direction, and similarly (xu3) is the length of x in the u3 direction. To understand more deeply why this is so, look at the definition of the cosine of an angle between vector and its relation to the dot product (Draw yourself the projection triangle).
