Find the orthogonal projection of b onto col A 
When finding the orthogonal projection for this problem, why were those vectors added? Aren't the vectors normally subtracted for Gram-Schmidt and finding projections?
Also, how do you carry out the Gram Schmidt process for doing part (a)?
 A: The column space of $A$ is $\operatorname{span}\left(\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}\right)$.
Those two vectors are a basis for $\operatorname{col}(A)$, but they are not normalized.
NOTE: In this case, the columns of $A$ are already orthogonal so you don't need to use the Gram-Schmidt process, but since in general they won't be, I'll just explain it anyway.
To make them orthogonal, we use the Gram-Schmidt process:
$w_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and $w_2 = \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix} - \operatorname{proj}_{w_1} \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}$, where $\operatorname{proj}_{w_1} \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}$ is the orthogonal projection of $\begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix}$ onto the subspace $\operatorname{span}(w_1)$.
In general, $\operatorname{proj}_vu = \dfrac {u \cdot v}{v\cdot v}v$.
Then to normalize a vector, you divide it by its norm:
$u_1 = \dfrac {w_1}{\|w_1\|}$ and $u_2 = \dfrac{w_2}{\|w_2\|}$.
The norm of a vector $v$, denoted $\|v\|$, is given by $\|v\|=\sqrt{v\cdot v}$.
This is how $u_1$ and $u_2$ were obtained from the columns of $A$.
Then the orthogonal projection of $b$ onto the subspace $\operatorname{col}(A)$ is given by $\operatorname{proj}_{\operatorname{col}(A)}b = \operatorname{proj}_{u_1}b + \operatorname{proj}_{u_2}b$.
A: Questions (a) and (b) turn out to be the same. By definition, the least squares solution is the $\DeclareMathOperator{\argmin}{\arg\!\min} \argmin_x \Vert Ax-b \Vert_2$. It's easy to prove that the minimum is attained for the orthogonal projection, i.e. for $x:(Ax-b)\perp col(A)$, or in matrix notation,
$$
A^t(Ax-b)=A^tAx-A^tb=0
$$
If the columns of $A$ are linearly independent, the solution is
$$x=(A^t A)^{-1}A^tb$$
It is both (b) the least squares solution and (a) the coordinates of the orthogonal projection in the basis of the columns-vectors of $A$, $Ax$ being the same vector given in the standard basis of the ambient space.
