Finding Least Square Solution for a System of Linear Equations

[Mindscrape]

This is a pretty basic question, but I just want to make sure. The question is to find the least square solution for
$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)} \left( \begin{array}{cc} 1 & 2\\ 2 & 4 \end{array} \right) x = \colv{2}{2}$$

I can just find the orthogonal projection of the two vectors, right? In other words, use the find $$w = a_1\mathbf{v_1} + a_2\mathbf{v_2}$$
$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)} a_1 {\colv{1}{2}} + a_2 {\colv{2}{4}} = {\colv{w_1}{w_2}}$$
where
$$a_1 = \frac{<b,v_1>}{||v_1||^2}$$
and
$$a_2 = \frac{<b,v_2>}{||v_2||^2}$$

I don't really want to use the method with positive definite because the numbers turn out sticky.

Then solution can be expressed as $$\mathbf{z} = \mathbf{b} - \mathbf{w}$$, such that z is orthogonal to the range of K?

 

[Mindscrape]

Two problems I just thought of though. If I want to do an orthogonal projection the basis vectors must be orthogonal. I suppose I could convert to an orthogonal basis.

Furthermore, K^T K = Kx will still not give an answer such that b is in the range of K. No wonder this problem was assigned, it's not as basic as I thought. :(

So if I convert the second vector to an orthogonal vector with Gram-Schmitt then I can use the orthogonal projection as the least squares answer?