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)}<br /> \left(<br /> \begin{array}{cc}<br /> 1 &amp; 2\\<br /> 2 &amp; 4<br /> \end{array}<br /> \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)}<br /> a_1 {\colv{1}{2}} + a_2 {\colv{2}{4}} = {\colv{w_1}{w_2}}
where
a_1 = \frac{&lt;b,v_1&gt;}{||v_1||^2}
and
a_2 = \frac{&lt;b,v_2&gt;}{||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?