Closest Point to a vector in a subspace Given v = [0 -3 -8 3], find the closest point to v in the subspace W spanned by [6 6 6 -1] and [6 5 -1 60]. This is web homework problem and I have used the formula (DotProduct(v, w.1)/DotProduct(w.1, w.1))*w.1 + (DotProduct(v, w.2)/DotProduct(w.2, w.2))*w.2 but the computer said the answer I got was wrong. If this isn't the formula than I'm not sure what is. I have triple checked my calculations as well. 
Any help would be greatly appreciated.

call the projection of $v$ in such space $p$, so the vector $v-p$ is normal to both $w_1$ and $w_2$, that is to say that the following equations hold
$$(v-p).w_1=0$$
$$(v-p).w_2=0$$
so
$$v.w_1=p.w_1$$
$$v.w_2=p.w_2$$
since $p$ is in the space spanned by $w_1$ and $w_2$, we now that, for some scalars $a$ and $b$
$$p=a w_1 + b w_2$$
replacing in the former equations
$$
\left(
\begin{array}{cc}
 v . w_1 \\
 v . w_2 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
   |w_1|^2 &    w_1.w_2  \\
   w_1.w_2 &    |w_2|^2 \\
\end{array}
\right)
.
\left(
\begin{array}{cc}
 a \\
 b \\
\end{array}
\right)
$$
from here you can solve for $a$ and $b$, and then find $p$ from  $p=a w_1 + b w_2$. This procedure does not require that the vectors $w_1$ and $w_2$ are orthogonal. 
