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.

$\def\\#1{{\bf#1}}$Let $\\v=(0,-3,-8,3)$, $\\w_1=(6,6,6,-1)$ and $\\w_2=(6,5,-1,60)$.  If we write
$$\\v=\lambda_1\\w_1+\lambda_2\\w_2+\\w\ ,\tag{$*$}$$
where $\\w$ is perpendicular to both $\\w_1$ and $\\w_2$, then
$$\\p=\lambda_1\\w_1+\lambda_2\\w_2\tag{$*\!*$}$$
will be the point in $W$ which is closest to $\\v$.  If you don't understand why, draw a picture of the situation - but make it in $\Bbb R^3$ rather than $\Bbb R^4$.
To do the calculations, take the dot product of $(*)$ with both $\\w_1$ and $\\w_2$, remembering that the dot products with $\\w$ will be zero.  We get
$$\eqalign{
  \\v\cdot\\w_1&=\lambda\\w_1\cdot\\w_1+\lambda\\w_2\cdot\\w_1\cr
  \\v\cdot\\w_2&=\lambda\\w_1\cdot\\w_2+\lambda\\w_2\cdot\\w_2\ .\cr}$$
Since $\\v,\\w_1$ and $\\w_2$ are known you can calculate all the coefficients, then solve the system to find $\lambda_1$ and $\lambda_2$, then substitute back into $(**)$ to find the answer.
BTW the $60$ in $\\w_2$ looks odd, are you sure it is correct?
