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.
 A: The projection of a vector $v$ over $\mathrm{span}(w_1, w_2)$ is the sum of the projections of $v$ over $w_1$ and $w_2$, if $w_1$ and $w_2$ are orthogonal. If they're not, you can find other vector that span the same subspace, using Gram-Schmidt's process. For example: $$\mathrm{proj}_{w_1} v = \frac{\langle v, w_1\rangle}{\langle w_1, w_1\rangle}w_1$$
The easy way to remember this formula is to think: if the projection goes in $w_1$'s direction, then it should be natural that $w_1$ appears the most in the formula. Think that $w_1$ guides $v$ through the right direction. Having this in mind, the exercise is just a calculation.
(also, I strongly suggest you look a bit about MathJax and LaTeX, so you can write formulas and stuff here right, otherwise, your question is not very visually appealing to people around here)
A: $\def\\#1{{\bf#1}}$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?
A: 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. 
