I am confused as to how to write $\mathbf{x}$ as the sum of the following vector in U and a vector in U-perp:
$$\mathbf{x} = (2,1,6), U=span{(3,-1,2), (2,0,-3)}.$$
Attempt:
I have row reduced U and found out that the reduced matrix gives (1,0, $\frac{-3}{2}$) and (0,1,$\frac{-13}{2}$). Does this mean that my basic solution of ($\frac{3}{2}$, $\frac{13}{2}$, 1) is my vector in U-perp? Then how do I continue on?
You have found a basis for $U^\perp$, call it $\{ v\}$
To find the vector along $U^\perp$, just compute $w= \frac{v^Tx}{\|v\|^2}v$.
$$x = (x-w) + w$$
$x-w$ is the component projected on $U$ and $w$ is the component along $v$.
Remark: Another possible way to find a basis for $U^\perp$ is to compute the cross product of the two elements in the basis of $U$.