Find a basis for orthogonal complement in R⁴ 
How do I approach part 2? I found the projection of 1. to be (6,-2,2,-2) but what do I do now?
 A: For vector $\mathbf v = (x_1, x_2, x_3,x_4)$, the dot products of $\mathbf v$ with the two given vectors respectively are zero.
$$\begin{align*}
\begin{bmatrix}1&2&3&4\\2&5&0&1\end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} &= \begin{bmatrix}0\\0\end{bmatrix}\\
\begin{bmatrix}1&2&3&4\\0&1&-6&-7\end{bmatrix}
\mathbf v &= \begin{bmatrix}0\\0\end{bmatrix}\\
\begin{bmatrix}1&0&15&18\\0&1&-6&-7\end{bmatrix}
\mathbf v &= \begin{bmatrix}0\\0\end{bmatrix}\\
\end{align*}$$
Let $x_3 = a$, $x_4 = b$, then $x_1 = -15a - 18b$, and $x_2 = 6a + 7b$.
$$\mathbf v = \begin{bmatrix}-15a - 18b\\6a+7b\\a\\b\end{bmatrix}
= a\begin{bmatrix}-15\\6\\1\\0\end{bmatrix} + b\begin{bmatrix}-18\\7\\0\\1\end{bmatrix}$$
So $(-15,6,1,0)$ and $(-18, 7,0,1)$ together is a basis.
Setting $a=1, b=-1$ gives $(3,-1,1,-1)$, which is one of the vectors in your basis above.
A: Alternatively, one could solve the linear system $Ax=0,$ where
$$A=\begin{bmatrix}1&2&3&4\\2&5&0&1\end{bmatrix}.$$
$A$ us equivalent to the matrix
$$A=\begin{bmatrix}1&0&15&18\\0&1&-6&-7\end{bmatrix}$$
which has the solution set $\operatorname{span}\{(-15,6,1,0),~(-18,7,0,1)\}.$
Note that since $\mathbb{R}^{4}=W\oplus W^{\perp}$ and $\operatorname{dim}W=2,$ you get that $\operatorname{dim}W^{\perp}=2,$ as well.
