Finding an orthonormal basis for subspace intersection, $W^{\perp}\cap V$ I am stuck trying to find an orthonormal basis for $W^{\perp}\cap V$.
I'm given V = span$\{v_1,v_2,v_3\}$ and that
$$
      v_1= \begin{bmatrix}
        1 \\
       1 \\
       0 \\
        1
        \end{bmatrix}
v_2= \begin{bmatrix}
        1 \\
       -1 \\
        2 \\
        -1
        \end{bmatrix}
v_3= \begin{bmatrix}
        2 \\
       1 \\
        1 \\
        -1
        \end{bmatrix}
$$
I'm also given that W = span$\{w_1,w_2\}$ and that 
$$
      w_1= \begin{bmatrix}
        1 \\
       1 \\
       -1 \\
        -1
        \end{bmatrix}
w_2= \begin{bmatrix}
        0 \\
       1 \\
        2 \\
        -1
        \end{bmatrix}
$$
Now I've calculated a basis for $W^{\perp}$, which  I find to be consisting of two vectors,
$$
      y_1= \begin{bmatrix}
        0 \\
       1 \\
       0 \\
        1
        \end{bmatrix}
y_2= \begin{bmatrix}
        3 \\
       -2 \\
        1 \\
        0
        \end{bmatrix}
$$
I'm now uncertain how to figure determine a basis for $W^{\perp}\cap V$. Once that is complete I will be able to find an orthonormal basis (Gram Schmidt if necessary and norm the vectors to unit vectors). Where do I go from here?
 A: You want to compute a basis for $W^\perp\cap V$ where 
$W^\perp=\DeclareMathOperator{Span}{Span}\Span\{y_1,y_2\}$ 
and
$V=\Span\{v_1,v_2,v_3\}$. To do so, form a matrix
$$
A=
\begin{bmatrix}
  y_1 & y_2 & \mid & v_1 & v_2 & v_3
\end{bmatrix}
$$
In our case
$$
A=
\left[
  \begin{array}{rr|rrr}
    0 & 3 & 1 & 1 & 2 \\
    1 & -2 & 1 & -1 & 1 \\
    0 & 1 & 0 & 2 & 1 \\
    1 & 0 & 1 & -1 & -1
  \end{array}
\right]
$$
We can determine relations between $\{y_1,y_2\}$ and $\{v_1,v_2,v_3\}$ by row
reducing $A$).  In this case.
$$
\DeclareMathOperator{rref}{rref}\rref A=
\left[\begin{array}{rr|rrr}
    1 & 0 & 0 & 0 & -4 \\
    0 & 1 & 0 & 0 & -1 \\
    0 & 0 & 1 & 0 & 4 \\
    0 & 0 & 0 & 1 & 1
    \end{array}\right]
$$
The number of free columns in $\rref{A}$ is the dimension of $W^\perp\cap V$. Each free column introduces a relation. In this case $\dim W^\perp\cap V=1$. The relation is
$$
-4\,y_1-y_2+4\,v_1+v_2=v_3
$$
which can be re-written as
$$
4\,v_1+v_2-v_3=4\,y_1+y_2=(3,2,1,4)
$$
Thus $W^\perp\cap V=\Span\{(3,2,1,4)\}$.
