The dimension of the intersection of two vector subspaces We're given: $$ V = Span\left\{ \begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}, \ \begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \ \begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix}
 \right\} \ \ \   \mathrm{and} \ \ \ W = Span \left\{\begin{bmatrix}1\\-3\\2\\1 \end{bmatrix} \right\} $$
and we're asked to find $\mathrm{dim}(V\cap W^{\bot})$
Here's my approach. First, by inspecting the basis of $W$, I managed to construct a basis for $W^{\bot}$, which is the following: $$W^{\bot} = Span \left\{\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\right\}$$ Then, for each vector $\textbf{w}$ in the basis of $W^{\bot}$, I tried to see if the system $A\textbf{x}=\textbf{w}$ was compatible or not. In this case, I found out that the system was only compatible with two of the vectors in $W^{\bot}$, thus indicating me that $\mathrm{dim}(V \cap W^{\bot} ) = 2$ (which is correct).
What I do not get, however, is  that since the dimension of the intersection is $2$, why aren't two of the vectors in the basis of $V$ orthogonal to the vector which spans $W$ (that was my initial approach, i.e try to see which vectors of the basis of $V$ are orthogonal to the vector that spans $W$).
Also, I was wondering if there was a simpler/quicker way of doing this.

At the first find the $\mathrm{dim}(V\cup W^{\bot})$.
And we know that $\mathrm{dim}(V\cup W^{\bot})=\mathrm{dim}(V)+\mathrm{dim}(W^{\bot})-\mathrm{dim}(V\cap W^{\bot})$
Finding the dimension of $(V\cup W^{\bot})$ is also an easy work.
For above one form a matrix $A$ like following one:
$$\begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix}\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}\begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$
Now find the rank of this matrix via using Gaussian Elimination process.
