Let $(V,\langle,\rangle)$ be the $\mathbb R^3$ with the standard bilinear-form and let $W \subset V$ be a two dimensional spanning set given by $v = (x_1,x_2,x_3)$ and $w = (y_1,y_2,y_3)$ and the cross product $v \times w := (x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1)$.
Why is the cross product $v \times w$ contained in the orthogonal complement $W^\perp$?
I know that the orthogonal complement contains all vectors that are orthogonal to both input vectors so I know it is valid, but I cannot prove it.
You can just compute that $v\times w$ is orthogonal to both $v$ and $w$, and hence it is orthogonal to the subspace $W$ generated by $v$ and $w$. This means it lies in the orthogonal complement of $W$.
There are a lot of ways to show $v\times w$ is orthogonal to both $v$ and $w$. If you have already proven the scalar triple product identity you could consider using that:
$(v\times w)\cdot w=(w\times w)\cdot v = 0$. etc.
If you are just very desperate for a proof using coordinates, you can just compute $(v\times w)\cdot w$ and $(v\times w)\cdot v$ directly from the coordinates.