$\textbf{Problem:}$ Let $v_1,...,v_{n-2} \in \mathbb{R}^n$ with $\{v_1,...,v_{n-2}\}$ is linearly independent. Let $B = (v_1,...,v_{n-2})$ and $C = \text{Col}(B)$. If $x \in \mathbb{R}^n$, show that $$\text{Proj}_{C^\perp}(x) = \frac{-1}{\det(B^TB)}X(v_1,...,v_{n-2},X(v_1,...,v_{n-2},x))$$ where $X$ represents the cross product.
The definition of cross product that I was given is $X: M_{n \times (n-1)}(\mathbb{R}) \to \mathbb{R}^n$ such that if $A = (u_1,...,u_{n-1}) \in M_{n \times (n-1)}(\mathbb{R}) $ where each $u_i \in \mathbb{R}^n$. Then $X(A)_j = (-1)^{n+j} \det(A^{(j)})$ where $A^{(j)}$ is obtained from A by removing the $j^{th}$ row.
I'm not quite sure how to proceed with this problem. Is there any suggestion?
If $x$ lies in the span of $v_1,\cdots,v_{n-2}$ then both sides of the equation are $0$ and we are done. Otherwise let $e_1,\cdots,e_{n-2}$ be an orthonormal basis for $C$ and let $e_{n-1}$ be ${\rm Proj}_{C^\perp}(x)$ normalised. Finally let $e_n$ extend this to an orthonormal basis of $\mathbb{R}^n$ with positive orientation.
If we define our operations in terms of this basis, then the result is clear. It remains to show that the right hand side of the equation is invariant under choice of orthonormal, positively oriented basis.
Let $A\in SO(n)$. Clearly $(AB)^T(AB)=B^T(A^TA)B=B^TB$.
For any $b\in \mathbb{R}^n$ we have: \begin{equation} X(Aa_1,\cdots, Aa_{n-1})\cdot Ab\\=|Aa_1\cdots Aa_{n-1}\,Ab|\\=|A||a_1\cdots a_{n-1}\, b|\\= |a_1\cdots a_{n-1}\, b|\\=X(a_1,\cdots,a_{n-1})\cdot b\\=AX(a_1,\cdots,a_{n-1})\cdot Ab \end{equation}
As $b$ was arbitrary and $A$ surjective, we have $AX(a_1,\cdots,a_{n-1})=X(Aa_1,\cdots, Aa_{n-1})$.
Thus the desired equation holds, independent of the choice of orthonormal, positively oriented basis.