Generalize an identity of cross products to $n$ dimensions

Question

For any vectors $x,y,a\in\Bbb R^3$, let \begin{align}f(x,y)&:=x⋅y\\ f(x,y,a)&:=f(x,y)f(a,a) - f(x,a)f(y,a) \end{align} then $$\tag1 f(x,y,a)=(x×a)⋅(y×a)$$ To generalize this to 4D, I found by calculations:

For any vectors $x,y,a,b\in\Bbb R^4$, let$$f(x,y,a,b):=f(x,y,b)f(a,a,b)-f(x,a,b) f(y,a,b)$$then $$\tag2 f(x,y,a,b)=C(x×a×b)⋅(y×a×b)$$ where $C=|b|^2$ is a scalar, independent of $x,y$.
If $b$ is a unit vector, $C=1$. For cross products in 4D, see this post.

To generalize this to 5D, I found by calculations:

For any vectors $x,y,a,b,c\in\Bbb R^5$, let$$f(x,y,a,b,c):=f(x,y,b,c)f(a,a,b,c)-f(x,a,b,c) f(y,a,b,c)$$then $$\tag3 f(x,y,a,b,c)=C(x×a×b×c)⋅(y×a×b×c)$$ where $C=|c|^4f(b,b,c)$ is a scalar, independent of $x,y$.
If $b,c$ are orthogonal unit vectors, $C=1$.

Can this be generalized to any dimension $n\ge3$? What is the geometric interpretation?

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I verified (1) using Mathematica:

I verified (2) using Mathematica:

I verified (3) using Mathematica:

Answer 1

This isn't a complete answer but it's too long for a comment.

I'm going to use the wedge product instead of the cross product and its higher-dimensional analogue (sometimes called the "external product"), but you can translate using the Hodge star / Poincaré dual operation.

Your first equation (1) is just a special case of the Lagrange identity: $$\langle x_1\wedge x_2,y_1\wedge y_2\rangle=\begin{vmatrix}\langle x_1,y_1\rangle&\langle x_1,y_2\rangle\\\langle x_2,y_1\rangle&\langle x_2,y_2\rangle\end{vmatrix}$$

More generally we have $$\langle x_1\wedge\cdots\wedge x_p,y_1\wedge\cdots\wedge y_p\rangle=\det\bigl(\langle x_i,y_j\rangle\bigr)\tag{1'}$$

To obtain your second equation (2), let $M=|m_{ij}|$ be the determinant in (1') with $p=3$ and $x_1=x$, $y_1=y$, $x_2=y_2=a$, and $x_3=y_3=b$. Let $M_{ij}$ denote the $(i,j)$-th $2\times 2$ rejector minor determinant of $M$. Then from Jacobi's identities we have $$\begin{vmatrix}M_{11}&M_{12}\\ M_{21}&M_{22}\end{vmatrix}=m_{33}M\tag{2'}$$ It's easy to verify that this is precisely your (2).

I think your (3) can be obtained in a similar way, and generalized further, using these results, but I don't have time to check right now. I'll update this post later if I find time.