Is there a deeper meaning behind the "determinant" formula for the cross product?

Solution 1:

The formula $A \cdot (B \times C) = \textrm{Det}(A,B,C)$ shows this the cross product can be thought of as the transpose of the linear map $\textrm{Det}(\cdot,B,C)$.

Using the notation of riemannian geometry (hodge star, sharps, and flats) another way to say this is that $A \times B=\star(A^\flat \wedge B^\flat)^\sharp$. This is the connection between the cross product and exterior product you were looking for.

Hmm. I actually just remembered that I answered a similar question here. I am thinking of the entries as columns rather than rows in the determinant here (because that is how my brain works), but I think you will see the connection.

You may also be interested in this question on MO.

Solution 2:

The coordinate expansion of a wedge product is

$$\begin{aligned}\mathbf{a} \wedge \mathbf{b} &= (\mathbf{e}_i a_i) \wedge (\mathbf{e}_j b_j) \\ &= \sum_{i, j} a_i b_j \mathbf{e}_i \wedge \mathbf{e}_j \\ &= \sum_{i < j} \begin{vmatrix}a_i & a_j \\ b_i & b_j\end{vmatrix}\mathbf{e}_i \wedge \mathbf{e}_j.\end{aligned} $$

In $\mathbb{R}^3$, this can be related to the wedge product by a duality transformation (i.e. factoring out of a pseudoscalar for the space which such as $ \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 $), as in

$$\begin{aligned}\mathbf{a} \wedge \mathbf{b} &=\begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2+\begin{vmatrix}a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\mathbf{e}_2 \mathbf{e}_3+\begin{vmatrix}a_3 & a_1 \\ b_3 & b_1\end{vmatrix}\mathbf{e}_3 \mathbf{e}_1 \\ &=\mathbf{e}_3\begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3+\mathbf{e}_1\begin{vmatrix}a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3+\mathbf{e}_2\begin{vmatrix}a_3 & a_1 \\ b_3 & b_1\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \\ &=(\mathbf{a} \times \mathbf{b}) \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3.\end{aligned} $$

The above factorization has made use of $ \mathbf{e}_i \mathbf{e}_i = 1 $, and $\mathbf{e}_i \mathbf{e}_j = -\mathbf{e}_j \mathbf{e}_i, i \ne j $.

Note that the wedge product has the geometric meaning of the area of the parallelopiped that are spanned by the two vectors, except that it is an oriented area, with a bivector orientation that equals the geometric product (or wedge product) of two perpendicular vectors that lie in the span of the plane.

In 3D you can have a mapping between the normal to that oriented area (i.e. the cross product). Which normal you get depends on the choice of the pseudoscalar used in the duality transformation.

The rationale for your choice of that multivector wedge product is not clear, so it is not suprising that it does not look anything like the cross product.