Is there a deeper meaning behind the "determinant" formula for the cross product? We all know that for all vectors $\mathbf{a}, \mathbf{b} \in \mathbb{R^3}$, if $(a_x,a_y,a_z)^\top$ is the component form of $\mathbf{a}$ and similarly $(b_x, b_y, b_z)^\top$ is the component form of $\mathbf{b}$ then, the cross product can be evaluated by the formal determinant
\begin{equation}
\mathbf{a} \times \mathbf{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 
 a_x & a_y & a_z \\ 
b_x & b_y & b_z 
\end{vmatrix}
\end{equation}
However, I'm not satisfied with such an understanding. I think the formula is too elegant to be a mere formal, memorisation aid. So is there a generalisation of the determinant, which has an obvious geometric meaning, the same way the determinant can be interpreted the area multiplication factor of a linear transformation?
In geometric algebra, one can generalise the determinant of the matrix using the wedge product. If $\mathbf{a,b,c}$ are vectors in $\mathbf{R}^3$ then the determinant of the matrix, which has the vectors $\mathbf{a,b,c}$ as columns is $|\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}|.$ However, the wedge product is defined for general multivectors, so I tried evaluating 
$(\mathbf{ii} + a_x\mathbf{j} + b_x\mathbf{k}) \wedge (\mathbf{ji} + a_y\mathbf{j} + b_y\mathbf{k}) \wedge (\mathbf{ki} + a_z\mathbf{j} + b_z\mathbf{k})$ to obtain,
$\left(a_y b_z-a_z b_y\right) \mathbf{jk} +  \left(a_y-b_z\right) \mathbf{ijk}.$ However, the result doesn't have any obvious connection to the cross product.

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.
