I'm wondering if there is another, possibly more efficient, way to get to the $3 \times 3 $ symmetric matrix $\mathbf{D}$ below from 3-vectors $\mathbf{a}$ and $\mathbf{b}$ then the straight forward calculation shown.
$\mathbf{c} = \mathbf{a} \times \mathbf{b}$
$\mathbf{D} = \mathbf{c} \otimes \mathbf{c}$
Where $\otimes$ is the outer product (or dyadic or tensor product).