Dot product and determinant?

Question

Why is it true that $$ (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{u} = \begin{array}{|rrr|} u_1&u_2&u_3\\ u_1&u_2&u_3\\ v_1&v2&v3 \end{array} $$? That is, the dot product of the cross product between $\mathbf{u}$ and $\mathbf{v}$ with $\mathbf{u}$ is equal to the above determinant. Why is that true?

I know that you can see the cross product as the determinant: $$ \begin{array}{|rrr|} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ u_1&u_2&u_3\\ v_1&v2&v3 \end{array} $$ but still I cannot see how the above is true.

Answer 1

The cross product $\mathbf u\times \mathbf v$ is $(u_2v_3-u_3v_2)\mathbf i+(u_3v_1-u_1v_3)\mathbf j+(u_1v_2-u_2v_1)\mathbf k$,

so $(\mathbf u\times \mathbf v )\cdot \mathbf u=(u_2v_3-u_3v_2)u_1+(u_3v_1-u_1v_3)u_2+(u_1v_2-u_2v_1)u_3,$

and the determinant is $u_1u_2v_3+u_2u_3v_1+u_3u_1v_2-u_3u_2v_1-u_2u_1v_3-u_1u_3v_2$,

so they're the same.