## Problem on determinants and the triple product

Let $\bfa = \langle a_1, a_2, a_3 \rangle, \bfb = \langle b_1, b_2, b_3\rangle, \bfc = \langle c_1, c_2, c_3 \rangle.$ Show that the triple product of the vectors equals the determinant of a matrix whose rows is given by the vectors:$$\bfa \cdot (\bfb \times \bfc) = \begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$
• ## Solution

To prove that the left hand and right hand sides are equal, we try to express them in a similar way. We will try to do so with as little work as possible.
Recall the relationship of the cross product to a determinant:
Because we did a minor and cofactor expansion along the first row of the determinant in the cross product, let's do the same thing for the determinant whose rows are $\bfa$, $\bfb$, and $\bfc$.
We conclude that for any vectors $\bfa, \bfb, \bfc$: $$\bfa \cdot (\bfb \times \bfc) = \begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$