Problem on determinants and the triple product - Leading Lesson

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:
Cross product definition

By definition \begin{align}\mathbf{b} \times \mathbf{c} &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \end{align}
Minor and cofactor expansions

This determinant can be computed by a minor and cofactor expansion along the top row:
\begin{align}\mathbf{b} \times \mathbf{c} = \begin{vmatrix}b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} \ \mathbf{i} - \begin{vmatrix}b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} \ \mathbf{j} + \begin{vmatrix}b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} \ \mathbf{k} \end{align}
Direct computation of dot product

Because we have the components of $\bfb \times \bfc$, we can now write down the dot product with $\bfa$: \begin{align} \bfa \cdot (\bfb \times \bfc) = & a_1 \begin{vmatrix}b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix}b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix}b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} \end{align}
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 $\bfc$.
Minor and cofactor expansions

By a minor and cofactor expansion along the first row, we get: \begin{align} \begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = & a_1 \begin{vmatrix}b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix}b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix}b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} \end{align}
We conclude that for any vectors $\bfa, \bfb, \bfc$: $$\bfa \cdot (\bfb \times \bfc) = \begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$

Related topics

Multivariable calculus
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