Prove the "cab minus bac" formula

The “cab minus bac” formula is the vector identity

$A \times (B \times C) = (C \cdot A)B - (B \cdot A)C.$

Let $B = (b_1, b_2, b_3)$ and $C = (c_1, c_2, c_3)$ . Prove that

$\mathbf{i} \times (B \times C) = c_1 B - b_1 C.$

This is the “cab minus bac” formula in the case $A = \mathbf{i}$ . Prove similar formulas for the special cases $A = \mathbf{j}$ and $A = \mathbf{k}$ . Put these three results together to prove the formula in general.

Proof. For the case $A = \mathbf{i}$ we have

$\begin{align*} \mathbf{i} \times (B \times C) &= \mathbf{i} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\ &= (0, -b_1 c_2 + b_2 c_1, b_3 c_1 - b_1 c_3) \\ &= c_1 B - b_1 C. \end{align*}$

Similarly, for $A = \mathbf{j}$ and $A = \mathbf{k}$ we have

$\begin{align*} \mathbf{j} \times (B \times C) &= \mathbf{j} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\ &= (b_1 c_2 - b_2 c_1, 0, -b_2 c_3 + b_3 c_2) \\ &= c_2 B - b_2 C \\ \mathbf{k} \times (B \times C) &= \mathbf{k} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\ &= (-b_3 c_1 + b_1 c_3, b_2 c_3 - b_3 c_2, 0) \\ &= c_3 B - b_3 C. \end{align*}$

So, if $A = a_i \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}$ is any vector in $\mathbb{R}^3$ then we have

$\begin{align*} A \times (B \times C) &= a_1 \mathbf{i} \times (B \times C) + a_2 \mathbf{j} \times (B \times C) + a_3 \mathbf{k} \times (B \times C) \\ &= a_1 (c_1 B - b_1 C) + a_2 (c_2 B - b_2 C) + a_3 (c_3 B - b_3 C) \\ &= (a_1 c_1 + a_2 c_2 + a_3 c_3) B - (a_1 b_1 + a_2 b_2 + a_3 b_3) C \\ &= (A \cdot C)B - (A \cdot B) C. \qquad \blacksquare \end{align*}$

Stumbling Robot

Prove the “cab minus bac” formula

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