Vector triple product

Statement

Let $a$, $b$ and $c$ be vectors in $\mathbb{R}^3$. Let $\times$ be the cross product and let $\cdot$ be the dot product. Then $$a \times (b \times c) = (a \cdot c) \, b - (a \cdot b) \, c.$$

Remarks

The formula is easy to remember as "BAC minus CAB": $a \times (b \times c) = b(a \cdot c) - c(a \cdot b)$. The cross product is not associative. The formula gives $(a \times b) \times c = (a \cdot c) \, b - (b \cdot c) \, a$, which is in general different from $a \times (b \times c)$. A consequence is the Jacobi identity $a \times (b \times c) + b \times (c \times a) + c \times (a \times b) = 0$.

Explanations

L1 human

The vector $b \times c$ is perpendicular to the plane of $b$ and $c$. A second cross product with $a$ gives a vector perpendicular to $b \times c$, so the result lies back in the plane of $b$ and $c$. The formula tells you exactly which combination of $b$ and $c$ it is.