Vector triple product

For vectors $\bar a,\bar b,\bar c$, the vector triple product $\bar a\times(\bar b\times\bar c)$ is

itself a vector (not a scalar, despite superficially resembling the scalar triple product), and is stated

(without proof at this level) to equal

Properties.

1. $\bar a\times(\bar b\times\bar c)=-(\bar b\times\bar c)\times\bar a=-(\bar c\times\bar b)\times\bar a$, from the anti-commutativity of the cross product.
2. Rewritten form: $\bar a\times(\bar b\times\bar c)=(\bar a\cdot\bar c)\bar b-(\bar a\cdot\bar b)\bar c$ (as above; note also the useful mnemonic "BAC minus CAB").
3. In general $\bar a\times(\bar b\times\bar c)\ne(\bar a\times\bar b)\times\bar c$ -- the cross product is not associative, so a vector triple product's bracket placement always matters.
4. $\hat\imath\times(\hat\jmath\times\hat k)=\bar 0$ (a degenerate case worth remembering: $\hat\jmath\times \hat k=\hat\imath$, and $\hat\imath\times\hat\imath=\bar 0$).
5. $\bar a\times(\bar b\times\bar c)$ is automatically a linear combination of $\bar b$ and $\bar c$ (from the expansion formula), so it is always coplanar with $\bar b$ and $\bar c$.

Worked examples.

• The expansion formula is verified directly for specific vectors: compute the left side by first finding $\bar b\times\bar c$ and then crossing with $\bar a$; compute the right side from the two dot products $\bar a\cdot\bar c$ and $\bar a\cdot\bar b$; check the two match component by component.
• Comparing $\bar a\times(\bar b\times\bar c)$ with $(\bar a\times\bar b)\times\bar c$ for the same three vectors typically gives two different vectors, which is the standard way this chapter demonstrates non-associativity concretely (rather than just asserting it). …