Rotating Indices Property of Vector Triple Product

Theorem

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Euclidean $3$-space.

Then:

$\mathbf a \times \paren {\mathbf b \times \mathbf c} + \mathbf b \times \paren {\mathbf c \times \mathbf a} + \mathbf c \times \paren {\mathbf a \times \mathbf b} = 0$

where $\mathbf a \times \paren {\mathbf b \times \mathbf c}$ denotes the vector triple product operator.

Proof

\(\ds \)

\(\)

\(\ds \mathbf a \times \paren {\mathbf b \times \mathbf c} + \mathbf b \times \paren {\mathbf c \times \mathbf a} + \mathbf c \times \paren {\mathbf a \times \mathbf b}\)

\(\ds \)

\(=\)

\(\ds \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf a \cdot \mathbf b} \mathbf c + \paren {\mathbf b \cdot \mathbf a} \mathbf c - \paren {\mathbf b \cdot \mathbf c} \mathbf a + \paren {\mathbf c \cdot \mathbf b} \mathbf a - \paren {\mathbf c \cdot \mathbf a} \mathbf b\)

Lagrange's Formula: $\mathbf a \cdot \mathbf b$ denotes dot product

\(\ds \)

\(=\)

\(\ds \paren {\mathbf a \cdot \mathbf c} \mathbf b - \paren {\mathbf a \cdot \mathbf b} \mathbf c + \paren {\mathbf a \cdot \mathbf b} \mathbf c - \paren {\mathbf b \cdot \mathbf c} \mathbf a + \paren {\mathbf b \cdot \mathbf c} \mathbf a - \paren {\mathbf a \cdot \mathbf c} \mathbf b\)

Dot Product Operator is Commutative

\(\ds \)

\(=\)

\(\ds 0\)

gathering terms and simplifying

$\blacksquare$

Sources

• 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 5$: $(12)$