Rotating Indices Property of Vector Triple Product
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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: $\S 5$: $(12)$