Vector Cross Product satisfies Jacobi Identity

Theorem

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

Let $\times$ denotes the cross product.

Then:

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

That is, the cross product operation satisfies the Jacobi identity.

Proof

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

\(=\)

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

\(\ds \)

\(\)

\(\, \ds + \, \)

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

\(\ds \)

\(\)

\(\, \ds + \, \)

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

Lagrange's Formula

\(\ds \)

\(=\)

\(\ds \mathbf 0\)

Dot Product Operator is Commutative

$\blacksquare$