Vector Proof for triple product How can I prove/disprove A x (B x C)=(A x B) x C +B x (A x C) ?
I know I could equate the right side to:
B(A dot C)-C(A dot B)
But I don't know where to go from there.
 A: The vector triple product satisfies the Jacobi identity:
$$ a \times (b \times c) + b \times ( c \times a) + c \times (a \times b) =0, $$
because Lagrange's identity implies that the left-hand side expands to
$$ b(c \cdot a)-c(a \cdot b) + c(a \cdot b)-a(b \cdot c)+a(b \cdot c)-b(c \cdot a), $$
and everything cancels. This is equivalent to your expression by the anticommutativity of the cross product, so it is always true.
A: This is the Jacobi identity for the vector cross product. Since you already have the identity:
$$
\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{a}\cdot\mathbf{b})
$$
applying this to both sides should show you they are equivalent. The left-hand side is:
$$
LHS=\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{a}\cdot\mathbf{b})
$$
and the right-hand side is:
$$
\begin{split} RHS&=(\mathbf{a}\times\mathbf{b})\times\mathbf{c}+\mathbf{b}\times(\mathbf{a}\times\mathbf{c}) \\ &=-\mathbf{c}\times(\mathbf{a}\times\mathbf{b})+\mathbf{a}(\mathbf{b}\cdot\mathbf{c})-\mathbf{c}(\mathbf{b}\cdot\mathbf{a}) \\ &=-\mathbf{a}(\mathbf{c}\cdot\mathbf{b})+\mathbf{b}(\mathbf{c}\cdot\mathbf{a})+\mathbf{a}(\mathbf{b}\cdot\mathbf{c})-\mathbf{c}(\mathbf{b}\cdot\mathbf{a}) \\ &=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{b}\cdot\mathbf{a})=LHS \end{split}
$$
