I believe the vector identity
$\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = 0$
is called the Jacobi identity and I know the proof.
Does anybody know of some elegant geometrical picture to illustrate why the identity is true?
Best Answer
Here is an abstract nonsense proof:
The vector $$\vec T(\vec{a},\vec{b},\vec{c}):=\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b})$$ is easily seen to be a skew trilinear function of the three vector variables $\vec{a}$, $\vec{b}$, $\vec{c}$. It follows that its coordinates $T_i$ are three real-valued such functions, whence are multiples of the determinant function (so-called triple vector product) $[\vec{a},\vec{b},\vec{c}]$ with factors $\lambda_i$ independent of $\vec{a}$, $\vec{b}$, $\vec{c}$. Putting $\vec{p}:=(\lambda_1,\lambda_2,\lambda_3)$ we therefore have $$\vec T(\vec{a},\vec{b},\vec{c})=[\vec{a},\vec{b},\vec{c}]\>\vec{p}$$ with a universal vector $\vec{p}$. This only makes sense when $\vec{p}=\vec{0}$.