Geometric proof for triple vector product Jacobi identity

Question

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?

Answer 1

This attempt at an answer is geometric in the sense that it is stated in terms of vectors and not components.

Geometrically the double-cross product is given by $$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b}- (\vec{a} \cdot \vec{b}) \vec{c}.$$ This shows three things:

1. $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b}- (\vec{a} \cdot \vec{b}) \vec{c}$ falls in the plane spanned by $\vec{b}$ and $\vec{c}$
2. $\vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c}- (\vec{b} \cdot \vec{c}) \vec{a}$ falls in the plane spanned by $\vec{c}$ and $\vec{a}$
3. $\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b}$ falls in the plane spanned by $\vec{a}$ and $\vec{b}$

Add these relations the terms in the $\vec{a},\vec{b}$ and $\vec{c}$ directions cancel thus revealing the Jacobi Identity.

We could visualize these in terms of three planes which intersect along the directions $\vec{a},\vec{b}, \vec{c}$. I illustrate as if they are orthogonal as to keep the picture manageable. The idea here is the lengths of the orange, purple and cyan arrows are indicative of the dot-products which appear in the spans.

Answer 2

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 $\vec{p}=\vec{0}$.