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?
$\endgroup$ 53 Answers
$\begingroup$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 $\vec{p}=\vec{0}$.
$\endgroup$ 4 $\begingroup$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:
- $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b}- (\vec{a} \cdot \vec{b}) \vec{c}$ falls in the plane spanned by $\vec{b}$ and $\vec{c}$
- $\vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c}- (\vec{b} \cdot \vec{c}) \vec{a}$ falls in the plane spanned by $\vec{c}$ and $\vec{a}$
- $\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b}$ 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.
Answering to Matthew Kvalheim. Let $R:\mathbb{R}^3\rightarrow \mathbb{R}^3$ be a rotation on $\mathbb{R}^3$ around a certain axis. Then $(R\vec{a}\times R\vec{b})=R(\vec{a}\times\vec{b})$. From here and the definition of the map $\vec{T}$ we have that
$$
\vec{T}(R\vec{a},R\vec{b},R\vec{c})=R\vec{T}(\vec{a},\vec{b},\vec{c}).
$$
From the expression in terms of $\vec{p}$ of the map $\vec{T}$ and the fact that $[R\vec{a},R\vec{b},R\vec{c}]=[\vec{a},\vec{b},\vec{c}]$, one obtains
$$
\vec{T}(R\vec{a},R\vec{b},R\vec{c})=[R\vec{a},R\vec{b},R\vec{c}]\vec{p}=[\vec{a},\vec{b},\vec{c}]\vec{p}=\vec{T}(\vec{a},\vec{b},\vec{c}).
$$
Comparing the above expressions, we obtain that
$$
R\vec{T}(\vec{a},\vec{b},\vec{c})=\vec{T}(\vec{a},\vec{b},\vec{c})
$$
for every $R$, which means that $\vec{T}(\vec{a},\vec{b},\vec{c})=0$ for arbitrary $\vec{a},\vec{b},\vec{c}$ and $\vec{p}=0$ always.
