Griffiths' Introduction to Electromagnetism -book has equations called 20.10 below.
I have proved this equation d) pretty much on the first mathematics -course I had but I have not yet understood a visual way to remember it. How can I visually explain it? Why does it have a negative sign?
$$\bar a\times (\bar b\times\bar c)=\bar b (\bar a\cdot\bar c)-\bar c (\bar a\cdot \bar b)$$
On
Not a completely satisfying explanation, but maybe I can help you understand why it looks the way it does, and why there needs to be a minus sign.
Assume $b,c$ to linearly independent and $a$ not simultaneously orthogonal to both $b$ and $c$ (otherwise the left hand side is $0$.) Then $a\times(b\times c)$ is orthogonal to the vector $b\times c$, so it lies in the linear span of $b$ and $c$, and so there are some $\lambda,\mu\in \mathbb R$ such that $a\times(b\times c)=\lambda.b+\mu.c$. But it is also orthogonal to $a$, so $\lambda(b\cdot a)+\mu(c\cdot a)=0$. Thus $(\lambda, \mu)$ is orthogonal (in $\mathbb R^2$) to the (non zero) vector $(b\cdot a,c\cdot a)$ (of $\mathbb R^2$), i.e. there is a real number $x$ such that $(\lambda, \mu)=x(c\cdot a,-b\cdot a)$ . Therefore there exists a real number $x$ such that $$a\times(b\times c)=x((c\cdot a)b-(b\cdot a)c).$$
On
Here's one way to think about it. Each term must contain $\overline{a}$, $\overline{b}$ and $\overline{c}$ (in fact it should be linear in each of those vectors). The result should be orthogonal to $\overline{a}$, so there is no term of the form $(\ldots \cdot \ldots) \overline{a}$. Switching $\overline{b}$ and $\overline{c}$ should multiply everything by $-1$ (which is why one of the terms has a $-$). To remember which term has the $-$, try it in the case $\overline{a} = \overline{b} = \overline{i}$, $\overline{c} = \overline{j}$: $\overline{i} \times (\overline{i} \times \overline{j}) = \overline{i} \times \overline{k} = - \overline{j} = (\overline{i} \cdot \overline{j}) \overline{i} - (\overline{i} \cdot \overline{i}) \overline{j}$, not the other way around.
On
My course book mentions here, page 184 in the book here, now translated to English.
Suppose $\bar a \times (\bar b \times \bar c)$. Now $\bar b\times\bar c$ is a normal to a plane when $b\not || \bar c$. So $\bar a \times (\bar b \times \bar c)$ must be parallel to the plane so $\bar a \times (\bar b \times \bar c)=\lambda\bar b+\mu\bar c:=\bar h$ where $\lambda,\mu\in\mathbb R$. The $\bar h$ is perpendicular to $\bar a$ so
$$\bar a\cdot(\lambda\bar b+\mu\bar c)=\lambda(\bar a\cdot\bar b)+\mu(\bar a\cdot\bar c)=0$$
which is satisfied when $\lambda=\gamma \bar a\cdot \bar c$ and $\mu=-\gamma\bar a\cdot\bar b$ (clarified by H1 and H3) where $\gamma\in\mathbb R$ so
$$\bar a\times(\bar b\times \bar c)=\gamma\left[(\bar a\cdot\bar c)\bar b-(\bar a\cdot\bar b)\bar c\right].$$
Let's select $\bar a=\bar b=\hat i$ and $\bar c=\hat j$ so $\overline{i} \times (\overline{i} \times \overline{j}) = \overline{i} \times \overline{k} = - \overline{j} = (\overline{i} \cdot \overline{j}) \overline{i} - (\overline{i} \cdot \overline{i}) \overline{j}$ (H2). Hence
$$\bar a\times (\bar b\times\bar c)=\bar b (\bar a\cdot\bar c)-\bar c (\bar a\cdot \bar b)$$
Notices
H1 clarified by Alex Jordan here.
H2 resonates with Robert Israel's point here.
H3 clarifies the orthogonality here, by Olivier Bégassat.
The result of the product $\vec{a}\times(\vec{b}\times\vec{c})$ should be orthogonal to $\vec{a}$, and the right-hand is indeed orthogonal to $\vec{a}$. If you compute it's dot product with $\vec{a}$ $$(\vec{a}\cdot\vec{b})(\vec{a}\cdot\vec{c})-(\vec{a}\cdot\vec{c})(\vec{a}\cdot\vec{b})$$ there is a symmetry that causes cancellation to $\vec{0}$. So this provides one way to remember the identity. $\vec{a}$, having the "least symmetric" role in $\vec{a}\times(\vec{b}\times\vec{c})$ is involved with both dot products on the right.
The identity should involve that minus sign, since swapping the roles of $\vec{b}$ and $\vec{c}$ should negate the resultant cross product vector. Also, as already noted, the presence of that negative sign helps confirm that $\vec{a}$ is orthogonal to the cross product.
At a more basic level, the result of a cross-product is a vector. So the right-hand side can be a linear combination of vectors (and not, say, of dot products). Here, we combine two vectors using scalar weights that are each obtained from dot products.