# Proof-without-words for $\bar a\times (\bar b\times\bar c)=\bar b (\bar a\cdot\bar c)-\bar c (\bar a\cdot \bar b)$ or some visual-biased explanation?

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?

$$\bar a\times (\bar b\times\bar c)=\bar b (\bar a\cdot\bar c)-\bar c (\bar a\cdot \bar b)$$

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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 $$a\times(b\times c)=x((c\cdot a)b-(b\cdot a)c).$$

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Quite similar to my teacher's explanation here but more succint. Now it is a challenge visually-formulate this. +1 for the clarification at the end. – hhh Sep 21 '12 at 2:02
Unless using some pre-knowledge such as $\overline{i} \times (\overline{i} \times \overline{j}) = \overline{i} \times \overline{k} = - \overline{j} = (\overline{i} \cdot \overline{j}) \overline{i} - (\overline{i} \cdot \overline{i}) \overline{j}$, I cannot understand why the minus is just there $\bar a\times (\bar b\times\bar c)=\bar b (\bar a\cdot\bar c)-\bar c (\bar a\cdot \bar b)$, why not before the other element in the right-hand-side? Why not $\bar a\times (\bar b\times\bar c)=-\bar b (\bar a\cdot\bar c)+\bar c (\bar a\cdot \bar b)$? – hhh Sep 21 '12 at 3:17
@hhh at this point it could be either way: the negaive sign could be on the left term or the right term; but there has to be a minus sign somewhere according to this reasoning. – Olivier Bégassat Sep 21 '12 at 6:21

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.

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+1 for the first two sentences, good clarification H1. Could you clarify the sentence "the presence of that negative sign helps confirm that a⃗ is orthogonal to the cross product."? – hhh Sep 21 '12 at 1:35
@hhh One of your questions was why does it have a negative sign. One very weak answer is that without that negative sign, there would be no cancellation in the third line of my answer. And that third line is where we are confirming that $\vec{a}$ is orthogonal to the triple cross product. That is what I am getting at. – alex.jordan Sep 21 '12 at 18:07

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.

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+1 for the last sentence, not yet fully understanding the beginning such as "Each term must contain $\bar a$, $\bar b$ and a $\bar c$ (in fact it should be linear in each of those vectors)." Linear? You mean some definition here? – hhh Sep 21 '12 at 1:41
Linear in the usual sense, as a function of $\overline{a}$ or $\overline{b}$ or $\overline{c}$ with the other two fixed: $f(s \overline{u} + t \overline{v}) = s f(\overline{u}) + t \overline{v})$ for any vectors $\overline{u}$, $\overline{v}$ and scalars $s,t$. – Robert Israel Sep 21 '12 at 3:00

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.

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