TITLE: Intuition behind the Triple Product identity
QUESTION [8 upvotes]: The well-known identity:
$$
\textbf{c}\times(\textbf{a}\times\textbf{b})=(\textbf{c}\cdot \textbf{b})\textbf{a}-(\textbf{c}\cdot \textbf{a})\textbf{b}
$$
and its counterpart for a curl:
$$
\nabla \times(\nabla \times A)=\nabla(\nabla\cdot A)-\nabla ^2 A
$$
are extremely useful, but I can't wrap my head around the intuition (geometrical or otherwise).
How exactly can these results be interpreted?

REPLY [2 votes]: I know this answer is $6$ years late, but I just thought I'd share another perspective on the vector triple product. So we want to show that:$$\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$$
Since this is just for the sake of conveying intuition, begin by assuming that the vector $\vec{b}$ lies purely along the $x$-axis (i.e. $\vec{b}=b_{1}\hat{i}$). Then let $\vec{c}$ lie somewhere in the $xy$ plane (i.e. $\vec{c}=c_{1}\hat{i}+c_{2}\hat{j}$). Finally, let $\vec{a}$ lie anywhere (i.e. $\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$). Intuitively one can just rotate this whole set-up and have the same basic dynamics play out, so it's generalizable (but again, this is just for intuition purposes). Now we make a few simple but crucial realizations: the $\hat{i}$-component of $\vec{c}$ (i.e. $c_{1}$) actually has no influence on the cross product $\vec{a}\times(\vec{b}\times\vec{c})$, since the (absolute value of the) signed area of the parallelogram spanned by $\vec{b}$ and $\vec{c}$ (which represents $||\vec{b}\times\vec{c}||$) only depends on the length of the parallelogram's base (i.e. $||\vec{b}||=|b_{1}|$) and the parallelogram's height (i.e. $|c_{2}|$). Furthermore, you can intuit that the cross product's direction also doesn't depend on what $c_{1}$ is. A similar line of reasoning allows us to neglect the $\hat{k}$-component of $\vec{a}$ (i.e. $a_{3}$ has no effect on $\vec{a}\times(\vec{b}\times\vec{c})$).
Therefore, for the sake of intuition, let us set $c_{1}=a_{3}=0$. As a result, we now have $\vec{b}$ lying purely along the $x$-axis, $\vec{c}$ lying orthogonal to $\vec{b}$ along the $y$-axis, and $\vec{a}$ lying somewhere on the $xy$ plane. In fact, for the moment, let us go even further and pretend $\vec{b}=\hat{i}$ (i.e. $b_{1}=1$) and $\vec{c}=\hat{j}$ (i.e. $c_{2}=1$). We know, out of context, that any vector $\vec{a}$ in $\mathbb{R^2}$ can be represented as:$$\vec{a}=(\vec{a}\cdot{\hat{i}})\hat{i}+(\vec{a}\cdot{\hat{j}})\hat{j}$$ When it is the case then that $\vec{b}=\hat{i}$ and $\vec{c}=\hat{j}$, then we have $\vec{b}\times\vec{c}=\hat{k}$. Then, from basic cross product intuition, we see that the vector $\vec{a}\times(\vec{b}\times\vec{c})$ will be proportional to a $90$-degree clockwise rotation of vector $\vec{a}$ (when viewed looking at the $xy$ plane from the head of vector $\vec{b}\times{\vec{c}}$). Furthermore, the magnitude of this cross product vector is equal to the magnitude of vector $\vec{a}$ (think of the rectangle's area). In this case, because $\vec{b}\times{\vec{c}}=\hat{k}$, the appropriate linear transformation that accomplishes the $90$-degree clockwise rotation is: $$\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}$$
(where we have pretended that all our vectors are in $\mathbb{R^2}$ instead of $\mathbb{R^3}$, though a similar rotation matrix exists for the $\mathbb{R^3}$ case). Thus: \begin{align}\vec{a}\times(\vec{b}\times\vec{c}) &=\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}\begin{bmatrix} \vec{a}\cdot{\hat{i}}\\ \vec{a}\cdot{\hat{j}}\\ \end{bmatrix} \\
& = \begin{bmatrix}\vec{a}\cdot{\hat{j}}\\-\vec{a}\cdot{\hat{i}}\\ \end{bmatrix}\\&=(\vec{a}\cdot{\hat{j}})\hat{i}-(\vec{a}\cdot{\hat{i}})\hat{j}\\&=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\end{align}
Now let us generalize a bit so that instead we have $\vec{b}=b_{1}\hat{i}$ and $\vec{c}=c_{2}\hat{j}$. First, suppose that $b_{1}$ is positive. Then if $c_{2}>0$ as well, we have (by the exact same logic as above): $$\vec{a}=\biggr(\vec{a}\cdot{\frac{\vec{b}}{||\vec{b}||}}\biggr)\frac{\vec{b}}{||\vec{b}||}+\biggr(\vec{a}\cdot{\frac{\vec{c}}{||\vec{c}||}}\biggr)\frac{\vec{c}}{||\vec{c}||}$$
But this time, the magnitude of the cross product $||\vec{a}\times(\vec{b}\times\vec{c})||$ will be the product of all $3$ of their magnitudes (can you see why?). So we can rotate vector $\vec{a}$ by $90$-degrees clockwise and then scale it up by $||\vec{b}||||\vec{c}||$ after (or before, since it's linear) the rotation: \begin{align}\vec{a}\times(\vec{b}\times\vec{c})\\&=||\vec{b}||||\vec{c}||\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}\begin{bmatrix}\vec{a}\cdot{\frac{\vec{b}}{||\vec{b}||}}\\ \vec{a}\cdot{\frac{\vec{c}}{||\vec{c}||}}\\ \end{bmatrix}\\&=||\vec{b}||(\vec{a}\cdot{\vec{c}})\hat{i}-||\vec{c}||({\vec{a}\cdot{\vec{b}}})\hat{j}\\&=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\end{align}
I'll leave the rest of the cases as an exercise to the reader.