TITLE: What is the relationship between $(u\times v)\times w$ and $u\times(v\times w)$?
QUESTION [10 upvotes]: Given three vectors $u$, $v$, and $w$, $(u\times v)\times w\neq u\times(v\times w)$. This has been a stated fact in my recent class. But what is the ultimate relationship between them? I would presume that one is a scalar multiple of the other?

REPLY [3 votes]: By vector triple product expansion:
$$\newcommand{\u}{\boldsymbol{u}}\newcommand{\v}{\boldsymbol{v}}\newcommand{\w}{\boldsymbol{w}}\newcommand{\n}{\boldsymbol{n}}\newcommand{\s}{\boldsymbol{s}}
(\u\times \v)\times \w - \u\times(\v\times \w) = (\u\cdot \v)\w - (\w\cdot \v)\u = \v\times(\w\times \u).
$$
Here I give two pure geometrical proof of the Jacobi identity based on the right hand rule of the cross product. You could draw a picture yourself to see the geometrical relation of them more clearly.

Without loss of generality, consider the normalized version where $\|\u\| = \|\v\| = \|\w\| = 1$.
Geometrical proof 1:  $\u\times \v$ is a vector perpendicular to the plane $S_{\u\v}$ spanned by $\u$ and $\v$, i.e., parallel to the unit normal vector $n$ of $S_{\u\v}$ pointing the same direction. We assume the angle between $\w$ and this plane is acute. Now cross product with $\w$:
$$
(\u\times \v)\times \w = \sin\alpha \,\n \times \w = \sin\alpha\,\n \times (\mathrm{Proj}_{S_{\u\v}} \w),\tag{1}
$$
where $\alpha$ is the angle between $\u$ and $\v$. 
The projection $\w_{\perp}$ can be decomposed into two parts if $\u$ and $\v$ are not parallel:
$$
\w_{\perp}:= \mathrm{Proj}_{S_{\u\v}} \w = a \u_{\perp} + b\v,
$$
where $$\u_{\perp} = \u - (\u\cdot \v)\v = \u - \cos\alpha \,\v\tag{2}$$ is perpendicular to $\v$. Dot product both sides with $\u_{\perp} $ and $\v$:
$$
\w_{\perp} \cdot \u_{\perp}  = a \sin^2 \alpha \quad \text{and}\quad  \w_{\perp} \cdot \v = b,
$$
and 
$$
\w_{\perp} \cdot \u_{\perp} = \|\w_{\perp}\| \|\u_{\perp}\|\cos(\alpha_u+\frac{\pi}{2}-\alpha) = \sin\theta \sin\alpha \cos(\frac{\pi}{2}-\alpha_v), \quad \text{and}\quad  \w_{\perp} \cdot \v=\|\w_{\perp}\| \|\v\| \cos\alpha_v = \sin\theta \cos\alpha_v,
$$
where $\theta$ is the angle between $\w$ and $\n$, $\alpha_u$ is the angle between $\u$ and $\w_{\perp}$, and $\alpha_v$ is the angle between $\v$ and $\w_{\perp}$, such that $$\alpha_u + \alpha_v = \alpha.\tag{3}$$
Now $$
\w_{\perp}=\frac{\sin\theta\sin(\alpha_v)}{\sin\alpha}\u_{\perp} + \sin\theta \cos\alpha_v\,\v ,
$$
Notice $\n$ cross product with a vector on the plane $S_{\u\v}$ is rotating the vector by $\pi/2$ counterclockwisely on the plane with respect to the normal $\n$:
$$
\n\times \u_{\perp} = \|\u_{\perp}\|\v = \sin\alpha\,\v
$$
and 
$$
\n\times \v= -\u_{\perp}/\|\u_{\perp}\| = -\u_{\perp}/\sin\alpha
$$
Finally by (2) and (3), (1) becomes
$$
\begin{aligned}
(\u\times \v)\times \w =&\sin\alpha\sin\theta\sin\alpha_v\,\v - \sin\theta \cos\alpha_v\, \u + \sin\theta \cos\alpha_v\cos\alpha \,\v
\\
=&-\sin\theta \cos\alpha_v\, \u+ \sin\theta \cos(\alpha-\alpha_v)\,\v = -\sin\theta \cos\alpha_v\, \u + \sin\theta \cos\alpha_u \,\v.\tag{4}
\end{aligned}$$
For $\theta$ is the angle between $\w$ and the normal of $\u\v$-plane, drawing a perpendicular line from the tip of $\w$ to $\v$ we can see
$$
\sin\theta \cos\alpha_v = \sin\theta\,  \w_{\perp}\cdot \v = (\w\cdot \v),
$$
similarly $\sin\theta \cos\alpha_u = (\w\cdot \u)$, (4) is
$$
(\u\times \v)\times \w = (\w\cdot \u)\v - (\w\cdot \v)\u.
$$

Geometrical proof 2: Let 
$$
\s = (\u\times \v)\times \w,
$$
It is easily to see from above argument, $\s$ lies in the plane of $\u\v$, while isperpendicular to $\w$. Hence we let again:
$$
\s = a\u+b\v,
$$
Dot product with $\w$ we see that:
$$
\s\cdot \w = 0 = a(\u\cdot \w)+b(\v\cdot \w),
$$
Hence
$$
a = c(\v\cdot \w)  \quad \text{and}\quad b = -c(\u\cdot \w),
$$
hence:
$$
\s = c(\v\cdot \w)\u-c(\u\cdot \w)\v \tag{5}
$$
Now cross product with $\u$ on both sides of (5), left hand side is:
$$\begin{aligned}
\s\times \u &=\big( (\u\times \v)\times \w \big)\times \u = (\sin\alpha\,\n\times \w)\times \u
\\
&=\sin\alpha( \n\times\w_{\perp})\times \u
\end{aligned}
$$
From above we know that $ \n\times\w_{\perp}$ is the counterclockwise rotation of $\w_{\perp}$ by $\pi/2$ on this plane with respect the normal $\n$, so the angle between $\n\times\w_{\perp}$ and $\u$ is $\pi/2 + \alpha_u$ (the definition see above (3)), and 
$$
\s\times \u  = -\sin\alpha \sin\theta \sin(\frac{\pi}{2} + \alpha_u)\n = -\sin\alpha \sin\theta \cos(\alpha_u)\,\n.
$$
Right hand side of (5) is:
$$
-c(\u\cdot \w)\v \times \u = c(\u\cdot \w)\sin\alpha\,\n = c\sin\theta \cos(\alpha_u)\sin\alpha \,\n. $$
Therefore $c = -1$, and (5) is:
$$
(\u\times \v)\times \w = (\w\cdot \u)\v - (\w\cdot \v)\u.
$$

Now for any three vectors in general we can use the formula for the normalized vectors and
$$
\left(\frac{\u}{\|\u\|}\times \frac{\v}{\|\v\|}\right)\times \frac{\w}{\|\w\|} = \left(\frac{\w}{\|\w\|} \cdot \frac{\u}{\|\u\|}\right)\frac{\v}{\|\v\|} - \left(\frac{\w}{\|\w\|} \cdot \frac{\v}{\|\v\|}\right)\frac{\u}{\|\u\|},
$$
cancelling their norms would give the result.