Why is the following not true: $(\vec{u} \times \vec{v}) \times \vec{u} = \vec{v}$ I don't believe the following is true. 
$(\vec u \times \vec v) \times \vec u = \vec{v}$
I'd like to know why it isn't true, in better terms than "v is probably not perpendicular to u" and what the product actually is. 
 A: I assume that you are asking about $(\vec u \times \vec v) \times \vec u = \vec{v}$ being true or not.
One way to see that your formula isn't (in general) correct would be to consider two orthogonal vectors $\vec{u}$ and $\vec{v}$ both of say length 2. Then the length of $\vec{u}\times \vec{v}$ is $\lvert \vec{u}\times \vec{v}\lvert = \lvert \vec{u} \lvert\lvert\vec{v}\lvert = 4$. And so $\lvert (\vec{u}\times \vec{v})\times \vec{u}\lvert$ is 8, but that is not the length of $\vec{v}$.
So I used the formula $\lvert \vec{u}\times \vec{v}\lvert = \lvert \vec{u}\lvert \lvert \vec{v}\lvert \sin(\theta)$ where $\theta$ is the angle between the two vectors ($0\leq \theta \leq \pi$). 
A: This equation is true if and only if $\vec{u},\vec{v}$  are perpendicular and $\vec{u}$ is a unit vector. (Or if $\vec{v}=\vec{0}.$) Since the cross-product of two vectors is perpendicular to both of them, the defining equation $(\vec{u}\times\vec{v})\times\vec{u}=\vec{v}$ implies that $\vec{u}$ and $\vec{v}$ are orthogonal. On the other hand, comparing the lengths of both sides, you get $|\vec{u}||\vec{v}||\vec{u}|=|\vec{v}|$, implying $|\vec u|=1$ or $\vec{v}=0$. 
On the other hand, if $\vec{u}$ is a unit vector orthogonal to $\vec{v}$, you can check using the right-hand rule that the equation is true.
A: The vector triple product identity says $$(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{v} \cdot \vec{w}) \vec{u}$$
Try this with $\vec{w}=\vec{u}$.
A: I think the easiest way to see that it's not true is to say that $v = u$. Then $v \times u \times v = v \times v \times v  = 0 \not = v$
That's a pretty serious problem, and more than a heuristic as to why it's not true.
