The magnitude of a triple product of two vectors So I was going through a past exam for electrodynamics and a question for radiation came up and within it was the following magnitude of a triple product
$ \lvert \hat{r} \times [\hat{r} \times \vec{\beta} ] \rvert^2 $
I looked in the solutions later to check I was correct and they gave the following for how to do that nasty magnitude: 
$ \lvert \hat{r} \times [\hat{r} \times \vec{\beta} ] \rvert^2 = \lvert \hat{r} \times \vec{\beta} \rvert^2$ 
I can't understand how they obtained this formula I've tried going through the triple product BAC - CAB identity but I just don't see it and it seems super useful to have. I was wondering if someone could give me how to obtain this.
 A: Since you want a magnitude, use the formula for the magnitude of a cross product. We know that the cross product of two vectors is perpendicular to each of the two vectors, so we know that the angle between $\hat r$ and $\hat r\times\hat\beta$ is $90°$. Therefore
$$|\hat r\times[\hat r\times\vec\beta]|^2
 =\left(|\hat r|\cdot |\hat r\times\vec\beta|\sin 90°\right)^2$$
$$=|\hat r|^2\cdot |\hat r\times\vec\beta|^2$$
Therefore the formula  you were given is true if and only if $\hat r$ is a unit vector.
A: First start with the double vector product identity :
$$A\times(A\times B) = (A \cdot B) A - (A \cdot A) B$$
Then, you get, using the cyclicity of the mixt product :
$$ \begin{array}{rcl}
\|A\times B\|^2 & = & (A\times B) \cdot (A \times B) \\
 & = & B \cdot ((A\times B) \times A) \\
 & = & B \cdot \left[ (A \cdot A) B - (A \cdot B) A \right] \\
 & = & A^2 B^2 - (A \cdot B)^2
\end{array}$$
Then, you have :
$$\begin{array}{rcl}
\|A\times(A\times B)\|^2 & = & \left[ (A \cdot B) A - (A \cdot A) B \right]^2 \\
 & = & (A\cdot B)^2 A^2 + A^4 B^2 -2 A^2(A \cdot B)^2 \\
 & = & A^2 \left[ A^2 B^2 - (A \cdot B)^2 \right] \\
 & = & A^2 \|A \times B\|^2
\end{array}$$
And, in your case, you conclude with $r$ being unitary.
A: You have
$$ \lvert a \times b \rvert^2 = \lvert a \rvert^2 \lvert b \rvert^2-(a \cdot b)^2, $$
which you can get from either definition of the cross product (geometric gives you a $\sin^2{\theta}$ which you can turn into a $1-\cos^2{\theta}$, or there's the $\epsilon_{ijk}\epsilon_{ilm}$ identity). Using the formula you mention,
$$ c \times (a \times b) = a(b \cdot c)-b(c \cdot a), $$
so the square of the norm is in general
$$ \lvert c \times (a \times b) \rvert^2 = (a(b \cdot c)-b(c \cdot a)) \cdot (a(b \cdot c)-b(c \cdot a)) \\
= \lvert a \rvert^2 (b \cdot c)^2 - 2(a \cdot b)(b \cdot c)(c \cdot a)+\lvert b \rvert^2(c \cdot a)^2 $$
Now set $c=a$, so
$$ \lvert a \times (a \times b) \rvert^2 = \lvert a \rvert^2 (a \cdot b)^2 - 2(a \cdot b)^2 \lvert a \rvert^2+\lvert a \rvert^4 \lvert b \rvert^2 \\
= \lvert a \rvert^2 \left( \lvert a \rvert^2 \lvert b \rvert^2 - (a \cdot b)^2 \right) = \lvert a \rvert^2 \lvert a \times b \rvert^2. $$
In your case, presumably $\lvert \hat{r} \rvert = 1$.
