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.

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.
