The magnitude of a triple product of two vectors

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\lvert \hat{r} \rvert = 1$.