A proof in vectors If it is given that:
$$ \vec{R} + \dfrac{\vec{R}\cdot(\vec{B}\times(\vec{B}\times\vec{A}))} {|\vec{A} \times \vec{B} |^2}\vec{A} + \dfrac{\vec{R}\cdot(\vec{A}\times(\vec{A}\times\vec{B}))} {|\vec{A} \times \vec{B} |^2}\vec{B} = \dfrac{K(\vec{A}\times\vec{B})} {|\vec{A} \times \vec{B} |^2} $$
then prove that $$K= [ \vec{R} \vec{A} \vec{B} ]$$
I don't know how to even begin. Any ideas?
 A: Let $\vec{r} = x \vec{a}+y \vec{b}+ z \left(\vec {a} \times \vec {b}\right)$
If we wish to compute, say $x$, we need to take dot product with a vector that is perpendicular to $\vec{b}$ as well as $\left(\vec {a} \times \vec {b}\right)$ and that vector is $\vec{b} \times \left(\vec {a} \times \vec {b}\right)$
We will need that $\left[\vec {a} \ \ \vec {b} \ \left(\vec {a} \times \vec {b}\right)\right] = \left[\left(\vec {a} \times \vec {b}\right) \ \ \vec {a} \ \ \vec {b} \ \right] = \left|\vec{a} \times \vec{b}\right|^2$.
and similarly $\left[\vec {b} \ \ \vec {a} \ \left(\vec {b} \times \vec {a}\right)\right] = \left|\vec{a} \times \vec{b}\right|^2$.
Taking dot product with  $\vec{b} \times \left(\vec {a} \times \vec {b}\right)$, we get
$\vec{r}.  \left(\vec{b} \times \left(\vec {a} \times \vec {b}\right)\right)=x \left|\vec{a} \times \vec{b}\right|^2$
In this way we have $\vec{r} = \dfrac{\left[\vec {r} \ \ \vec {b} \ \left(\vec {a} \times \vec {b}\right)\right] }{\left|\vec{a} \times \vec{b}\right|^2} \vec{a}+ \dfrac{\left[\vec {r} \ \ \vec {b} \ \left(\vec {b} \times \vec {a}\right)\right] }{\left|\vec{a} \times \vec{b}\right|^2} \vec{b}+\dfrac{\left[\vec{r} \ \ \vec{a} \ \ \vec{b} \right]}{\left|\vec{a} \times \vec{b}\right|^2} \vec{c}$
Its now clear that $K = \left[\vec{r} \ \ \vec{a} \ \ \vec{b} \right]$
A: Start from dot product of both sides by $\vec{A}\times\vec{B}$.
