Question

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=1$ and $\vec{a} \cdot(\vec{b} \times(\vec{a} \times \vec{b}))=8$. If the angle between $\vec{a}$ and $\vec{b}$ is $\text{cosec}^{-1} \sqrt{2}$, then magnitude of $\vec{b}$ is

Answer 1

$\vec{a} \cdot(\vec{b} \times(\vec{a} \times \vec{b})) =\vec{a} \vec{b}(\vec{a} \times \vec{b})=|\vec{a} \times \vec{b}|$
$\Rightarrow|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta =8$
$\Rightarrow|\vec{b}|=4$