Let $\vec a,\vec b$ and $\vec c$ be non-zero vectors such that no two are collinear and $(\vec a×\vec b)×\vec c =\frac{1}{3}|\vec b||\vec c|\vec a$. If θ is the acute angle between the vectors b and c, then $\sin θ$ equals

$\frac{2\sqrt{2}}{3}$

We have,

$(\vec a×\vec b)×\vec c =\frac{1}{3}|\vec b||\vec c|\vec a$

$⇒(\vec a.\vec c)\vec b − (\vec b ·\vec c) \vec a=\frac{1}{3}|\vec b||\vec c|\vec a$

$⇒(\vec a.\vec c)\vec b −\left\{(\vec b ·\vec c)+\frac{1}{3}|\vec b||\vec c|\right\}\vec a=\vec 0$

$⇒\vec a.\vec c=0$ and $\vec b.\vec c+\frac{1}{3}|\vec b||\vec c|=0$  [$\vec a,\vec b$ non-collinear]

$⇒|\vec b||\vec c|\cos θ+\frac{1}{3}|\vec b||\vec c|=0$

$⇒\cos θ=-\frac{1}{3}⇒\sin θ=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$