1)

Let a,b and c be three non-zero vectors such that no two of them are collinear and  (a xb)xc =   $\frac{1}{3}|b||c|a.$   . If θ is the angle between vectors b and c, then a value of   $\sin\theta$  is


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

B) $\frac{-\sqrt{2}}{3}$

C) $\frac{{2}}{3}$

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

Answer:

Option A

Explanation:

 Given,

   (a x b)x c=  $\frac{1}{3}|b||c|a.$

   $\Rightarrow$            -c x (a x b) = $\frac{1}{3}|b||c|a.$

     $\Rightarrow$      $-(c.b).a+(c.a)b= \frac{1}{3}|b||c|a$

             $\left[ \frac{1}{3}|b||c|+(c.b)\right]a=(c.a)b$

  Since a  and b are not collinear

          $c.b +\frac{1}{3}|b||c|=0 $   and c.a=0

  $\Rightarrow$       $|c||b|\cos \theta +\frac{1}{3}|b||c|=0 $

$\Rightarrow$      $|b||c|\left(\cos \theta +\frac{1}{3}\right)=0 $

$\Rightarrow$     $\cos \theta +\frac{1}{3}=0, (\therefore |b|\neq0,|c|\neq0)$

$\Rightarrow$      $\cos \theta =-\frac{1}{3}$

$\Rightarrow$      $\sin \theta =\frac{\sqrt{8}}{3}=\frac{2\sqrt{2}}{3}$