Discussion Forum

Let $\widetilde{a},\widetilde{b}$ and  $\widetilde{c}$ be three unit vectors such that  $\widetilde{a} \times (\widetilde{b}\times\widetilde{c})=\frac{\sqrt{3}}{2}(\widetilde{b}+\widetilde{c})$ . If  $\widetilde{b} $ is not parallel to $\widetilde{c} $ , then the angle between $\widetilde{a} $ and $\widetilde{b} $ is

Answer:

Explanation:

Given  $\mid\widetilde{a}\mid=\mid \widetilde{b}\mid=\mid \widetilde{c}\mid =1$

 and   $\widetilde{a}\times(\widetilde{b}\times \widetilde{c}) =\frac{\sqrt{3}}{2}(\widetilde{b}+\widetilde{c})$

Now consider $\widetilde{a}\times(\widetilde{b}\times \widetilde{c}) =\frac{\sqrt{3}}{2}(\widetilde{b}+\widetilde{c})$

$\Rightarrow (\widetilde{a}.\widetilde{c})\widetilde{b}-(\widetilde{a}.\widetilde{b})\widetilde{c}=\frac{\sqrt{3}}{2}\widetilde{b}+\frac{\sqrt{3}}{2}\widetilde{c}$

On comparing , we get

                   $\widetilde{a}.\widetilde{b}=-\frac{\sqrt{3}}{2}$

$\Rightarrow  \mid\widetilde{a}\mid \mid\widetilde{b}\mid  \cos\theta =-\frac{\sqrt{3}}{2}$

$\Rightarrow   \cos\theta =-\frac{\sqrt{3}}{2}[\because \mid\widetilde{a}\mid=\mid\widetilde{b}\mid=1]$

$\Rightarrow   \cos\theta =\cos(\pi-\frac{\pi}{6})$

$\Rightarrow   \theta =\frac{5\pi}{6}$