Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{2\pi }{3}$
• C. $\frac{5\pi }{6}$
• D. $\frac{3\pi }{4}$

Answer 1

Answer: (C)

$\frac{5\pi }{6}$

We have,

$\begin{array}{c}\overline{a}\times \left(\overline{b}\times \overline{c}\right)=\frac{\sqrt{3}}{2}\left(\overline{b}+\overline{c}\right)\\ \left(\overline{a}\cdot \overline{c}\right)\overline{b}-\left(\overline{a}\cdot \overline{b}\right)\overline{c}=\frac{\sqrt{3}}{2}\left(\overline{b}+\overline{c}\right)\\ \overline{a}\cdot \overline{c}=\frac{\sqrt{3}}{2},\overline{a}\cdot \overline{b}=-\frac{\sqrt{3}}{2}\\ \left|a\right|\left|b\right|\cos \theta =-\frac{\sqrt{3}}{2}\\ 1\times 1\cos \theta =-\frac{\sqrt{3}}{2}\\ \cos \theta =-\frac{\sqrt{3}}{2}\\ \theta =\pi -\frac{\pi }{6}\\ \theta =\frac{5\pi }{6}\end{array}$

Hence, option $\left(C\right)$ is correct answer.