Question

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

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

Answer 1

The correct option is D

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

Given, $|\hat{a}|=|\hat{b}|=|\hat{c}|=1\text{and}\hat{a}\times (\hat{b}\times \hat{c})=\frac{\sqrt{3}}{2}(\hat{b}+\hat{c})$
Now, consider $\hat{a}\times (\hat{b}\times \hat{c})=\frac{\sqrt{3}}{2}(\hat{b}+\hat{c})$
$\Rightarrow (\hat{a}.\hat{c})\hat{b}-(\hat{a}.\hat{b})\hat{c}=\frac{\sqrt{3}}{2}\hat{b}+\frac{\sqrt{3}}{2}\hat{c}$
On comparing, we get
$\hat{a}.\hat{b}=-\frac{\sqrt{3}}{2}\Rightarrow |\hat{a}||\hat{b}|\text{cos}\theta =-\frac{\sqrt{3}}{2}\Rightarrow \cos \theta =-\frac{\sqrt{3}}{2}[\because |\hat{a}|=|\hat{b}|=1]\Rightarrow \text{cos}\theta =\text{cos}(\pi -\frac{\pi }{6})\Rightarrow \theta =\frac{5\pi }{6}$