Vectors 4 Question 3
3. Let $\hat{\mathbf{a}}, \hat{\mathbf{b}}$ and $\hat{\mathbf{c}}$ be three unit vectors such that $\hat{\mathbf{a}} \times(\hat{\mathbf{b}} \times \hat{\mathbf{c}})=\frac{\sqrt{3}}{2}(\hat{\mathbf{b}}+\hat{\mathbf{c}})$. If $\hat{\mathbf{b}}$ is not parallel to $\hat{\mathbf{c}}$, then the angle between $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ is
(2016 Main)
(a) $\frac{3 \pi}{4}$
(b) $\frac{\pi}{2}$
(c) $\frac{2 \pi}{3}$
(d) $\frac{5 \pi}{6}$
Show Answer
Answer:
Correct Answer: 3. (d)
Solution:
- Given, $|\hat{\mathbf{a}}|=|\hat{\mathbf{b}}|=|\hat{\mathbf{c}}|=1$
and $\quad \hat{\mathbf{a}} \times(\hat{\mathbf{b}} \times \hat{\mathbf{c}})=\frac{\sqrt{3}}{2}(\hat{\mathbf{b}}+\hat{\mathbf{c}})$
Now, consider
$$ \hat{\mathbf{a}} \times(\hat{\mathbf{b}} \times \hat{\mathbf{c}})=\frac{\sqrt{3}}{2}(\hat{\mathbf{b}}+\hat{\mathbf{c}}) $$
$\Rightarrow(\hat{\mathbf{a}} \cdot \hat{\mathbf{c}}) \hat{\mathbf{b}}-(\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}) \hat{\mathbf{c}}=\frac{\sqrt{3}}{2} \hat{\mathbf{b}}+\frac{\sqrt{3}}{2} \hat{\mathbf{c}}$
On comparing, we get
$$ \begin{array}{rlrl} & & \hat{\mathbf{a}} \cdot \hat{\mathbf{b}}=-\frac{\sqrt{3}}{2} \Rightarrow|\hat{\mathbf{a}}||\hat{\mathbf{b}}| \cos \theta=-\frac{\sqrt{3}}{2} \\ \Rightarrow \quad & \cos \theta=-\frac{\sqrt{3}}{2} \quad[\because|\hat{\mathbf{a}}|=|\hat{\mathbf{b}}|=1] \\ \Rightarrow \quad & \cos \theta=\cos \left(\pi-\frac{\pi}{6}\right) \Rightarrow \theta=\frac{5 \pi}{6} \end{array} $$
