Question

Let $\hat{\alpha },\hat{\beta },\hat{\gamma }$ be three unit vectors such that $\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=\frac{1}{2}(\hat{\beta }+\hat{\gamma })$ where $\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=(\hat{\alpha }.\hat{\gamma })\hat{\beta }-(\hat{\alpha }.\hat{\beta })\hat{\gamma }$. If $\hat{\beta }$ is not parallel to $\hat{\gamma }$, then the angle between $\hat{\alpha }$ and $\hat{\beta }$ is

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

Answer 1

The correct option is D $\frac{2\pi }{3}$

$\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=\frac{1}{2}(\hat{\beta }+\hat{\gamma })$
$\Rightarrow (\hat{\alpha }.\hat{\gamma })\hat{\beta }-(\hat{\alpha }.\hat{\beta })\hat{\gamma }=\frac{1}{2}\hat{\beta }+\frac{1}{2}\hat{\gamma }$
$\Rightarrow (\hat{\alpha }.\hat{\gamma }-\frac{1}{2})\hat{\beta }=(\frac{1}{2}+\hat{\alpha }.\hat{\beta })\hat{\gamma }$
The only way $k_1\hat{\beta }=k_2\hat{\gamma }$ is when $|k_1|=|k_2|$ and direction is same.

Since, $\hat{\beta }$ is not parallel to $\hat{\gamma }$. That means $k_1=k_2=0$.
$\therefore \hat{\alpha }.\hat{\beta }=-\frac{1}{2}$
$\Rightarrow$ Angle between $\hat{\alpha }$ and $\hat{\beta }$ is ${\cos }^{-1}\left(\frac{-1}{2}\right)=\frac{2\pi }{3}$

Alternate solution:
$\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=\frac{1}{2}(\hat{\beta }+\hat{\gamma })$
$\Rightarrow (\hat{\alpha }.\hat{\gamma })\hat{\beta }-(\hat{\alpha }.\hat{\beta })\hat{\gamma }=\frac{1}{2}\hat{\beta }+\frac{1}{2}\hat{\gamma }$
On comparing, we get
$\hat{\alpha }.\hat{\gamma }=\frac{1}{2}$ and $\hat{\alpha }.\hat{\beta }=-\frac{1}{2}$
$\Rightarrow$ Angle between $\hat{\alpha }$ and $\hat{\beta }$ is ${\cos }^{-1}\left(\frac{-1}{2}\right)=\frac{2\pi }{3}$