Question

If $a,b$ and $c$ are three unit vectors equally inclined to each other at angle $\theta$. Then, angle between $a$ and the plane of $b$ and $c$ is

• A. ${\cos }^{-1}\left(\frac{\cos \theta }{\cos \left(\theta /2\right)}\right)$
• B. ${\sin }^{-1}\left(\frac{\sin \theta }{\sin \left(\theta /2\right)}\right)$
• C. ${\sin }^{-1}\left(\frac{\cos \theta }{\cos \left(\theta /2\right)}\right)$
• D. ${\cos }^{-1}\left(\frac{\sin \theta }{\sin \left(\theta /2\right)}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{\cos \theta }{\cos \left(\theta /2\right)}\right)$

Given,

a, b and c are unit vectors.

$\therefore |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$

Let $\alpha$ be the angle between a and the angular bisector of b and c.

We know that,

$\cos \alpha =\left(\frac{\overrightarrow{a}\cdot (\overrightarrow{b}+\overrightarrow{c})}{|\overrightarrow{a}||\overrightarrow{b}+\overrightarrow{c}|}\right)$

and

$\cos \theta =2{\cos }^2\frac{\theta }{2}-1$

Solution:

$\Rightarrow \cos \alpha =\left(\frac{|\overrightarrow{a}||\overrightarrow{b}|\cos \theta +|\overrightarrow{a}||\overrightarrow{c}|\cos \theta }{|\overrightarrow{a}|\sqrt{|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2|\overrightarrow{b}||\overrightarrow{c}|\cos \theta }}\right)$

$\Rightarrow \cos \alpha =\frac{2\cos \theta }{\sqrt{1+1+2\cos \theta }}$

$\Rightarrow \cos \alpha =\frac{2\cos \theta }{\sqrt{2+2\cos \theta }}$

$\Rightarrow \cos \alpha =\frac{2\cos \theta }{\sqrt{2+2(2{\cos }^2\frac{\theta }{2}-1)}}$

$\Rightarrow \cos \alpha =\frac{2\cos \theta }{\sqrt{2+4{\cos }^2\frac{\theta }{2}-2}}$

$\Rightarrow \cos \alpha =\frac{2\cos \theta }{2\cos \frac{\theta }{2}}$

$\Rightarrow \alpha ={\cos }^{-1}\left(\frac{\cos \theta }{\cos \frac{\theta }{2}}\right)$