Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two unit vectors and $\theta$ is the angle between them, then the unit vector along the angular bisector of $\overrightarrow{a}$ and $\overrightarrow{b}$ will be given by

• A. $\frac{\overrightarrow{a}-\overrightarrow{b}}{2\cos \left(\theta /2\right)}$
• B. $\frac{\overrightarrow{a}+\overrightarrow{b}}{2\cos \left(\theta /2\right)}$
• C. $\frac{\overrightarrow{a}-\overrightarrow{b}}{\cos \left(\theta /2\right)}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\overrightarrow{a}+\overrightarrow{b}}{2\cos \left(\theta /2\right)}$

Vector in the direction of angular bisector of $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\overrightarrow{a}+\overrightarrow{b}}{2}$
Unit vector in this direction is $\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}$
From the figure, position vector of $E$ is $\frac{a+b}{2}$

Now in triangle $AEB$, $AE=AB\cos \frac{\theta }{2}$
$\Rightarrow \left|\frac{\overrightarrow{a}+\overrightarrow{b}}{2}\right|=\cos \frac{\theta }{2}$

Hence, unit vector along the bisector is $\frac{\overrightarrow{a}+\overrightarrow{b}}{2\cos \left(\frac{\theta }{2}\right)}$