If $\vec{a}$ and $\vec{b}$ are two unit vectors and $\theta$ is the angle between them, then the unit vector…

We have,

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1$

Hence, required unit vector in this direction is $\frac{\overrightarrow{a}+\overrightarrow{b}}{\left|\overrightarrow{a}+\overrightarrow{b}\right|}$

Now, we know that resultant of $\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ is given by

$R=\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}·\overrightarrow{b}}$

$\Rightarrow \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}·\overrightarrow{b}}$

$\Rightarrow \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{2+2\cos \theta }$

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

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