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If $\mathbf{a}$ and $\mathbf{b}$ are two unit vectors and $\theta$ is the angle between them, then the unit vector along the angular bisector of $\mathbf{a}$ and $\mathbf{b}$ is given by
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Verified Answer
The correct answer is:
$\frac{a+b}{2 \cos (\theta / 2)}$
Since $\mathbf{a}$ and $\mathbf{b}$ are unit vectors and angle between them is $\theta$, so unit vector along the angle bisector of $\mathbf{a}$ and $\mathbf{b}$ is $\mathbf{p}=\lambda(\mathbf{a}+\mathbf{b})$ where $|\mathbf{p}|=\mathbf{l}$.
Now, since $p$ and a are inclined at angle $\frac{\theta}{2}$, so
$$
\begin{aligned}
& \mathbf{p} \cdot \mathbf{a}=|\mathbf{p} \| \mathbf{a}| \cos \frac{\theta}{2} \\
& \Rightarrow \lambda(\mathbf{a}+\mathbf{b}) \cdot \mathbf{a}=1 \times 1 \cos \frac{\theta}{2} \\
& \Rightarrow \lambda(1+\cos \theta)=\cos \frac{\theta}{2}\{\because \mathbf{a} \cdot \mathbf{b}=1 \times 1 \cos \theta=\cos \theta\} \\
& \Rightarrow \lambda 2 \cos ^2 \frac{\theta}{2}=\cos \frac{\theta}{2} \Rightarrow \lambda=\frac{1}{2 \cos \frac{\theta}{2}} \Rightarrow \mathbf{p}=\frac{\mathbf{a}+\mathbf{b}}{2 \cos \frac{\theta}{2}}
\end{aligned}
$$
Now, since $p$ and a are inclined at angle $\frac{\theta}{2}$, so
$$
\begin{aligned}
& \mathbf{p} \cdot \mathbf{a}=|\mathbf{p} \| \mathbf{a}| \cos \frac{\theta}{2} \\
& \Rightarrow \lambda(\mathbf{a}+\mathbf{b}) \cdot \mathbf{a}=1 \times 1 \cos \frac{\theta}{2} \\
& \Rightarrow \lambda(1+\cos \theta)=\cos \frac{\theta}{2}\{\because \mathbf{a} \cdot \mathbf{b}=1 \times 1 \cos \theta=\cos \theta\} \\
& \Rightarrow \lambda 2 \cos ^2 \frac{\theta}{2}=\cos \frac{\theta}{2} \Rightarrow \lambda=\frac{1}{2 \cos \frac{\theta}{2}} \Rightarrow \mathbf{p}=\frac{\mathbf{a}+\mathbf{b}}{2 \cos \frac{\theta}{2}}
\end{aligned}
$$
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