Question

If $e_{1}, e_{2}$ are two unit vectors and $\theta$ is the angle between them, then $\cos \frac{\theta}{2}$ is

• A. $\frac{1}{2}\left|e_{1}+e_{2}\right|$
• B. $\frac{1}{2}\left|e_{1}-e_{2}\right|$
• C. $\frac{\left|e_{1} e_{2}\right|}{2}$
• D. $\frac{\left|e_{1}+e_{2}\right|}{2\left|e_{1}\right|\left|e_{2}\right|}$

Answer 1

Answer: (A)

$\left(e_{1}+e_{2}\right)^{2}=e_{1} \cdot e_{1}+2 e_{1} \cdot e_{2}+e_{2} \cdot e_{2}$
$\Rightarrow \left|e_{1}+e_{2}\right|=|e|^{2}+2\left|e_{1}\right|\left|e_{2}\right| \cos \theta+\left|e_{2}\right|^{2}$
$\Rightarrow \left|e_{1}+e_{2}\right|^{2}=1+2 \cdot 1 \cdot 1 \cos \theta+1$
$\left(\because \left|e_{1}\right|=\left|e_{2}\right|=1\right)$
$\Rightarrow \left|e_{1}+e_{2}\right|^{2}=2(1+\cos \theta)$
$=2\left(2 \cos ^{2} \frac{\theta}{2}\right)$
$\Rightarrow \left|e_{1}+e_{2}\right|^{2}=4 \cos ^{2} \frac{\theta}{2} $
$\Rightarrow \cos \frac{\theta}{2}=\frac{1}{2}\left|e_{1}+e_{2}\right|$