Question

If $\theta$ is the angle between unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, then $\sin \left(\frac{\theta }{2}\right)$ equals

• A. $\frac{1}{2}|\overrightarrow{a}+\overrightarrow{b}|$
• B. $\frac{1}{2}|\overrightarrow{a}+\times \overrightarrow{b}|$
• C. $\frac{1}{2}\sqrt{1-\overrightarrow{a}.\overrightarrow{b}}$
• D. $\frac{1}{2}|\overrightarrow{a}-\overrightarrow{b}|$

Answer 1

The correct option is D $\frac{1}{2}|\overrightarrow{a}-\overrightarrow{b}|$

As $|\overrightarrow{a}-\overrightarrow{b}{|}^2=(\overrightarrow{a}-\overrightarrow{b})\cdot (\overrightarrow{a}-\overrightarrow{b})=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{b}$

$=1+1-2\left(1\right)\left(1\right)\cos \theta$

$\Rightarrow |\overrightarrow{a}-\overrightarrow{b}{|}^2=2(1-\cos \theta )$

$\Rightarrow |\overrightarrow{a}-\overrightarrow{b}{|}^2=2\left(2{\sin }^2\theta /2\right)$

$\Rightarrow {\sin }^2\theta /2=\frac{|\overrightarrow{a}-\overrightarrow{b}{|}^2}{4}$

$\Rightarrow \sin \frac{\theta }{2}=\frac{|\overrightarrow{a}-\overrightarrow{b}|}{2}$

$\Rightarrow \left(D\right)$.