Question

If $\overrightarrow{e_1}$ and $\overrightarrow{e_2}$ are two unit vectors and $\theta$ is the angle between them, then $sin\left(\begin{array}{c}\frac{\theta }{2}\end{array}\right)$ is:

• A. $\frac{1}{2}\left|\begin{array}{c}\overrightarrow{e_1}+\overrightarrow{e_2}\end{array}\right|$
• B. $\frac{1}{2}\left|\begin{array}{c}\overrightarrow{e_1}-\overrightarrow{e_2}\end{array}\right|$
• C. $\frac{\overrightarrow{e_1}.\overrightarrow{e_2}}{2}$
• D. $\frac{\left|\begin{array}{c}\overrightarrow{e_1}\times \overrightarrow{e_2}\end{array}\right|}{2\left|\begin{array}{c}\overrightarrow{e_1}\end{array}\right|\left|\begin{array}{c}\overrightarrow{e_2}\end{array}\right|}$

Answer 1

The correct option is B $\frac{1}{2}\left|\begin{array}{c}\overrightarrow{e_1}-\overrightarrow{e_2}\end{array}\right|$

For A: $|\overrightarrow{e_1}+\overrightarrow{e_2}{|}^2=1^2+1^2+2cos\theta =2(1+\cos \theta )=2\left(2{\cos }^2\frac{\theta }{2}\right)$ (as magnitude of unit vector is 1)

or $|\overrightarrow{e_1}+\overrightarrow{e_2}|=2\cos \frac{\theta }{2}$

For B: $|\overrightarrow{e_1}-\overrightarrow{e_2}{|}^2=1^2+1^2+2cos\theta =2(1-\cos \theta )=2\left(2{\sin }^2\frac{\theta }{2}\right)$

or $|\overrightarrow{e_1}-\overrightarrow{e_2}|=2\sin \frac{\theta }{2}$

or $\sin \frac{\theta }{2}=\frac{1}{2}|\overrightarrow{e_1}-\overrightarrow{e_2}|$

For C: $\frac{\overrightarrow{e_1}.\overrightarrow{e_2}}{2}=\frac{\cos \theta }{2}$

For D: $\frac{|\overrightarrow{e_1}\times \overrightarrow{e_2}|}{2|\overrightarrow{e_1}||\overrightarrow{e_2}|}=\frac{\sin \theta }{2}$