Question

If $\overrightarrow{e_1}$ and $\overrightarrow{e_2}$ are unit vectors and $\theta$ is the angle between them, then $|\overrightarrow{e_1}-\overrightarrow{e_2}|$ is equal to :

• A. $2cos\left(\frac{\theta }{2}\right)$
• B. $2tan\left(\frac{\theta }{2}\right)$
• C. $2sin\left(\frac{\theta }{2}\right)$
• D. $tan\left(\frac{\theta }{2}\right)$

Answer 1

The correct option is C $2sin\left(\frac{\theta }{2}\right)$

Let $e_1$ and $e_2$ are two unit vector

angle between them is $\theta$

We know $|A-B|=\sqrt{|A{|}^2+|B{|}^2+2|A||B|\cos \theta }$

$|e_1-e_2|=\sqrt{|e_1{|}^2+|e_2{|}^2-2|e_1||e_2|\cos \theta }$

Here $e_1$ and $e_2$ are unit vector

So, $|e_1|=1|e_2|=1$

$|e_1-e_2|=\sqrt{1^2+1^2-2\times 1\times 1\cos \theta }=\sqrt{2-2\cos \theta }=\sqrt{2(1-\cos \theta )}=\sqrt{2\times 2{\sin }^2\frac{\theta }{2}}=2\sin \frac{\theta }{2}$

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

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