Question

A vector of magnitude $b$ is rotated through angle $\theta$. What is the change in magnitude of the vector?

• A. $2bsin\frac{\theta }{2}$
• B. $2bcos\frac{\theta }{2}$
• C. $2bsin\theta$
• D. $2bcos\theta$

Answer 1

The correct option is A $2bsin\frac{\theta }{2}$

$\overrightarrow{PQ}+\overrightarrow{QR}=\overrightarrow{PR}$
$\therefore \overrightarrow{QR}=\overrightarrow{b_1}-\overrightarrow{b}$

Now $|\overrightarrow{b_1}-\overrightarrow{b}{|}^2$ $=(\overrightarrow{b_1}-\overrightarrow{b}).(\overrightarrow{b_1}-\overrightarrow{b})$
$=(b_1{)}^2-2b_{1}bcos\theta +b^2$
$=2b^2(1-cos\theta )$
$\overrightarrow{b_1}-\overrightarrow{b}=\sqrt{2}b\sqrt{1-cos\theta }$ $=\sqrt{2}b\left(\sqrt{2}sin\frac{\theta }{2}\right)$ $=2bsin\frac{\theta }{2}$