Question

A vector of length l is turned through the angle $\theta$ about its tail. What is the magnitude change in the position vector of its head?

• A. $lcos\left(\theta /2\right)$
• B. $2lsin\left(\theta /2\right)$
• C. $2lcos\left(\theta /2\right)$
• D. $lsin\left(\theta /2\right)$

Answer 1

Answer: (B)

$2lsin\left(\theta /2\right)$

Using parallelogram law:
$|\overrightarrow{R}|=\sqrt{(l^2+l^2-2l^{2}cos\theta )}$

$|\overrightarrow{R}|=l\sqrt{2(1-cos\theta )}$

$|\overrightarrow{R}|=l\sqrt{2\left(2si{n}^2\right(\frac{\theta }{2}\left)\right)}$

$\therefore |\overrightarrow{R}|=2lsin\left(\frac{\theta }{2}\right)$