Question

A vector $\vec{A}$ is rotated by a small angle $\Delta \theta$ radians $\left(\Delta \theta <<1\right)$ to get a new vector $\vec{B}.$ In that case $\left|\vec{B}-\vec{A}\right|$ is ;

• A. $0$
• B. $\left|\vec{A}\right|\left(1-\frac{\Delta {\theta }^2}{2}\right)$
• C. $\left|\vec{A}\right|\Delta \theta$
• D. $\left|\vec{B}\right|\Delta \theta -\left|\vec{A}\right|$

Answer 1

Answer: (C)

$|\vec{B}-\vec{A}|=\sqrt{A^2+B^2+2AB\cos (\pi -\Delta \theta )}$
$\Rightarrow$ since $|\vec{A}|=|\vec{B}|=A,$
$\cos (\pi -\Delta \theta )=-\cos \Delta \theta$
$=\sqrt{2A^2(1+\cos (\pi -\Delta \theta )]}$
$=\sqrt{2A^{2}2{\cos }^2\left(\frac{\pi -\Delta \theta }{2}\right)}=2A\cos \left(\frac{\pi }{2}-\frac{\Delta \theta }{2}\right)=2A\sin \left(\frac{\Delta \theta }{2}\right)$
$\sin \Delta \theta <\sin \left(\frac{\Delta \theta }{2}\right)\approx \frac{\Delta \theta }{2}$ $\Rightarrow |\vec{B}-\vec{A}|=A\Delta \theta$