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)

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