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

The Correct Option is C

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