Question 38 of 121

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

A$\left|\vec{A}\right|\left[1-\dfrac{{\left(∆\theta \right)}^{2}}{2}\right]$B$0$C$\left|\vec{A}\right|∆\theta$D$\left|\vec{B}\right|∆\theta -\left|\vec{A}\right|$

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Answer: (C) $\left|\vec{A}\right|∆\theta$

$\left|\vec{B}\right|=\left|\vec{A}\right|$
$\left|\vec{B}-\vec{A}\right|=\sqrt{{\left|\vec{A}\right|}^{2}+{\left|\vec{A}\right|}^{2}+2\left|\vec{A}\right|\left|\vec{A}\right|\cos \left(180^{\circ}-∆\theta \right)}$
$=\sqrt{{\left|\vec{A}\right|}^{2}+{\left|\vec{A}\right|}^{2}-2{\left|\vec{A}\right|}^{2}\cos ∆\theta }$
$=\sqrt{2{\left|\vec{A}\right|}^{2}-2{\left|\vec{A}\right|}^{2}\cos ∆\theta }$
$=\sqrt{2{\left|\vec{A}\right|}^{2}\left(1-\cos ∆\theta \right)}$
$∵1-\cos ∆\theta =2{\sin }^{2}\left(\dfrac{∆\theta }{2}\right)$
$=\sqrt{2{\left|\vec{A}\right|}^{2}\times 2{\sin }^{2}\left(\dfrac{∆\theta }{2}\right)}$
$=2\left|\vec{A}\right|\sin \left(\dfrac{∆\theta }{2}\right)$
For small $∆\theta ,\sin \left(\dfrac{∆\theta }{2}\right)≃\left(\dfrac{∆\theta }{2}\right)$
$\left|\vec{B}-\vec{A}\right|=2\left|\vec{A}\right|\left(\dfrac{∆\theta }{2}\right)$
$=\left|\vec{A}\right|∆\theta$

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