QJEEAdv18Phea49: JEE Advanced 2018 Physics - Class 11 JEEAdv18Phea49 Solution: Two vectors and are defined as and where is a constant and If…

The vector $\overrightarrow A = \alpha \widehat i$ has a magnitude a and its direction is fixed. The vector $\overrightarrow B = a(\cos \omega t\widehat i + \sin \omega t\widehat j)$ rotates with an angular speed $\omega$ = $\pi$ / 6 rad/s in a circle of radius a. The magnitude of the sum and the difference of these vectors are given by

$\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {(a + a\cos \omega t)\widehat i + a\sin \omega t\widehat j} \right|$

$= a\sqrt {{{(1 + \cos \omega t)}^2} + {{\sin }^2}\omega t}$

$= 2a\cos (\omega t/2)$ .

$\left| {\overrightarrow A - \overrightarrow B } \right| = \left| {(a - a\cos \omega t)\widehat i - a\sin \omega t\widehat j} \right|$

$= 2a\sin (\omega t/2)$

Hence,

${{\left| {\overrightarrow A - \overrightarrow B } \right|} \over {\left| {\overrightarrow A + \overrightarrow B } \right|}} = \tan {{\omega t} \over 2} = {1 \over {\sqrt 3 }}$ ,

which gives $t = \tau = (2/\omega )(\pi /6) = 2$ s.