Question 22 of 121

The angle between vector $(\vec{A})$ and $(\vec{A}-\vec{B})$ is :

A${\tan }^{-1}\left(\dfrac{B\cos \theta }{A-B\sin \theta }\right)$B${\tan }^{-1}\left(\dfrac{\sqrt{3}B}{2A-B}\right)$C${\tan }^{-1}\left(\dfrac{-\dfrac{B}{2}}{A-B\dfrac{\sqrt{3}}{2}}\right)$D${\tan }^{-1}\left(\dfrac{A}{0.7B}\right)$

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Answer: (B) ${\tan }^{-1}\left(\dfrac{\sqrt{3}B}{2A-B}\right)$

$\tan \phi =\dfrac{B\sin 60^{\circ}}{A-B\cos 60^{\circ}}$
$\phi ={\tan }^{-1}\left(\dfrac{B\sin 60^{\circ}}{A-B\cos 60^{\circ}}\right)={\tan }^{-1}\left(\dfrac{\sqrt{3}B}{2A-B}\right)$

JEE Main Physics practice question on Mathematics in Physics (Addition and Subtraction of Vectors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.