JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 5)
The angle between vector $$\left( {\overrightarrow A } \right)$$ and $$\left( {\overrightarrow A - \overrightarrow B } \right)$$ is :
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$${\tan ^{ - 1}}\left( {{{ - {B \over 2}} \over {A - B{{\sqrt 3 } \over 2}}}} \right)$$
$${\tan ^{ - 1}}\left( {{A \over {0.7B}}} \right)$$
$${\tan ^{ - 1}}\left( {{{\sqrt 3 B} \over {2A - B}}} \right)$$
$${\tan ^{ - 1}}\left( {{{B\cos \theta } \over {A - B\sin \theta }}} \right)$$
Explanation
_26th_August_Evening_Shift_en_5_2.png)
Angle between $${\overrightarrow A }$$ and $${\overrightarrow B }$$, $$\theta$$ = 60$$^\circ$$
Angle between $${\overrightarrow A }$$ and -$${\overrightarrow B }$$, $$\theta$$ = 120$$^\circ$$
If angle between $${\overrightarrow A }$$ and $${\overrightarrow A }$$ $$-$$ $${\overrightarrow B }$$ is $$\alpha $$
then $$\tan \alpha = $$$${{\left| { - \overrightarrow B } \right|\sin \theta } \over {\overrightarrow A + \left| { - \overrightarrow B } \right|\cos \theta }}$$
= $${{B\sin 120^\circ } \over {A + B\cos 120^\circ }}$$
=$${{B{{\sqrt 3 } \over 2}} \over {A - {B \over 2}}}$$
$$ \Rightarrow $$ $$\tan \alpha = {{\sqrt 3 B} \over {2A - B}}$$
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