JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 24)
Two vectors $$\overrightarrow A $$ and $$\overrightarrow B $$ have equal magnitudes. The magnitude of $$\left( {\overrightarrow A + \overrightarrow B } \right)$$ is 'n' times the magnitude of $$\left( {\overrightarrow A - \overrightarrow B } \right)$$ . The angle between $${\overrightarrow A }$$ and $${\overrightarrow B }$$ is -
$${\sin ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$$
$${\sin ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$$
$${\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]$$
$${\cos ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]$$
Explanation
$$\left| {\overrightarrow A + \overrightarrow B } \right| = 2a\cos \theta /2$$ . . . (1)
$$\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2$$ . . . (2)
$$ \Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}$$
$$ \Rightarrow \,\,\,\tan {\theta \over 2} = n$$
$$\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2$$ . . . (2)
$$ \Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}$$
$$ \Rightarrow \,\,\,\tan {\theta \over 2} = n$$
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