JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 16)
Two vectors $${\overrightarrow P }$$ and $${\overrightarrow Q }$$ have equal magnitudes. If the magnitude of $${\overrightarrow P + \overrightarrow Q }$$ is n times the magnitude of $${\overrightarrow P - \overrightarrow Q }$$, then angle between $${\overrightarrow P }$$ and $${\overrightarrow Q }$$ is :
$${\sin ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$$
$${\cos ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$$
$${\sin ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$
$${\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$
Explanation
$$\left| {\overrightarrow P } \right| = \left| {\overrightarrow Q } \right| = x$$ ..... (i)
$$\left| {\overrightarrow P + \overrightarrow Q } \right| = n\left| {\overrightarrow P - \overrightarrow Q } \right|$$
$${P^2} + {Q^2} + 2PQ\cos \theta = {n^2}({P^2} + {Q^2} - 2PQ\cos \theta )$$
Using (i) in above equation
$$\cos \theta = {{{n^2} - 1} \over {1 + {n^2}}}$$
$$\theta = {\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$
$$\left| {\overrightarrow P + \overrightarrow Q } \right| = n\left| {\overrightarrow P - \overrightarrow Q } \right|$$
$${P^2} + {Q^2} + 2PQ\cos \theta = {n^2}({P^2} + {Q^2} - 2PQ\cos \theta )$$
Using (i) in above equation
$$\cos \theta = {{{n^2} - 1} \over {1 + {n^2}}}$$
$$\theta = {\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$
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