JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 17)
Two vectors $$\overrightarrow X $$ and $$\overrightarrow Y $$ have equal magnitude. The magnitude of ($$\overrightarrow X $$ $$-$$ $$\overrightarrow Y $$) is n times the magnitude of ($$\overrightarrow X $$ + $$\overrightarrow Y $$). The angle between $$\overrightarrow X $$ and $$\overrightarrow Y $$ is :
$${\cos ^{ - 1}}\left( {{{ - {n^2} - 1} \over {{n^2} - 1}}} \right)$$
$${\cos ^{ - 1}}\left( {{{{n^2} - 1} \over { - {n^2} - 1}}} \right)$$
$${\cos ^{ - 1}}\left( {{{{n^2} + 1} \over { - {n^2} - 1}}} \right)$$
$${\cos ^{ - 1}}\left( {{{{n^2} + 1} \over {{n^2} - 1}}} \right)$$
Explanation
Given X = Y
$$\sqrt {{X^2} + {Y^2} - 2 \times Y\cos \theta } $$
$$ = n\sqrt {{X^2} + {Y^2} + 2 \times Y\cos \theta } $$
Square both sides
$$2{X^2}(1 - \cos \theta ) = {n^2}.2{X^2}(1 + \cos \theta )$$
$$1 - \cos \theta = {n^2} + {n^2}\cos \theta $$
$$\cos \theta = {{1 - {n^2}} \over {1 + {n^2}}}$$
$$\theta = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} \right]$$
$$\sqrt {{X^2} + {Y^2} - 2 \times Y\cos \theta } $$
$$ = n\sqrt {{X^2} + {Y^2} + 2 \times Y\cos \theta } $$
Square both sides
$$2{X^2}(1 - \cos \theta ) = {n^2}.2{X^2}(1 + \cos \theta )$$
$$1 - \cos \theta = {n^2} + {n^2}\cos \theta $$
$$\cos \theta = {{1 - {n^2}} \over {1 + {n^2}}}$$
$$\theta = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} \right]$$
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