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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 :

JEE · Physics · previous-year question

  1. A.$${\cos ^{ - 1}}\left( {{{ - {n^2} - 1} \over {{n^2} - 1}}} \right)$$
  2. B.$${\cos ^{ - 1}}\left( {{{{n^2} - 1} \over { - {n^2} - 1}}} \right)$$correct
  3. C.$${\cos ^{ - 1}}\left( {{{{n^2} + 1} \over { - {n^2} - 1}}} \right)$$
  4. D.$${\cos ^{ - 1}}\left( {{{{n^2} + 1} \over {{n^2} - 1}}} \right)$$

Answer

B. $${\cos ^{ - 1}}\left( {{{{n^2} - 1} \over { - {n^2} - 1}}} \right)$$

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 $$2{X^2}(1 - \cos \theta ) = {n^2}.2{X^2}(1 + \cos \theta )$$ $$1 - \cos \theta = {n^2} + {n^2}\cos \theta $$ $$\cos \theta = {{1 - {n^2}} \over {1 + {n^2}}}$$ $$\theta = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} \right]$$

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