JEE Main 202125 Jul, Evening shiftPhysicsMediumKinematics177× since 2002Two vectors X→\overrightarrow XX and Y→\overrightarrow YY have equal magnitude. The magnitude of (X→\overrightarrow XX −-− Y→\overrightarrow YY) is n times the magnitude of (X→\overrightarrow XX + Y→\overrightarrow YY). The angle between X→\overrightarrow XX and Y→\overrightarrow YY is :AOption A. cos−1(−n2−1n2−1){\cos ^{ - 1}}\left( {{{ - {n^2} - 1} \over {{n^2} - 1}}} \right)cos−1(n2−1−n2−1)BOption B. cos−1(n2−1−n2−1){\cos ^{ - 1}}\left( {{{{n^2} - 1} \over { - {n^2} - 1}}} \right)cos−1(−n2−1n2−1)COption C. cos−1(n2+1−n2−1){\cos ^{ - 1}}\left( {{{{n^2} + 1} \over { - {n^2} - 1}}} \right)cos−1(−n2−1n2+1)DOption D. cos−1(n2+1n2−1){\cos ^{ - 1}}\left( {{{{n^2} + 1} \over {{n^2} - 1}}} \right)cos−1(n2−1n2+1)CheckSkip · show solution
