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JEE PYQ: Vector Algebra - Question ID d8759381ea9f (JEE Main 2021)

ID: d8759381ea9fJEE Main 2021Single Correct MCQ
Two vectors X→\overrightarrow X and Y→\overrightarrow Y have equal magnitude. The magnitude of (X→\overrightarrow X −- Y→\overrightarrow Y) is n times the magnitude of (X→\overrightarrow X + Y→\overrightarrow Y). The angle between X→\overrightarrow X and Y→\overrightarrow Y is :

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Step-by-step Explanation

Given X = Y

X2+Y2−2×Ycos⁡θ\sqrt {{X^2} + {Y^2} - 2 \times Y\cos \theta }

=nX2+Y2+2×Ycos⁡θ= n\sqrt {{X^2} + {Y^2} + 2 \times Y\cos \theta }

Square both sides

2X2(1−cos⁡θ)=n2.2X2(1+cos⁡θ)2{X^2}(1 - \cos \theta ) = {n^2}.2{X^2}(1 + \cos \theta )

1−cos⁡θ=n2+n2cos⁡θ1 - \cos \theta = {n^2} + {n^2}\cos \theta

cos⁡θ=1−n21+n2\cos \theta = {{1 - {n^2}} \over {1 + {n^2}}}

θ=cos⁡−1[n2−1−n2−1]\theta = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} \right]
View Related Chapter: Vector AlgebraPractice Chapter Test (CBT Mode)

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