Mathematics in Physics

JEE Main level, single correct, medium

2 of 121

Question 2 of 121

Two vectors X⃗\vec{X} and Y⃗\vec{Y} have equal magnitude. The magnitude of (X⃗−Y⃗)(\vec{X}-\vec{Y}) is nn times the magnitude of (X⃗+Y⃗)(\vec{X}+\vec{Y}). The angle between X⃗\vec{X} and Y⃗\vec{Y} is :

Show answer and solution

Answer: (B) cos⁡−1(n2−1−n2−1){\cos }^{-1}\left(\dfrac{{n}^{2}-1}{-{n}^{2}-1}\right)

Given X=YX=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}\cdot 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 =\dfrac{1-{n}^{2}}{1+{n}^{2}}
θ=cos⁡−1[n2−1−n2−1]\theta ={\cos }^{-1}\left[\dfrac{{n}^{2}-1}{-{n}^{2}-1}\right]

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