Mathematics in Physics

JEE Main level, single correct, medium

51 of 121

Question 51 of 121

Two vectors A⃗\vec{A} and B⃗\vec{B} have equal magnitudes. The magnitude of (A⃗+B⃗)\left(\vec{A}+\vec{B}\right) is ' nn ' times the magnitude of (A⃗−B⃗)\left(\vec{A}-\vec{B}\right) . The angle between A⃗\vec{A} and B⃗\vec{B} is:

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Answer: (A) cos−1[n2−1n2+1]{cos}^{-1}\left[\dfrac{{n}^{2}-1}{{n}^{2}+1}\right]

Given: ∣A⃗+B⃗∣=n∣A⃗−B⃗∣\left|\vec{A}+\vec{B}\right|=n\left|\vec{A}-\vec{B}\right|, so, the magnitudes of this will be given as,
A2+B2+2ABcosθ=n(A2+B2−2ABcosθ)\sqrt{{A}^{2}+{B}^{2}+2ABcos\theta }=n\left(\sqrt{{A}^{2}+{B}^{2}-2ABcos\theta }\right);where θ\theta is the angle between A⃗\vec{A} and B⃗\vec{B}.
Also, ∣A⃗∣=∣B⃗∣\left|\vec{A}\right|=\left|\vec{B}\right|,
2A2+2A2cosθ=n2A2−2A2cosθ\sqrt{2{A}^{2}+2{A}^{2}cos\theta }=n\sqrt{2{A}^{2}-2{A}^{2}cos\theta }
Squaring both sides:

2A2(1+cosθ)=n22A2(1−cosθ)2{A}^{2}\left(1+cos\theta \right)={n}^{2}2{A}^{2}\left(1-cos\theta \right)
cosθ=n2−1n2+1cos\theta =\dfrac{{n}^{2}-1}{{n}^{2}+1}
i.e., θ=cos−1[n2−1n2+1]\theta ={cos}^{-1}\left[\dfrac{{n}^{2}-1}{{n}^{2}+1}\right]

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