Two vectors X and Y have equal magnitude. The magnitude of ( X - Y) is n times the magnitude of ( X + Y). The angle between X and Y is -

(C) \cos ^{-1}\left(\frac{-n^{2}-1}{n^{2}-1}\right)

(D) \cos ^{-1}\left(\frac{n^{2}+1}{n^{2}-1}\right)

JEE Mains · Physics · STD 11 - Vectors

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 -

Pick an option to check your answer.

Verified Solution

Correct Answer

(B) \(\cos ^{-1}\left(\frac{{n}^{2}-1}{-{n}^{2}-1}\right)\)

Detailed 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

\(2 X ^{2}(1-\cos \theta)= n ^{2} \cdot 2 X ^{2}(1+\cos \theta)\)

\(1-\cos \theta= n ^{2}+ n ^{2} \cos \theta\)

\(\cos \theta=\frac{1- n ^{2}}{1+ n ^{2}}\)

\(\theta=\cos ^{-1}\left[\frac{ n ^{2}-1}{- n ^{2}-1}\right]\)

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