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JEE Mains · Physics · STD 11 - 3.1 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 -

  1. A \(\cos ^{-1}\left(\frac{n^{2}+1}{n^{2}-1}\right)\)
  2. B \(\cos ^{-1}\left(\frac{{n}^{2}-1}{-{n}^{2}-1}\right)\)
  3. C \(\cos ^{-1}\left(\frac{-n^{2}-1}{n^{2}-1}\right)\)
  4. D \(\cos ^{-1}\left(\frac{n^{2}+1}{n^{2}-1}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

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