JEE Mains · Physics · STD 11 - 3.1 vectors
Two vectors \(\vec A\) and \(\vec B\) have equal magnitudes. The magnitude of \((\vec A + \vec B)\) is \(‘n’\) times the magnitude of \((\vec A - \vec B)\). The angle between \( \vec A\) and \(\vec B\) is
- A \({\cos ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]\)
- B \({\cos ^{ - 1}}\left[ {\frac{{n - 1}}{{n + 1}}} \right]\)
- C \({\sin ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]\)
- D \({\sin ^{ - 1}}\left[ {\frac{{n - 1}}{{n + 1}}} \right]\)
Answer & Solution
Correct Answer
(A) \({\cos ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]\)
Step-by-step Solution
Detailed explanation
\begin{array}{*{20}{l}} {\left| {\vec A + \vec B} \right| = n\left| {\vec A - \vec B} \right|} \\ { \Rightarrow \,{A^2} + {B^2} + 2AB\,\cos \,\theta } \\ {\,\,\,\,\,\,\, = {n^2}\left( {{A^2} + {B^2} - 2AB\,\cos \,\theta } \right)} \\ { \Rightarrow \,\cos \,\theta \,\left( {1 +…
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