Two vectors A and B have equal magnitudes. The magnitude of ( A + B) is ‘n’ times the magnitude of ( A - B). The angle between A and B is

(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]

enEnglishhiहिन्दी guગુજરાતી

JEE Mains · Physics · STD 11 - Vectors

JEE Mains 2019, 2021
Hard
MCQ

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

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\begin{array}{*{20}{l}} {\left| {\vec A + \vec B} \right| = n\left| {\vec A - \vec B} \right|} \\ { \Rightarrow \,{A^2} + {B^2} + 2AB\,\cos \,\theta } \\ {\,\,\,\,\,\,\, = {n^2}\left( {{A^2} + {B^2} - 2AB\,\cos \,\theta } \right)} \\ { \Rightarrow \,\cos \,\theta \,\left( {1 + {n^2}} \right) = \frac{{2{a^2}\left( {{n^2} - 1} \right)}}{{2{a^2}}}\,\,} \\ {[A = B = c]} \\ {\cos \,\theta = \frac{{{n^2} - 1}}{{{n^2} + 1}}} \end{array}\)

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app