ExamBro
Web Login Download App
ExamBro
Download App
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Physics · STD 11 - 3.1 vectors

બે સદિશ \(\vec A\) અને \(\vec B\) સમાન માન ધરાવે છે. \((\vec A + \vec B)\) નું માન એ \((\vec A - \vec B)\) ના માન કરતા \(n\) ગણું છે. \(\vec A\) અને \(\vec B\) વચ્ચેનો ખૂણો કેટલો હશે?

  1. A \({\cos ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]\)
  2. B \({\cos ^{ - 1}}\left[ {\frac{{n - 1}}{{n + 1}}} \right]\)
  3. C \({\sin ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]\)
  4. D \({\sin ^{ - 1}}\left[ {\frac{{n - 1}}{{n + 1}}} \right]\)
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 +…

Same subject

More questions from Physics

Explore more questions on app
From JEE Mains

More PYQs from JEE Mains

Explore more questions on app