दो सदिशों overrightarrow{ A } तथा overrightar | vedclass

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Question

$\begin{array}{*{20}{l}} {\left| {\vec A + \vec B} ight| = n\left| {\vec A - \vec B} ight|} \ { \Rightarrow \,{A^2} + {B^2} + 2AB\,\cos \,	heta

Vedclass · Accepted Answer

${\cos ^{ - 1}}\left[ {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right]$

Vedclass · Answer

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

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