A) \[{{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
B) \[{{\left( {{A}^{2}}+{{B}^{2}}+\frac{AB}{\sqrt{3}} \right)}^{1/2}}\]
C) \[A+B\]
D) \[{{({{A}^{2}}+{{B}^{2}}+\sqrt{3}AB)}^{1/2}}\]
Correct Answer: A
Solution :
Key Idea: \[\vec{A}\times \vec{B}=AB\,\sin \theta \] and \[\vec{A}\,\,.\,\,\vec{B}=AB\,\cos \theta \]. Given, \[\left| \vec{A}\times \vec{B} \right|=\sqrt{3}\,\vec{A}\,\,.\,\vec{B}\] ... (i) but \[\left| \vec{A}\times \vec{B} \right|=\left| {\vec{A}} \right|\left| {\vec{B}} \right|\,\sin \theta =AB\sin \theta \] and \[\vec{A}\,\,.\,\,\vec{B}=\left| {\vec{A}} \right|\left| {\vec{B}} \right|\,\cos \theta =AB\cos \theta \] Putting these values in Eq. (i), we get \[AB\sin \theta =\sqrt{3}AB\cos \theta \] or \[\tan \theta =\sqrt{3}\] \[\therefore \] \[\theta ={{60}^{o}}\] The addition of vector \[\vec{A}\] and \[\vec{B}\] can be given by the law of parallelogram. \[\therefore \] \[\left| \vec{A}+\vec{B} \right|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\,\cos {{60}^{o}}}\] \[=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\times \frac{1}{2}}\] \[={{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
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