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, \[|\vec{A}\times \vec{B}|=\sqrt{3}\vec{A}.\vec{B}\] but\[|\vec{A}\times \vec{B}|=|\vec{A}|\,\,|\vec{B}|sin\theta =AB\sin \theta \] and \[\vec{A}.\vec{B}=|\vec{A}|\,\,|\vec{B}|\cos \theta =AB\cos \theta \] Make these substitutions 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 vectors \[\text{\vec{A}}\] and \[\text{\vec{B}}\] can be given by the law of parallelogram. \[\text{ }\!\!|\!\!\text{ \vec{A}}\,\text{+}\,\text{\vec{B} }\!\!|\!\!\text{ }\,\text{=}\,\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos {{60}^{o}}}\] \[=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\times \frac{1}{2}}\] \[={{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
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