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: \[\overrightarrow{A}\times \overrightarrow{B}=AB\text{ }sin\,\theta \] and \[\overrightarrow{A}\,.\,\overrightarrow{B}=AB\text{ cos}\,\theta \] Given, \[|\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{3}\overrightarrow{A}.\overrightarrow{B}\] ...(i) but \[|\overrightarrow{A}\times \overrightarrow{B}|=|\overrightarrow{A}||\overrightarrow{B}\text{ }\!\!|\!\!\text{ sin}\,\theta =AB\sin \theta \] and \[\overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}\text{ }\!\!|\!\!\text{ cos}\,\theta =AB\cos \theta \] Make these substitution in Eq. (i), we get \[AB\sin \theta =\sqrt{3}AB\cos \theta \] Or \[\tan \theta =\sqrt{3}\] \[\therefore \] \[\theta =60{}^\circ \] The addition of vector\[\overrightarrow{A}\]and\[\overrightarrow{B}\]can be given by the law of parallelogram. \[\therefore \]\[|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos 60{}^\circ }\] \[=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\times \frac{1}{2}}\] \[={{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
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