A) \[{{\left( {{A}^{2}}+{{B}^{2}}+\frac{AB}{\sqrt{3}} \right)}^{1/2}}\]
B) \[A+B\]
C) \[{{({{A}^{2}}+{{B}^{2}}+\sqrt{3}AB)}^{1/2}}\]
D) \[{{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
Correct Answer: D
Solution :
\[|\,\overrightarrow{A}\times \overrightarrow{B}|\,=\sqrt{3}(\overrightarrow{A}.\overrightarrow{B})\] \[AB\sin \theta =\sqrt{3}AB\cos \theta \]\[\Rightarrow \] \[\tan \theta =\sqrt{3}\]\\[\theta =60{}^\circ \] Now \[|\overrightarrow{R}|\,=\,|\overrightarrow{A}+\overrightarrow{B}|\,=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] \[=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\left( \frac{1}{2} \right)}\]\[={{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
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