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 :
\[A\times B=AB\sin \theta \] \[A.B=AB\cos \theta \] \[|A\times B|=\sqrt{3}A.B\] \[|A\times B|=|A|\,\,|B|\sin \theta \] \[=AB\sin \theta \] \[A.B=|A|\,|B|\cos \theta =AB\cos \theta \] \[AB\sin \theta =\sqrt{2}AB\cos \theta \] \[\tan \theta =\sqrt{3},\theta =60{}^\circ \] Now \[{{(A+B)}^{2}}={{A}^{2}}+{{B}^{2}}+2A.B\] \[={{A}^{2}}+{{B}^{2}}+2AB\] \[={{A}^{2}}+{{B}^{2}}+AB\frac{1}{2}\] \[={{A}^{2}}+{{B}^{2}}+2A{{B}^{2}}\] or \[|A+B|=({{A}^{2}}+{{B}^{2}}+A{{B}^{1/2}})\]
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