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