Question

If $ |A\times B|=\sqrt{3}A.B, $ then the value of $ |A+B| $ is

• 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}} $

Answer 1

Answer: (A)

$ 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}}) $