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: \[\overset{\to }{\mathop{A}}\,\,\times \,\overset{\to }{\mathop{B}}\,=AB\sin \theta \] |
| and \[\overset{\to }{\mathop{A}}\,\,.\,\overset{\to }{\mathop{B}}\,=AB\cos \theta \] |
| Given, \[|\overset{\to }{\mathop{A}}\,\,\,\times \,\overset{\to }{\mathop{B}}\,|=\sqrt{3}\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,\] ...(i) |
| but \[|\overset{\to }{\mathop{A}}\,\,\,\times \,\overset{\to }{\mathop{B}}\,|=|\overset{\to }{\mathop{A}}\,||\overset{\to }{\mathop{B}}\,|\sin \theta =AB\sin \theta \] |
| and \[\overset{\to }{\mathop{A}}\,.\,\overset{\to }{\mathop{B}}\,=|\overset{\to }{\mathop{A}}\,||\overset{\to }{\mathop{B}}\,|\cos \theta =AB\cos \theta \] |
| Make these substitution in Eq. (i), we get |
| \[AB\sin \theta =\sqrt{3}AB\cos \theta \] |
| or \[\tan \theta =\sqrt{3}\] |
| \[\therefore \] \[\theta ={{60}^{0}}\] |
| The addition of vector \[\overset{\to }{\mathop{A}}\,\] and \[\overset{\to }{\mathop{B}}\,\] can be given by the law of parallelogram. |
| \[\therefore \] \[|\overset{\to }{\mathop{A}}\,+\overset{\to }{\mathop{B}}\,|=\sqrt{{{A}^{1}}+{{B}^{2}}+2AB\cos \,{{60}^{0}}}\] |
| \[|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\times \frac{1}{2}}\] |
| \[{{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\] |
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