A) \[\frac{a}{\sqrt{{{v}^{2}}+v_{1}^{2}}}\]
B) \[\sqrt{\frac{{{a}^{2}}}{{{v}^{2}}-v_{1}^{2}}}\]
C) \[\frac{a}{(v-{{v}_{1}})}\]
D) \[\frac{a}{(v+{{v}_{1}})}\]
Correct Answer: A
Solution :
Given\[|\vec{A}\times \vec{B}|=\sqrt{3}\vec{A}.\vec{B}\] but \[|\vec{A}\times \vec{B}|=|\vec{A}||\vec{B}|sin\theta =AB\sin \theta \] and \[\vec{A}.\vec{B}=|\vec{A}||\vec{B}|cos\theta =AB\cos \theta \] Make these substitution in Eq. (i), we get \[AB\sin \theta =\sqrt{3}\,AB\cos \theta \] or \[\tan \theta =\sqrt{3}\Rightarrow \theta ={{60}^{o}}\] The resultant of vector \[\vec{A}\]and \[\vec{B}\]can be given by the law of parallelogram. \[\therefore \] \[|\vec{A}+\vec{B}|=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos {{60}^{o}}}\] \[=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\times \frac{1}{2}}\] \[{{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
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