A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: A
Solution :
Let \[\vec{A}=\sqrt{3\hat{i}}+\hat{j}\]and \[\vec{B}=\hat{i}\] (unit vector along X-axis) \[=\vec{A}.\vec{B}=(\sqrt{3}\hat{i}+\hat{j}).\hat{i}\] \[=\sqrt{3}+0=\sqrt{3}\] \[(\because \,\,\hat{i}.\hat{i}=1,\,\hat{j}.\hat{i}=0)\] Also, \[|\vec{A}|=\sqrt{{{(\sqrt{3})}^{2}}+{{(1)}^{2}}}=\sqrt{3+1}=2\] \[|\vec{B}|=\sqrt{{{(1)}^{2}}}=1\] \[\because \] \[AB\,\cos \theta =\vec{A}.\vec{B}\]where\[\theta \] is the angle between vector \[\vec{A}\]and X - axis. \[\cos \theta =\frac{\vec{A}.\vec{B}}{AB}=\frac{\sqrt{3}}{2.1}=\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}\] \[\theta =\frac{\pi }{6}\,\text{ rad}\]
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