Question

Angle (in rad) made by the vector $ \sqrt{3\widehat{i}}+\widehat{j} $ with the X-axis:

• A. $ \frac{\pi }{6} $
• B. $ \frac{\pi }{4} $
• C. $ \frac{\pi }{3} $
• D. $ \frac{\pi }{2} $

Answer 1

The Correct Option is A

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} $