Question

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

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

Answer 1

Answer: (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. $\cos \theta =\frac{\vec{A}.\vec{B}}{AB}=\frac{\sqrt{3}}{2.1}=\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$ $\theta =\frac{\pi }{6}\text{ rad}$