+3 Votes

The angle made by the vector \(\sqrt{3} \hat{i}+\hat{j}\) with \(x\)-axis is

(a) \(30^{\circ}\)

(b) \(45^{\circ}\)

(c) \(60^{\circ}\)

(d) \(90^{\circ}\)

in Physics by | 44 Views

1 Answer

+2 Votes
Best Answer

The correct option of this question will be (a).

Solution —

Let $\theta$ be the angle made by $\sqrt{3} \hat{i}+\hat{j}$ with $x$-axis.

If $\hat{i}$ is the unit vector along $x$-axis, then

$\cos \theta=\frac{(\sqrt{3} \hat{i}+\hat{j}) \cdot \hat{i}}{|\sqrt{3} \hat{i}+\hat{j}||\hat{i}|}$

=$\frac{\sqrt{3}}{\sqrt{\sqrt{3}^{2}+1^{2}} \sqrt{1^{2}}}$



$\theta=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=30^{\circ}$


Related Questions

Welcome To Informesia

Informesia Helps You To Prepare India's All States Board Exams (CBSE, ICSE, UP Board, BSEB, HPBOSE, RBSE, MSBSHSE) and Competitive Exam Like JEE (Main + Advanced), AIIMS, NEET, KVPY, NTSE, BITSAT, Olympiad, CLAT...etc.
For Any Query or Suggestion, Please Contact us on :
[email protected]