38 views

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) $90^{\circ}$

## 1 Answer

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

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

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

=$\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}$

or

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

by
5.0k Points

## Related Questions

3 Votes
1 Answer 61 Views
61 views
3 Votes
1 Answer 41 Views
41 views
3 Votes
1 Answer 28 Views
28 views
1 Vote
1 Answer 3 Views
3 views
1 Vote
1 Answer 3 Views
3 views
2 Votes
1 Answer 22 Views
22 views
3 Votes
1 Answer 47 Views
47 views
1 Vote
1 Answer 18 Views
18 views
1 Vote
1 Answer 17 Views
17 views
4 Votes
1 Answer 20 Views
20 views