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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) $90^{\circ}$

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

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

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