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The vector $\hat{a} = \hat{i}$ rotates anticlockwise in xy plane through an angle of $60^{o}$ and the new vector becomes $\hat{b}$ . Then $(\hat{a} + \hat{b})$ is

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$\frac{3}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}$

$\frac{1}{2} \hat{i} + \frac{3}{2} \hat{j}$

$\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}$

$\frac{\sqrt{3}}{2} \hat{i} + \frac{3}{2} \hat{j}$

answer is A.

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Detailed Solution

$\hat{a} = \hat{i}, \hat{b} = 1 \cos 60^{o} \hat{i} + 1 \sin 60^{o} \hat{j} = (\frac{\hat{i}}{2} + \frac{\sqrt{3} \hat{j}}{2})$

$∴ \hat{a} + \hat{b} = (\frac{3}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})$

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