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(C) Given $a = i$ and $|b| = 1$. The angle between $a$ and $b$ is $120^\circ$.Since $a$ is along the $x$-axis,we can represent $b$ as $b = \cos(120^\c

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

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(C) Given $a = i$ and $|b| = 1$. The angle between $a$ and $b$ is $120^\circ$.Since $a$ is along the $x$-axis,we can represent $b$ as $b = \cos(120^\circ)i + \sin(120^\circ)j$.Calculating the components: $b = -\frac{1}{2}i + \frac{\sqrt{3}}{2}j$.Now,find the vector $a + b$: $a + b = i + (-\frac{1}{2}i + \frac{\sqrt{3}}{2}j) = \frac{1}{2}i + \frac{\sqrt{3}}{2}j$.Since $|a+b| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$,the vector $\frac{1}{2}i + \frac{\sqrt{3}}{2}j$ is already a unit vector.

Let $a = i$ be a vector which makes an angle of $120^\circ$ with a unit vector $b$. Then the unit vector $(a + b)$ is

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