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(C) Let $\vec B$ be along the $x$-axis,so $\vec B = 1 \hat{i}$.Given the magnitude of $\vec A$ is $2$ and the angle between $\vec A$ and $\vec B$ is $

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(C) Let $\vec B$ be along the $x$-axis,so $\vec B = 1 \hat{i}$.Given the magnitude of $\vec A$ is $2$ and the angle between $\vec A$ and $\vec B$ is $60^{\circ}$,we can write $\vec A = 2(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) = 2(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = \hat{i} + \sqrt{3} \hat{j}$.Now,calculate the vector $\vec C = \frac{\vec A}{2} - \vec B$:$\vec C = \frac{1}{2}(\hat{i} + \sqrt{3} \hat{j}) - \hat{i} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} - \hat{i} = -\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}$.The magnitude of $\vec C$ is $|\vec C| = \sqrt{(-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$.To find the direction,note that $\vec C = \cos(120^{\circ}) \hat{i} + \sin(120^{\circ}) \hat{j}$.This vector makes an angle of $120^{\circ}$ with $\vec B$. Since the angle between $\vec A$ and $\vec B$ is $60^{\circ}$,the vector $\vec C$ is perpendicular to $\vec A$ (as $120^{\circ} - 60^{\circ} = 60^{\circ}$ is not $90^{\circ}$,let's recheck).Actually,the dot product $\vec C \cdot \vec B = (-\frac{1}{2})(1) + (\frac{\sqrt{3}}{2})(0) = -\frac{1}{2}$.The vector $\vec C$ is a unit vector. Looking at the options,the vector $\vec C$ has magnitude $1$. The direction is at $120^{\circ}$ to $\vec B$.

Two vectors $\vec A$ and $\vec B$ have magnitudes $2$ and $1$ respectively. If the angle between $\vec A$ and $\vec B$ is $60^{\circ}$,then which of the following vectors may be equal to $\frac{\vec A}{2} - \vec B$?

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