We are given a regular hexagon ABCDEF. We are given that $\vec{AB} = \vec{u}$ and $\vec{BC} = \vec{v}$. We want to find $\vec{CD}$ in terms of $\vec{u}$ and $\vec{v}$.

1. Problem Description

We are given a regular hexagon ABCDEF. We are given that $\vec{AB} = \vec{u}$ and $\vec{BC} = \vec{v}$. We want to find $\vec{CD}$ in terms of $\vec{u}$ and $\vec{v}$.

2. Solution Steps

Since the hexagon is regular, we can relate the vectors around the hexagon.

We can express $\vec{CD}$ as the negative of $\vec{FA}$. Thus $\vec{CD} = - \vec{FA}$.

We also know that $\vec{FA} = \vec{CB}$. Since $\vec{BC} = \vec{v}$, then $\vec{CB} = -\vec{v}$.

Also, $\vec{CD}$ is parallel to $\vec{BA}$, but with a length ratio equal to the ratio of the length of $BC$ to the length of $AB$. Since the hexagon is regular, all sides are equal.

Consider $\vec{AC} = \vec{AB} + \vec{BC} = \vec{u} + \vec{v}$.

Let the center of the hexagon be O.

Since the hexagon is regular, the angle between two consecutive vectors representing the sides of the hexagon is 120 degrees.

We know that $\vec{CD}$ is parallel to $\vec{BA}$, and $\vec{BA} = -\vec{AB} = -\vec{u}$.

However, the direction is not $-\vec{u}$. $\vec{CD}$ is not $-\vec{u}$, it is pointing in a similar direction but not directly opposite of $\vec{u}$.

Since the interior angle of a hexagon is $120^{\circ}$, consider a coordinate system such that $A = (0,0)$, $B = (1,0)$, and $C = (1 + \cos(120^{\circ}), \sin(120^{\circ})) = (1 - \frac{1}{2}, \frac{\sqrt{3}}{2}) = (\frac{1}{2}, \frac{\sqrt{3}}{2})$.

Then $\vec{u} = (1,0)$ and $\vec{v} = (\frac{1}{2}, \frac{\sqrt{3}}{2})$. We can find the coordinates of D to be $(\frac{1}{2} - \frac{1}{2}, \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = (0, \sqrt{3})$.

Then $\vec{CD} = (0 - \frac{1}{2}, \sqrt{3} - \frac{\sqrt{3}}{2}) = (-\frac{1}{2}, \frac{\sqrt{3}}{2})$.

We are looking for $a$ and $b$ such that $\vec{CD} = a\vec{u} + b\vec{v} = a(1,0) + b(\frac{1}{2}, \frac{\sqrt{3}}{2}) = (a + \frac{b}{2}, \frac{b\sqrt{3}}{2}) = (-\frac{1}{2}, \frac{\sqrt{3}}{2})$.

Then $\frac{b\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$, so $b=1$.

And $a + \frac{1}{2} = -\frac{1}{2}$, so $a = -1$.

Thus $\vec{CD} = -\vec{u} + \vec{v} = \vec{v} - \vec{u}$.

3. Final Answer

D. $\vec{v} - \vec{u}$