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

1. Problem Description

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

2. Solution Steps

Since ABCDEF is a regular hexagon, we have $AB = BC = CD = DE = EF = FA$. Also, the angle between consecutive sides is $120^{\circ}$.

Since ABCDEF is a regular hexagon, we can express $\overrightarrow{CD}$ in terms of other vectors. We know $\overrightarrow{BC} = \vec{v}$.

Consider the parallelogram formed by the vectors $\vec{u}$ and $\vec{v}$. Then $\overrightarrow{AC} = \vec{u} + \vec{v}$. Since the hexagon is regular, $\overrightarrow{AC}$ is parallel to $\overrightarrow{ED}$. Furthermore, $AC = \sqrt{AB^2 + BC^2 - 2(AB)(BC)\cos(120^\circ)} = \sqrt{AB^2 + AB^2 - 2(AB)^2(-\frac{1}{2})} = \sqrt{3AB^2} = AB\sqrt{3}$.

We can observe that the angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$ is $120^{\circ}$. The angle between $\overrightarrow{BC}$ and $\overrightarrow{CD}$ is also $120^{\circ}$. The angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is $240^{\circ}$ or $-120^{\circ}$.

Consider $\overrightarrow{AD}$. Since the hexagon is regular, $\overrightarrow{AD} = 2 \overrightarrow{BC} + \overrightarrow{AB} - \overrightarrow{AB} = 2\overrightarrow{BC} = 2 \vec{v}$.

We know $\overrightarrow{AC} = \vec{u} + \vec{v}$. Consider $\overrightarrow{BD}$. Notice that $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = -\vec{u} + 2\vec{v}$.

Now consider the vector sum $\overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} = \overrightarrow{BE}$. We also have $\overrightarrow{BE} = \overrightarrow{BA} + \overrightarrow{AE} = -\vec{u} + \overrightarrow{AE}$.

Since ABCDEF is a regular hexagon, $\overrightarrow{AE} = 2\overrightarrow{BC} = 2\vec{v}$. Then $\overrightarrow{BE} = -\vec{u} + 2\vec{v}$. So $\vec{v} + \overrightarrow{CD} + \overrightarrow{DE} = -\vec{u} + 2\vec{v}$.

Then $\overrightarrow{CD} + \overrightarrow{DE} = -\vec{u} + \vec{v}$. We also know $\overrightarrow{DE} = -\overrightarrow{AB} = -\vec{u}$. So $\overrightarrow{CD} - \vec{u} = -\vec{u} + \vec{v}$. Therefore, $\overrightarrow{CD} = \vec{v} - \vec{u}$.

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

3. Final Answer

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