We are given a regular hexagon $ABCDEF$. We are given that $\vec{AB} = p$ and $\vec{BC} = q$. We need to calculate the vectors $\vec{CD}$, $\vec{DE}$, $\vec{EF}$, $\vec{FA}$, $\vec{AD}$, $\vec{EA}$, and $\vec{AC}$ in terms of $p$ and $q$.

1. Problem Description

We are given a regular hexagon $ABCDEF$. We are given that $\vec{AB} = p$ and $\vec{BC} = q$. We need to calculate the vectors $\vec{CD}$, $\vec{DE}$, $\vec{EF}$, $\vec{FA}$, $\vec{AD}$, $\vec{EA}$, and $\vec{AC}$ in terms of $p$ and $q$.

2. Solution Steps

Since $ABCDEF$ is a regular hexagon, all sides have the same length, and all interior angles are equal to $120^{\circ}$.

First, consider $\vec{CD}$. Since $\vec{AB} + \vec{BC} + \vec{CD} = \vec{AD}$ and $\vec{AD} = 2 \vec{BC}$, we have $p + q + \vec{CD} = 2q$, so $\vec{CD} = 2q - p - q = q - p$.

Next, consider $\vec{DE}$. Since $\vec{DE}$ is parallel to $\vec{AB}$ but in the opposite direction, $\vec{DE} = -\vec{AB} = -p$.

Now, consider $\vec{EF}$. Since $\vec{EF}$ is parallel to $\vec{BC}$ but in the opposite direction, $\vec{EF} = -\vec{BC} = -q$.

Next, consider $\vec{FA}$. $\vec{FA}$ is parallel to $\vec{CD}$ but in the opposite direction, so $\vec{FA} = -\vec{CD} = -(q-p) = p-q$.

Now, consider $\vec{AD}$. Since $ABCDEF$ is a regular hexagon, $\vec{AD}$ is twice the length of the altitude of the equilateral triangle with side length equal to that of the hexagon.

We have $\vec{AD} = 2\vec{BC}$, so $\vec{AD} = 2q$.

Next, consider $\vec{EA}$. We have $\vec{EA} = -\vec{AE}$. Also, we have $\vec{EA} = \vec{ED} + \vec{DA} = p - 2q$.

Finally, consider $\vec{AC}$. We have $\vec{AC} = \vec{AB} + \vec{BC} = p + q$.

3. Final Answer

$\vec{CD} = q - p$

$\vec{DE} = -p$

$\vec{EF} = -q$

$\vec{FA} = p - q$

$\vec{AD} = 2q$

$\vec{EA} = p - 2q$

$\vec{AC} = p + q$