We are given a regular hexagon $ABCDEF$. We are given that $\vec{AB} = p$ and $\vec{BC} = q$. We want to find 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 want to find 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, we know that all sides have equal length and all interior angles are equal to 120 degrees.

Also, in a regular hexagon $ABCDEF$, $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = 0$.

We have $\vec{AB}=p$ and $\vec{BC}=q$. Then $|\vec{AB}| = |\vec{BC}|$, and the angle between $\vec{AB}$ and $\vec{BC}$ is $120^\circ$.

From properties of regular hexagon we know that $\vec{CD} = \vec{BA} + \vec{BC} = -p + q$.

$\vec{DE} = \vec{CB} = -q$

$\vec{EF} = \vec{DC} = p - q$

$\vec{FA} = \vec{ED} = q$

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q + (q-p) = 2q$

$\vec{AC} = \vec{AB} + \vec{BC} = p + q$

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

3. Final Answer

$\vec{CD} = q - p$

$\vec{DE} = -q$

$\vec{EF} = p - q$

$\vec{FA} = q$

$\vec{AD} = 2q$

$\vec{EA} = -2q$

$\vec{AC} = p + q$