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

1. Problem Description

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

2. Solution Steps

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

Also, $\vec{AB} = \vec{EF} = \vec{CD}$ and $\vec{BC} = \vec{DE} = \vec{FA}$.

Therefore,

$\vec{CD} = \vec{AB} = p$

$\vec{DE} = \vec{BC} = q$

$\vec{EF} = \vec{AB} = p$

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

Now, let's find $\vec{AD}$. In a regular hexagon, $AD$ is twice the length of $AB \sin(60^\circ)$ and $AD$ is parallel to $BC$. $\vec{AD} = 2\vec{MB} = 2(\vec{AB}+\vec{BC}) = 2 (\vec{AM} - \vec{CM})$. However, a simpler approach is $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p+q+p = 2p+q$.

Therefore, $\vec{AD} = 2p+q$.

Now, let's find $\vec{EA}$.

$\vec{EA} = -(\vec{AE}) = -(\vec{AD} + \vec{DE}) = -(\vec{AD} - \vec{ED}) = -\vec{AD} - \vec{DE}$

However $\vec{AE} = -\vec{FA} - \vec{FE} = -q - p$.

Alternatively, $\vec{EA} = -\vec{AE} = -(\vec{AB}+\vec{BC}+\vec{CD}+\vec{DE}+\vec{EF}) = -(\vec{AD}+\vec{DE}) = -(\vec{AF} + \vec{FE}) = -(p+q)-(-p) = -p - q - p =-q-p$.

$\vec{EA} = -\vec{FA} - \vec{EF} = -q - p$

Thus, $\vec{EA} = -p-q$.

Now, let's find $\vec{AC}$.

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

3. Final Answer

$\vec{CD} = p$

$\vec{DE} = q$

$\vec{EF} = p$

$\vec{FA} = q$

$\vec{AD} = 2p+q$

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

$\vec{AC} = p+q$