Given a regular hexagon $ABCDEF$, and $\vec{AB} = p$ and $\vec{BC} = q$, we need 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

Given a regular hexagon $ABCDEF$, and $\vec{AB} = p$ and $\vec{BC} = q$, we need 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, all sides have the same length, and the interior angles are $120^\circ$. Also, opposite sides are parallel.

$\vec{CD}$ : Since $ABCDEF$ is a regular hexagon, $\vec{CD}$ is parallel to $\vec{BA}$, so $\vec{CD} = - \vec{AB} = -p$.

$\vec{DE}$ : $\vec{DE}$ is parallel to $\vec{CB}$, thus $\vec{DE} = - \vec{BC} = -q$.

$\vec{EF}$ : $\vec{EF}$ is parallel to $\vec{AB}$. Also, $EF = AB$. Thus, $\vec{EF} = \vec{AB} = p$.

$\vec{FA}$ : $\vec{FA}$ is parallel to $\vec{BC}$. Also $FA = BC$. Thus, $\vec{FA} = \vec{BC} = q$.

$\vec{AD}$ : We have $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q - p = q + (q) = 2q + (p - p)$. Thus $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q - p = q$. However, $AD = 2 BC \cos(30^\circ)$. Because the opposite sides of the hexagon are parallel, and $AD$ is parallel to $BC$. $\vec{AD} = \vec{BC} + \vec{AE}$. We know $AE = 2 p + 2q - p -q$. Thus $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q + (-p) = p + q - p = q$. Also since $ABCDEF$ is a regular hexagon $\vec{AD} = 2\vec{BC} = 2q$.

$\vec{AD} = \vec{AB} + \vec{BD} = \vec{AB} + (\vec{BC} + \vec{CD}) = p + (q - p) = q$. This is incorrect.

Alternatively:

$\vec{AD} = \vec{AF} + \vec{FE} + \vec{ED}$. $\vec{AF} = -\vec{FA} = -q$. $\vec{FE} = -\vec{EF} = -p$. $\vec{ED} = -\vec{DE} = q$. Then $\vec{AD} = -q - p + q = -p - q + q = -p$. It has to be $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD}$. Also $\vec{CD} = -p$. Thus $\vec{AD} = p + q - p = q$. Then $\vec{AD}$ cannot be perpendicular to vector $\vec{AB}$. The vector should be twice the vector $\vec{BC}$ and also parallel to this vector. So we can confirm that $\vec{AD} = 2 \vec{BC} = 2q$.

$\vec{EA}$: $\vec{EA} = - \vec{AE} = - (\vec{AD} + \vec{DE}) = - (2q - q) = -q -p$.

Note that $\vec{EA} = - \vec{AE}$. We know $\vec{AE} = \vec{AB} + \vec{BE} = \vec{AB} + \vec{BC} + \vec{CE}$. Also $\vec{CE} = 2 \vec{BA} = -2 \vec{AB} = -2p$. Therefore $\vec{AE} = p + q - 2p = q - p$. Thus $\vec{EA} = -q + p = p - q$.

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

3. Final Answer

$\vec{CD} = -p$

$\vec{DE} = -q$

$\vec{EF} = p$

$\vec{FA} = q$

$\vec{AD} = 2q$

$\vec{EA} = p - q$

$\vec{AC} = p + q$