We are given a regular hexagon $ABCDEF$. We are given that $\vec{AB} = p$ 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

We are given a regular hexagon $ABCDEF$. We are given that $\vec{AB} = p$ 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

In a regular hexagon, all sides are of equal length and all interior angles are equal to 120 degrees. We know that $\vec{AB} = p$ and $\vec{BC} = q$.

Since it is a regular hexagon, we have $|\vec{AB}| = |\vec{BC}| = |\vec{CD}| = |\vec{DE}| = |\vec{EF}| = |\vec{FA}|$. Also, the angle between $\vec{AB}$ and $\vec{BC}$ is 120 degrees.

$\vec{CD}$: Since it is a regular hexagon, $\vec{CD}$ will have the same length as $\vec{AB}$ and $\vec{FA}$ and its direction makes an angle of 120 degrees with $\vec{BC}$. Hence, $\vec{CD} = -p$.

$\vec{DE}$: The vector $\vec{DE}$ has the same length as $\vec{BC}$ but points in the opposite direction. Hence, $\vec{DE} = -q$.

$\vec{EF}$: The vector $\vec{EF}$ is parallel to $\vec{AB}$ but in the opposite direction. Thus, $\vec{EF} = -p$.

$\vec{FA}$: $\vec{FA}$ is parallel to $\vec{BC}$ but in the opposite direction. Thus, $\vec{FA} = -q$.

$\vec{AD}$: In a regular hexagon, the distance between opposite vertices is twice the length of a side. Also $\vec{AD}$ is parallel to $\vec{BC}$. Therefore, $\vec{AD} = 2\vec{BC} = 2q$.

$\vec{EA}$: $\vec{EA} = \vec{ED} + \vec{DA}$. We know $\vec{ED} = q$ and $\vec{DA} = -\vec{AD} = -2q$. Thus, $\vec{EA} = q - 2q = -q-p$.

Alternatively, $\vec{EA} = -\vec{AE}$. Also $\vec{AE} = \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = p + q -p -q =0$.

Thus, $\vec{AE} = \vec{AB} + \vec{BE}$. $\vec{AE} = -p -2q$

$\vec{EA} = -\vec{AE} = -(\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE}) = -(p+q+(-p)+(-q)) = q - p$

Since the hexagon is regular, consider the center $O$. We have $\vec{OA} = -\vec{DO}$. Thus, $\vec{AD} = 2\vec{OD} = 2\vec{BC} = 2q$. $\vec{AE} = \vec{AO}+\vec{OE} = -\vec{OA}+\vec{OE}$. Also, $\vec{OE} = \vec{BA} = -p$. Also, $\vec{OA} = \vec{p}+\vec{q}$. Then $\vec{AE} = -p-q-p = -2p-q$ and $\vec{EA} = 2p+q$.

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

3. Final Answer

$\vec{CD} = -p$

$\vec{DE} = -q$

$\vec{EF} = -p$

$\vec{FA} = -q$

$\vec{AD} = 2q$

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

$\vec{AC} = p+q$