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

(1) $\vec{CD}$: Since $ABCDEF$ is a regular hexagon, $\vec{CD}$ is parallel to $\vec{BA}$ and has the same length. Thus, $\vec{CD} = -\vec{AB} = -p$.

(2) $\vec{DE}$: Since $ABCDEF$ is a regular hexagon, $\vec{DE}$ is parallel to $\vec{CB}$ and has the same length. Thus, $\vec{DE} = -\vec{BC} = -q$.

(3) $\vec{EF}$: Since $ABCDEF$ is a regular hexagon, $\vec{EF}$ is parallel to $\vec{AB}$ and has the same length. Thus, $\vec{EF} = \vec{AB} = p$.

(4) $\vec{FA}$: Since $ABCDEF$ is a regular hexagon, $\vec{FA}$ is parallel to $\vec{BC}$ and has the same length. Thus, $\vec{FA} = \vec{BC} = q$.

(5) $\vec{AD}$: In a regular hexagon, $\vec{AD} = 2\vec{BC} + \vec{AB} - \vec{AB} = \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = p + q - p - q$. So $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{AC}+\vec{CD}$.

We know that $\vec{AC} = \vec{AB} + \vec{BC} = p + q$. Also, the distance $AD$ is twice the distance from the center to any vertex. Also, the angle between $\vec{AB}$ and $\vec{AD}$ is $60^{\circ}$, and the distance from $A$ to $D$ is twice the height of an equilateral triangle with side length equal to the hexagon's side length. Thus, $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD}$.

Also we have $\vec{AD} = 2(\vec{BC} - \vec{AB}\cos{120} ) = 2\vec{BE}$. $\vec{AD} = 2 \text{ times the distance from the center } O \text{ to } D$. Hence, $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD}$. We have that $\vec{CD} = -p$ from our work above.

$\vec{AD} = p + q -p = q$. However, this is incorrect. We know $\vec{AD}$ must have a magnitude of

2. We notice that the angle between $\vec{AB}$ and $\vec{BC}$ is $120^\circ$.

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = p + q - p - q$. $\vec{AD} = 2\vec{AO} = 2\vec{BE}$. So, $\vec{AO} = \vec{AE}$.

Consider the vector $\vec{AD}$. Note that the length of AD is twice the length of AB. The vector from A to the center O is equal to the vector from O to D. Let O be the center of the hexagon. Then $\vec{AO} = \vec{AB} + \vec{BO}$. Also, $\vec{DO} = -\vec{AO}$. $\vec{BO} = \vec{BC} + \vec{CO}$ where $\vec{CO} = \vec{AB}$.

Thus, $\vec{AO} = \vec{AB} + \vec{BC}$. Thus, $\vec{AD} = 2\vec{AO} = 2(\vec{AB} + \vec{BC}) = 2(p + q)$.

(6) $\vec{EA}$:

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

$\vec{EA} = - \vec{AE} = - (-\vec{FA}-\vec{FE})$. $\vec{EA} = \vec{FA} + \vec{EF}$.

Since $\vec{FA} = q$ and $\vec{EF} = p$, $\vec{EA} = q+p$. Also, $\vec{EA} = -\vec{AE} = - \vec{DE} - \vec{AD}$. We calculated $\vec{AD} = 2(p+q)$.

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

However, since we also know that $\vec{AO}=\vec{OE} = \vec{AB} + \vec{BC} = p+q$, $\vec{EA} = -\vec{OE} = q+p$.

$\vec{EA} = \vec{CD}$.

So we have $\vec{EA} = p+q$. Also $\vec{DE} = -q$, and $\vec{AD} = -\vec{DA} = \vec{AO} + \vec{OD} = 2(p+q)$.

So, this means $\vec{AD}=\vec{AO} + \vec{OD}$.

So, the $A$. $\vec{FA}=\vec{BC}$.

(7) $\vec{AC}$:

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

3. Final Answer

$\vec{CD} = -p$

$\vec{DE} = -q$

$\vec{EF} = p$

$\vec{FA} = q$

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

$\vec{EA} = p + q$

$\vec{AC} = p + q$