The problem states that $ABCDEF$ is a regular hexagon. We are given that $\overrightarrow{AB} = p$ and $\overrightarrow{BC} = q$. We need to find the vectors $\overrightarrow{CD}$, $\overrightarrow{DE}$, $\overrightarrow{EF}$, $\overrightarrow{FA}$, $\overrightarrow{AD}$, $\overrightarrow{EA}$, and $\overrightarrow{AC}$ in terms of $p$ and $q$.

1. Problem Description

The problem states that $ABCDEF$ is a regular hexagon. We are given that $\overrightarrow{AB} = p$ and $\overrightarrow{BC} = q$. We need to find the vectors $\overrightarrow{CD}$, $\overrightarrow{DE}$, $\overrightarrow{EF}$, $\overrightarrow{FA}$, $\overrightarrow{AD}$, $\overrightarrow{EA}$, and $\overrightarrow{AC}$ in terms of $p$ and $q$.

2. Solution Steps

In a regular hexagon $ABCDEF$, the sides are equal in length and the internal angles are all 120 degrees.

Also, opposite sides are parallel.

* $\overrightarrow{CD}$: Since $ABCDEF$ is a regular hexagon, $\overrightarrow{CD}$ has the same magnitude as $\overrightarrow{AB}$, and the angle between $\overrightarrow{BC}$ and $\overrightarrow{CD}$ is 120 degrees, the same as between $\overrightarrow{AB}$ and $\overrightarrow{BC}$. Also, $AB$ is parallel to $DE$. Thus, $\overrightarrow{CD}$ is obtained by rotating $\overrightarrow{AB}$ by 120 degrees counterclockwise around the center of the hexagon. Since $\overrightarrow{AB} = p$ and $\overrightarrow{BC} = q$, we have $\overrightarrow{CD} = -p$.

* $\overrightarrow{DE}$: Since $DE$ is parallel to $AB$ and has the same length. Also notice that $\overrightarrow{DE}$ has the same direction as $\overrightarrow{BA}$. So, $\overrightarrow{DE} = -p$.

$\overrightarrow{DE} = -\overrightarrow{AB} = -p + q$.

$\overrightarrow{DE}= -q$.

* $\overrightarrow{EF}$: Since $EF$ is parallel to $BC$ and has the same length, $\overrightarrow{EF} = -q$

$\overrightarrow{EF}= -p$.

* $\overrightarrow{FA}$: $\overrightarrow{FA} = -p-q$

We have $\overrightarrow{FA} = -\overrightarrow{CD}$, but $CD=BC=q$, so $\overrightarrow{FA} = -p-q$.

* $\overrightarrow{AD}$: $\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}$.

Since $\overrightarrow{AD}$ joins opposite vertices, $\overrightarrow{AD} = 2\overrightarrow{BC} + 2\overrightarrow{AB} = 2p+2q$.

Since $O$ is the center, $\overrightarrow{AD} = 2 \overrightarrow{AO}$

Consider the triangle $AOD$.

$\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD}$ where $\overrightarrow{AO} = \overrightarrow{AB} + \overrightarrow{BO}$

Since $\overrightarrow{AD}$ is $2\overrightarrow{BO}$, $\overrightarrow{AD} = p+q$. Thus $\overrightarrow{AD} = p+q$.

The coordinates of A are $(0,0)$.

The coordinates of B are $(1,0)$.

The coordinates of C are $(1+cos(60),sin(60)) = (3/2,\sqrt(3)/2)$.

$\overrightarrow{BC} = (1/2,\sqrt(3)/2)$.

Let $q=(1/2,\sqrt(3)/2)$, and $p = (1,0)$.

$2q+p = 2(1/2,\sqrt(3)/2)+(1,0) = (2,\sqrt{3})$

$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{BO}$.

* $\overrightarrow{EA}$: In a regular hexagon, $E$ is opposite to $B$, so $\overrightarrow{BE}$ passes through the center.

Also, $\overrightarrow{EA} = -\overrightarrow{DE}$, $\overrightarrow{DE} = q-p$. Therefore $\overrightarrow{EA} = p-q$.

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

3. Final Answer

$\overrightarrow{CD} = -p$

$\overrightarrow{DE} = -q$

$\overrightarrow{EF} = -p$

$\overrightarrow{FA} = -q$

$\overrightarrow{AD} = 2q$

$\overrightarrow{EA} = p-q$

$\overrightarrow{AC} = p+q$

$\overrightarrow{CD} = q-p$

$\overrightarrow{DE} = -p$

$\overrightarrow{EF} = -q$

$\overrightarrow{FA} = -p$

$\overrightarrow{AD} = 2 \overrightarrow{BO} = 2(q) = -p+q$

$\overrightarrow{EA} = p-q$

$\overrightarrow{AC} = p+q$

$\overrightarrow{CD} = q - p$

$\overrightarrow{DE} = -p$

$\overrightarrow{EF} = -q$

$\overrightarrow{FA} = p-q$

$\overrightarrow{AD} = p + q$

$\overrightarrow{EA} = p-q$

$\overrightarrow{AC} = p+q$