Given a regular hexagon $ABCDEF$, with $\vec{AB} = p$ and $\vec{BC} = q$, we need to express the following vectors in terms of $p$ and $q$: $\vec{CD}$, $\vec{DE}$, $\vec{EF}$, $\vec{FA}$, $\vec{AD}$, $\vec{EA}$, and $\vec{AC}$.

1. Problem Description

Given a regular hexagon $ABCDEF$, with $\vec{AB} = p$ and $\vec{BC} = q$, we need to express the following vectors in terms of $p$ and $q$: $\vec{CD}$, $\vec{DE}$, $\vec{EF}$, $\vec{FA}$, $\vec{AD}$, $\vec{EA}$, and $\vec{AC}$.

2. Solution Steps

Since $ABCDEF$ is a regular hexagon, all sides have the same length, and each interior angle is $120^\circ$. We have the following relationships:

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

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

$\vec{CD}$: Since $ABCDEF$ is a regular hexagon, $\vec{CD}$ has the same length as $\vec{AB}$, but rotated $120^\circ$ counterclockwise relative to $\vec{AB}$, or $60^\circ$ clockwise relative to $\vec{BC}$. $\vec{CD} = \vec{BA} + \vec{AF}$, so $\vec{CD} = -\vec{p} + \vec{q} - \vec{p} = -\vec{p}$.

Since $\vec{CD}$ has same length as $\vec{AB}$ and forms 60 degrees with $\vec{BC}$, $\vec{CD} = -\vec{p}$.

$\vec{DE}$: $\vec{DE}$ has the same length as $\vec{BC}$ and is in the opposite direction, so $\vec{DE} = -\vec{q}$.

$\vec{EF}$: $\vec{EF}$ is parallel and equal in length to $\vec{AB}$ and points in the opposite direction. Thus $\vec{EF} = -\vec{p}$.

$\vec{FA}$: $\vec{FA}$ is parallel and equal in length to $\vec{BC}$ and points in the opposite direction. Thus $\vec{FA} = -\vec{q}$.

$\vec{AD}$: $\vec{AD}$ is twice the length of the altitude of an equilateral triangle with side length equal to the hexagon's side length. Alternatively, $\vec{AD}$ is parallel to $\vec{BC}$ rotated by 60 degrees and length twice that of height from $\vec{AC}$. Also $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} + (-\vec{p}) = \vec{p} + \vec{q} - \vec{p} = \vec{q} + \vec{BC}$. Therefore, $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} - \vec{p} = \vec{BC}$. So $\vec{AD}=2\vec{AE}$ and is twice the length of the altitude of the triangle. This is incorrect. $\vec{AD} = 2\vec{BC}+2\vec{AE}$ or $\vec{AD}$ is parallel to $\vec{BC}$ and is twice the height above.

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} + (-\vec{p}) = \vec{q} + (\vec{q})$. Therefore, $\vec{AD} = \vec{p} + \vec{q} + (-\vec{p}) = \vec{q}$. Since the hexagon is regular, $AD = 2BE$. The vector $\vec{AD} = \vec{AB}+\vec{BC}+\vec{CD} = \vec{p}+\vec{q}-\vec{p} = \vec{q}$.

$\vec{AD} = 2(\vec{q})$ which is $2\vec{q}$.

$\vec{EA}$: $\vec{EA} = -(\vec{AD} + \vec{DE}) = - (\vec{q}-\vec{p})= \vec{q}$. $\vec{EA}=\vec{qp}$. Also $\vec{EA}= \vec{FA}-\vec{FE}$. Thus $\vec{FE} = -\vec{EF}= \vec{p}$. So $\vec{EA}= \vec{q}+\vec{p}$

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

3. Final Answer

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

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

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

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

$\vec{AD} = \vec{BC}+\vec{CD} = \vec{p} + \vec{q}-\vec{p} = \vec{AB} + \vec{BC}= \vec{p}+\vec{q}$ so $\vec{AD}=\vec{AC}+\vec{CD}= \vec{AC}+\vec{BA} =$

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} + (-\vec{p}) = \vec{q}$, but must be length equals two. $AD = 2$.

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} - \vec{p} = \vec{q}$ $\vec{AD} = \vec{q} + (\vec{AB} + \vec{BA}) =$

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

$\vec{AD} = \vec{AB} + \vec{BC} = \vec{AD} = 2\vec{AN}= + \vec{FA}$.

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

$\vec{AC} = \vec{p} + \vec{q}$

Therefore:

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

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

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

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

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

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

$\vec{AC} = \vec{p} + \vec{q}$

Final Answer:

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

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

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

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

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

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

$\vec{AC} = \vec{p} + \vec{q}$