We are given a regular hexagon $ABCDEF$. We are also given that $\vec{AB} = \vec{p}$ and $\vec{BC} = \vec{q}$. We want to express the vectors $\vec{CD}$, $\vec{DE}$, $\vec{EF}$, $\vec{FA}$, $\vec{AD}$, $\vec{EA}$, and $\vec{AC}$ in terms of $\vec{p}$ and $\vec{q}$.

1. Problem Description

We are given a regular hexagon $ABCDEF$. We are also given that $\vec{AB} = \vec{p}$ and $\vec{BC} = \vec{q}$. We want to express the vectors $\vec{CD}$, $\vec{DE}$, $\vec{EF}$, $\vec{FA}$, $\vec{AD}$, $\vec{EA}$, and $\vec{AC}$ in terms of $\vec{p}$ and $\vec{q}$.

2. Solution Steps

Since $ABCDEF$ is a regular hexagon, all sides have equal length, and each interior angle is $120^\circ$. Therefore $|AB| = |BC| = |CD| = |DE| = |EF| = |FA|$. Also, $AB \parallel DE$, $BC \parallel EF$, $CD \parallel FA$, $AD \parallel BE \parallel CF$.

$\vec{CD} = \vec{BA} + \vec{BC} + \vec{CD} = \vec{BA} + \vec{BC} + \vec{AF} = \vec{BA} + \vec{AF}$

Since $ABCDEF$ is regular hexagon: $\vec{CD} = -\vec{p}$.

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

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

$\vec{FA} = \vec{ -BA} + \vec{-BC} + \vec{-CD} = -(-\vec{CD})$

$FA= -CD$.

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

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

Since $AD = 2 \cdot BC = 2 \cdot BE = q$

Therefore: $\vec{AD} = \vec{BC} + \vec{CF} -(\vec{q} ) = 2 \vec{BE}$

Since opposite sides are equal. $\vec{AD} = \vec{BC} + \vec{AF} = \vec{BC} - \vec{BC}$

Also. $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} + (-p) = \vec{q} + \vec{EF} + \vec{FA}$

In regular hexagon AD = 2BE

$AD = AB+BC+CD = \vec{p}+\vec{q}+CD$

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

$\vec{AD} = \vec{BC} + 2\vec{BM}$

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = (\vec{p} + \vec{q}) + (-\vec{p}) = \vec{q} + BE = \vec{BC}+\vec{CE}= \vec{BC}+\vec{AF}=\vec{CD}= 2q$

$\vec{AD}= \vec{p}+\vec{q}+ (-\vec{p}) = \vec{q} +(\vec{q}+\vec{p}-\vec{p}) \vec{2q} \implies \vec{AD} = \vec{CD} -2\vec{p}$

Since a regular hexagon can be divided into six equilateral triangles.

$\vec{AD} = 2\vec{BC} = 2\vec{q}$

Now we find $\vec{EA}$:

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

$\vec{AE} = \vec{AD} + \vec{DE} = 2\vec{q} - \vec{q} = \vec{q}+\vec{FA} = \vec{p}+\vec{q}=\vec{ED}$

$\vec{EA} = -\vec{AE}=-\vec{AE} = \vec{EF}+\vec{FA}$

$\vec{EA} = - (\vec{DE}+\vec{EF}+\vec{FA})= (\vec{-2AC})$.

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

Consider $\vec{AC}$:

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

$\vec{CD} = -p$

$\vec{DE} = -q$

$\vec{EF} = -p$

$\vec{FA} = -q$

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

$\vec{EA} = -(\vec{q})$

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

3. Final Answer

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

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

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

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

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

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

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