Given a regular hexagon $ABCDEF$, where $\vec{AB} = \vec{p}$ and $\vec{BC} = \vec{q}$, 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

Given a regular hexagon $ABCDEF$, where $\vec{AB} = \vec{p}$ and $\vec{BC} = \vec{q}$, 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

In a regular hexagon, we know that all sides are equal in length, and all interior angles are 120 degrees. Also, opposite sides are parallel.

Since $ABCDEF$ is a regular hexagon:

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

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

$\vec{CD}$: Since $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = \vec{0}$ and $\vec{AB}$ is parallel to $\vec{ED}$ and equal in length, then $\vec{ED} = -\vec{p}$. Also $\vec{BC}$ is parallel to $\vec{FE}$ and equal in length, then $\vec{FE} = -\vec{q}$. We can see that $\vec{CD} = \vec{FE} - \vec{BA} = \vec{p} - \vec{q}$.

$\vec{DE}$: Because $ABCDEF$ is a regular hexagon, the sides are of equal length, and $\angle ABC = 120^\circ$. Thus, $\angle BCD = 120^\circ$. We also have $\vec{CD} = \vec{BC} - \vec{AB} = \vec{q} - \vec{p}$. Since $AB || DE$ and the hexagon is regular, $\vec{DE} = -\vec{AB} = -\vec{p}$.

$\vec{EF}$: $\vec{EF} = -\vec{BC} = -\vec{q}$

$\vec{FA}$: $\vec{FA} = -\vec{CD} = -(\vec{q} - \vec{p}) = \vec{p} - \vec{q}$

$\vec{AD}$: $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} + (\vec{q} - \vec{p}) = 2\vec{q} = \vec{p}+\vec{q}+\vec{q}-\vec{p} = 2\vec{q}$.

Or $\vec{AD} = 2 \vec{BC}$ and $\vec{BC} = \vec{q}$.

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

$\vec{EA}$: $\vec{EA} = -\vec{DE} = -(-\vec{p}) = \vec{p}$. Also $\vec{EA} = \vec{BA} + \vec{FA}$. So $\vec{FA} = \vec{EA} - \vec{BA} = \vec{p} - (-\vec{p}) = \vec{EA}$ then $\vec{FA} = - \vec{CD} = \vec{p}-\vec{q}$ and $-\vec{AB} = -\vec{p}$ then $\vec{EA} = -\vec{AB} = \vec{p} -(\vec{p}-\vec{q}) = -\vec{AC}$

$\vec{EA} = -\vec{DE}=\vec{q}-\vec{p}$. Therefore $\vec{EA} = -\vec{AC}$. Since $\vec{AC}=\vec{p} + \vec{q}$ $\vec{EA} = - (\vec{AC}) = - (\vec{p} + \vec{q} - \vec{p})$.

Since ABCDEF is a regular hexagon, then $\vec{EA} = \vec{CB} + \vec{BA} = -\vec{q} -\vec{p}$. Therefore $\vec{EA} = -\vec{p}-\vec{q}$

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

3. Final Answer

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

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

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

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

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

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

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