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

1. Problem Description

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

2. Solution Steps

Since $ABCDEF$ is a regular hexagon, we can deduce the relationships between the sides.

$\vec{CD}$ is parallel to $\vec{FA}$ and equal in length to $\vec{AB}$, but pointing in the opposite direction to $\vec{AB}$ rotated 60 degrees counter-clockwise. We can also say $\vec{CD}$ is equal to a negative rotation of $\vec{AB}$. But we need to express it in terms of $\vec{p}$ and $\vec{q}$. Also, $\vec{CD}$ is parallel to $\vec{BA}$ shifted by the vector $\vec{BC}$.

$\vec{CD} = \vec{BA} + \vec{BC} + \vec{BC} - \vec{BC} = \vec{BA}$.

Since $\vec{BA} = -\vec{AB} = -\vec{p}$.

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

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

$\vec{CD}$ can be expressed as $\vec{BA}$ translated by $\vec{BC} = \vec{q}$. The angle between $\vec{p}$ and $\vec{q}$ is 120 degrees.

$\vec{CD} = \vec{BA} + \vec{q} = \vec{EF}$

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

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

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

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

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

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

$\vec{DE}=-\vec{p}-\vec{q}+\vec{p} = \vec{CD}$ rotated by 60

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

$\vec{DE} = \vec{CA}=-\vec{AC} = -(\vec{p}+\vec{q})=-\vec{p}-\vec{q}$

$\vec{EF} = \vec{DB} = \vec{DA}+\vec{AB}= -2\vec{q} + \vec{p} = \vec{p}-2\vec{q}$

$\vec{FA} = \vec{EC} = -2\vec{p}-\vec{p} = -\vec{DE} = \vec{CD} = 2(\vec{q} - \vec{p}) - 2\vec{p}+\vec{q} = 3\vec{q} - 2\vec{p}$

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

$\vec{AD} = \vec{AC} + \vec{CD} = \vec{AB}+\vec{BC}+\vec{CD}$

Since $ABCDEF$ is a regular hexagon, the distance from $A$ to $D$ is twice the length of $BC$. The direction of $AD$ is the same as $BC$. Therefore, $\vec{AD} = 2\vec{BC} = 2\vec{q}$.

$\vec{EA} = -\vec{AE}$.

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

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

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

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

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

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

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

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

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

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

3. Final Answer

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

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

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

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

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

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

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