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

1. Problem Description

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

2. Solution Steps

Since $ABCDEF$ is a regular hexagon, all sides have equal length, and each interior angle is 120 degrees. Also, opposite sides are parallel.

* $\vec{CD}$: Since $ABCDEF$ is a regular hexagon, $\vec{CD}$ has the same length as $\vec{AB}$. The angle between $\vec{BC}$ and $\vec{CD}$ is $120^\circ$. We can express $\vec{CD}$ as $\vec{CD} = -p + q$

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

* $\vec{DE}$: Since $ABCDEF$ is a regular hexagon, $\vec{DE}$ has the same length as $\vec{BC}$. The angle between $\vec{CD}$ and $\vec{DE}$ is $120^\circ$. We can express $\vec{DE}$ as $-p$

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

* $\vec{EF}$: Since $ABCDEF$ is a regular hexagon, $\vec{EF}$ has the same length as $\vec{AB}$. The angle between $\vec{DE}$ and $\vec{EF}$ is $120^\circ$.

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

* $\vec{FA}$: Since $ABCDEF$ is a regular hexagon, $\vec{FA}$ has the same length as $\vec{BC}$.

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

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

Alternatively, since $AD$ connects opposite vertices in the hexagon, it goes through the center. Since the hexagon is regular, $AD = 2 BC$, hence $\vec{AD} = 2q$.

* $\vec{EA}$:

$\vec{EA} = \vec{ED} + \vec{DC} + \vec{CB} + \vec{BA} = p + (p-q) + (-q) + (-p) = p-2q$

Alternatively, $\vec{EA} = -\vec{AE} = -(\vec{AB}+\vec{BE}) = -(\vec{AB} + \vec{AD}) = -p - 2q$. Also, $\vec{EA} = - \vec{DE} + \vec{DA}$. Since $\vec{DA} = -\vec{AD}=-2q$, and $\vec{DE}=-p$ $\vec{EA}=-p-2q$.

So, $\vec{EA}= \vec{EF} + \vec{FA} = -q + p -q = p-2q$

* $\vec{AC}$:

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

3. Final Answer

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

$\vec{DE} = -p$

$\vec{EF} = -q$

$\vec{FA} = p - q$

$\vec{AD} = 2q$

$\vec{EA} = p - 2q$

$\vec{AC} = p + q$