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

1. Problem Description

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

2. Solution Steps

In a regular hexagon, all sides are of equal length, and all interior angles are $120^\circ$.

The opposite sides are parallel.

$\vec{AB} = p$

$\vec{BC} = q$

Since $ABCDEF$ is a regular hexagon, we have:

$|AB| = |BC| = |CD| = |DE| = |EF| = |FA|$.

Also, $\angle ABC = 120^\circ$.

$\vec{CD}$: Since $CD$ is parallel to $FA$, we can say that $\vec{CD}$ is equivalent to a rotation of $\vec{AB}$ by $120^\circ$ clockwise from $\vec{AB}$, which is a rotation of $\vec{BC}=q$ by $60^\circ$. We can also say that it is $-p+q$.

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

$\vec{DE}$: Since $DE$ is parallel to $AB$ but in the opposite direction, with length $|DE|=|AB|$, $\vec{DE} = -p$

$\vec{EF}$: Since $EF$ is parallel to $BC$ but in the opposite direction, with length $|EF|=|BC|$, $\vec{EF} = -q$

$\vec{FA}$: Since $FA$ is parallel to $CD$ but in the opposite direction, with length $|FA|=|CD|$, $\vec{FA} = p - q$.

$\vec{AD}$: We can write $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q + (-p+q) = 2q+0p = -p + q+p +q +(-p+q)=p+q+(-p+q)=p+q-p+q = 2q+0p$, also $\vec{AD} = \vec{BC} + \vec{EF}=2q$. $\vec{AD}=2q$

$\vec{EA}$: We can write $\vec{EA} = - \vec{AE}$. Also, $\vec{AE} = \vec{AD} + \vec{DE} = 2q + (-p) = -p + 2q$. So, $\vec{EA} = p - 2q$

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

3. Final Answer

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

$\vec{DE} = -p$

$\vec{EF} = -q$

$\vec{FA} = p-q$

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

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

$\vec{AC} = p + q$