Given a regular hexagon $ABCDEF$, we are given that $\vec{AB} = p$ and $\vec{BC} = q$. We need to 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 $ABCDEF$, we are given that $\vec{AB} = p$ and $\vec{BC} = q$. We need to 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 are equal in length and each interior angle is $120^\circ$.

$\vec{AB} = p$, $\vec{BC} = q$.

* $\vec{CD}$: Since the hexagon is regular, $\vec{CD}$ is parallel to $\vec{FA}$ and has the same length as $\vec{AB}$. The angle between $\vec{BC}$ and $\vec{CD}$ is $120^\circ$. We can express $\vec{CD}$ as $\vec{CD} = -p + q$. This is obtained by considering the parallelogram BCDG, where BG=CD. $\vec{CD} = \vec{BG} = \vec{BC}+\vec{CG} = \vec{BC}-\vec{AB}=q-p$. Thus $\vec{CD}= q-p$.

* $\vec{DE}$: $\vec{DE}$ is parallel to $\vec{AB}$, and has the same length. $\vec{DE} = -p$.

* $\vec{EF}$: $\vec{EF}$ is parallel to $\vec{BC}$, and has the same length, but points in the opposite direction. $\vec{EF} = -q$.

* $\vec{FA}$: $\vec{FA} = p-q$, since $\vec{FA}$ is parallel to $\vec{CD}$, opposite in direction, same magnitude.

* $\vec{AD}$: The vector $\vec{AD}$ is twice the length of the height of an equilateral triangle formed by two adjacent sides of the hexagon. It is also equal to $\vec{AB} + \vec{BC} + \vec{CD}$. Substituting the values, $\vec{AD} = p + q + (q-p) = 2q+0p$. Thus, $\vec{AD} = p+q+(q-p) = 2q$.

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

Thus, $\vec{EA} = p-2q$.

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

3. Final Answer

$\vec{CD} = q-p$

$\vec{DE} = -p$

$\vec{EF} = -q$

$\vec{FA} = p-q$

$\vec{AD} = 2q$

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

$\vec{AC} = p+q$