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, all sides have the same length, and all interior angles are equal to $120^{\circ}$.

Also, since $ABCDEF$ is a regular hexagon, we know that $\vec{AB}$, $\vec{BC}$, $\vec{CD}$, $\vec{DE}$, $\vec{EF}$, and $\vec{FA}$ have the same magnitude.

$\vec{CD}$ is parallel to $\vec{FA}$ and has the same magnitude. Also $\vec{CD} = -\vec{AB} = -\vec{p}$

$\vec{DE}$ is parallel to $\vec{BC}$ and has the same magnitude. Also $\vec{DE} = -\vec{BC} = -\vec{q}$

$\vec{EF}$ is parallel to $\vec{AB}$ and has the same magnitude. Also $\vec{EF} = -\vec{CD} = \vec{p}$

$\vec{FA}$ is parallel to $\vec{CD}$ and has the same magnitude. Also $\vec{FA} = -\vec{DE} = \vec{q}$

Now, consider $\vec{AD}$. Since $ABCDEF$ is a regular hexagon, $AD$ is twice the length of $AB\cos(30^{\circ}) = \sqrt{3}AB$. But since the hexagon is regular, we can express $\vec{AD}$ as the sum of vectors:

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} + (-\vec{p}) = \vec{p} + \vec{q} - \vec{p} = 2\vec{BC} = \vec{p} + \vec{q} + \vec{EF} = \vec{p} + \vec{q} - \vec{CD}$ But $\vec{CD}$ can be expressed as $-\vec{p} - \vec{q} + \vec{BC}$

$\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} - \vec{p} = \vec{q} + \vec{FA} + ...$

Since $ABCDEF$ is a regular hexagon, $\vec{AD} = 2\vec{BC} + 2\vec{0}= \vec{BC}+ \vec{BC}$ , so we can derive that $\vec{AD} = \vec{BC} - \vec{CD}$. We also have $\vec{FA} = \vec{BC}$ so $\vec{FA} = \vec{q}$.

Since $AD$ is made up of two parallel vectors of length $AB = p$, We can assume that it equals to $AD =2\vec{BC}$ because this creates a parrallelogram. Thus $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD}$

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

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

We know that the coordinates of a regular hexagon can be thought of as composed of 6 equliateral triangles.

Also, $\vec{AD}$ is $2\vec{BC}$ because of congruent triangles and similar vector transformations. Thus, $\vec{AD} = \vec{p}+\vec{q}-\vec{p} = 2\vec{BC}$.

Then $\vec{AD} = \vec{p} + \vec{q} -\vec{EF}$ where $\vec{EF} = p$. So we know from geometry or using $AD = 2(BC) = \vec{2q}$

$\vec{EA} = -\vec{AE}$. We have $\vec{AE} = \vec{AD} + \vec{DE} = 2\vec{BC} + \vec{DE} = 2\vec{q}-\vec{q} = \vec{q}$. Therefore, $\vec{EA} = -\vec{q}$

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

3. Final Answer

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

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

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

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

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

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

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