Given a regular hexagon $ABCDEF$ with $\vec{AB} = p$ 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 $ABCDEF$ with $\vec{AB} = p$ 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 the same length, and all interior angles are equal to $120^\circ$.

We have $\vec{AB} = p$ and $\vec{BC} = q$.

Then $\vec{CD}$ has the same length as $AB$ and $BC$ and is at an angle of $120^\circ$ with respect to $\vec{BC}$. The sum of interior angles in a hexagon is $(6-2) \times 180^\circ = 720^\circ$. Therefore, each interior angle is $720^\circ / 6 = 120^\circ$.

Since it's a regular hexagon, we have:

$\vec{CD} = -p$

$\vec{DE} = -q$

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

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

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

$\vec{AD} = 2\vec{BC}-\vec{AB} = 2q-p =BC+BC=-p+2q$

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

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

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

Now consider $\vec{AD}$:

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

Alternatively $\vec{AD} = \vec{AF} + \vec{FE} + \vec{ED} = q + p + q = p+2q$

Since $BC=q$ and the vector along the axis of symmetry through B and opposite E is BE and AD = BE.

AD=BE= AB+AE = AB +2BC

If O is centre $\vec{AO}$ = $\vec{BC}$+$\vec{CD}$

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

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

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

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

$\vec{AD} = 2\vec{BC} - \vec{AB} = 2q+1\vec{AB}=(2q-1p)$

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

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

3. Final Answer

$\vec{CD} = -p$

$\vec{DE} = -q$

$\vec{EF} = -p$

$\vec{FA} = -q$

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

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

$\vec{AC} = p + q$