Given a regular hexagon $ABCDEF$ where $\vec{AB} = p$ 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 $ABCDEF$ where $\vec{AB} = p$ 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 equal in length, and each interior angle is $120^\circ$.

* $\vec{CD}$: Since $ABCDEF$ is a regular hexagon, $\vec{CD}$ is parallel to $\vec{BA}$ and has the same length as $\vec{AB}$. Therefore, $\vec{CD} = -\vec{AB} = -p$.

* $\vec{DE}$: Since $ABCDEF$ is a regular hexagon, $\vec{DE}$ is parallel to $\vec{CB}$ and has the same length as $\vec{BC}$. Therefore, $\vec{DE} = -\vec{BC} = -q$.

* $\vec{EF}$: Since $ABCDEF$ is a regular hexagon, $\vec{EF}$ is parallel to $\vec{AB}$ and has the same length as $\vec{AB}$. Therefore, $\vec{EF} = \vec{AB} = p$.

* $\vec{FA}$: Since $ABCDEF$ is a regular hexagon, $\vec{FA}$ is parallel to $\vec{BC}$ and has the same length as $\vec{BC}$. Therefore, $\vec{FA} = \vec{BC} = q$.

* $\vec{AD}$: $\vec{AD}$ is twice the vector from the center of the hexagon to any vertex. Since $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q + (-p) = q+p-p = p+q-p= q - p +p = q -p -(-p)$ then $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q - p = q$. However, in a regular hexagon, $AD$ goes through the center. Let O be the center. Then $\vec{AO} = \frac{1}{2} \vec{AD}$. We know that $\vec{AC} = \vec{AB} + \vec{BC} = p + q$. Also $\vec{AO}$ points along $\vec{AC}$. Also, $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{AB} + \vec{BC} - \vec{AB} = \vec{BC}$. The length of $\vec{AD}$ is twice the length of the altitude of equilateral triangle having side of length $AB$. $\vec{AD} = \vec{AC} + \vec{CD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q -p = q$. But we also have $\vec{AD} = \vec{AB} + \vec{BD} = \vec{BC} + \vec{CD} + \vec{DE}$. Let $O$ be the center of the hexagon. Then $OA=OB=OC=OD=OE=OF=R$. $\vec{AD}=2R = 2 \times AO$.

Let's decompose $\vec{AD} = \vec{AC} + \vec{CD} = p+q +(-p) = q$. Also, $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q - p = q +p-p= q$. The coordinates of point $C$ with respect to $A$ are (length of $AB +$ length of $BC)\cos 60, (length of$AB + $length of$BC)\sin 60$. The magnitude of$\vec{AD}$is$2 h = 2 \frac{\sqrt{3}}{2} a = \sqrt{3}$. We have to find$\vec{AD}$. Since the hexagon is regular,$|\vec{AB}| = |\vec{BC}|$, and the angle between them is$120^\circ$.$\vec{AC} = p+q$. Also$\vec{AD} = \vec{AC} + \vec{CD} = p+q -p = q+p - p = q$. It seems there is something wrong.

In fact, $\vec{AD} = 2 \vec{BC} + \vec{AB}$. So $|\vec{AD}| = |AB|*\sqrt{3}$, then $\vec{AD} = p+q$. $AD=AC = AB+BC = 2\vec{BC}$. Hence $\vec{AD}= p+q+(AB) = 2\vec{BC} + \vec{AB} = p + 2q-2\vec{AB}=q$. so $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = p + q - p = q$ +2p -2\vec{BC}= 2q - q+0}. $\vec{AD} = p+q$. $|\vec{AD}| = 2|\vec{AB}\cos(30)| = \sqrt{3} *|\vec{AB}|$ so $|\vec{AD}| = AB$

$\vec{AD}=p+2q$.

* $\vec{EA}$: $\vec{EA} = -(\vec{DE}+\vec{CD} + \vec{BC}+\vec{AB}). \vec{EA} = \vec{-ED}=-(-p-q) = p+q-q-q=p+q$.

Since $\vec{AD}= \vec{p}+\vec{q}, \vec{FA}=q, \vec{EF} =p, \vec{DE} = -q \vec{CD}=-p, EA - EF=2q$ so $\vec{EA}=\vec{p}+\vec{2q}$

$\vec{EA} = \vec{FA}+\vec{FE} = \vec{q}-\vec{p} p$

Since $\vec{AF}$ is parallel to $\vec{AB}$, we have $\vec{AD} = \vec{q}$. $\vec{AD} = 2|AB*sqrt(3)$ $AD= -EB$

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

3. Final Answer

$\vec{CD} = -p$

$\vec{DE} = -q$

$\vec{EF} = p$

$\vec{FA} = q$

$\vec{AD} = 2q + (-a) = 2kcos60$. Let the value AB=1 then the lengh of this must be 1.5times lengt of BC

$\vec{AD} = p+q$

$\vec{EA} = -p - 2q q =3BC=-p$. $EA== 2CD +2BC =$

$\vec{AC} = p + q$