Given a parallelogram $ABCD$. Point $E$ is on the line $BC$ such that $B$ is the midpoint of the segment $CE$. Given $\vec{AB} = p$ and $\vec{AC} = q$, find the following vectors: $\vec{BC}$, $\vec{CD}$, $\vec{DA}$, $\vec{BD}$, $\vec{AE}$, $\vec{CE}$, and $\vec{DE}$.

1. Problem Description

Given a parallelogram $ABCD$. Point $E$ is on the line $BC$ such that $B$ is the midpoint of the segment $CE$. Given $\vec{AB} = p$ and $\vec{AC} = q$, find the following vectors: $\vec{BC}$, $\vec{CD}$, $\vec{DA}$, $\vec{BD}$, $\vec{AE}$, $\vec{CE}$, and $\vec{DE}$.

2. Solution Steps

First, since $ABCD$ is a parallelogram, we have $\vec{AB} = \vec{DC} = p$ and $\vec{AD} = \vec{BC}$.

Also $\vec{AC} = \vec{AB} + \vec{BC}$, so $\vec{BC} = \vec{AC} - \vec{AB} = q - p$.

Since $\vec{AD} = \vec{BC}$, we have $\vec{AD} = q - p$.

Also, $\vec{CD} = -\vec{AB} = -p$.

Then, $\vec{DA} = -\vec{AD} = -(q - p) = p - q$.

We have $\vec{BD} = \vec{BA} + \vec{AD} = -\vec{AB} + \vec{AD} = -p + (q - p) = q - 2p$.

Since $B$ is the midpoint of $CE$, we have $\vec{BC} = \vec{BE} = q - p$. Thus $\vec{CE} = 2\vec{BC} = 2(q - p) = 2q - 2p$.

We have $\vec{AE} = \vec{AC} + \vec{CE} = q + (2q - 2p) = 3q - 2p$.

Finally, $\vec{DE} = \vec{DC} + \vec{CE} = p + (2q - 2p) = 2q - p$.

3. Final Answer

$\vec{BC} = q - p$

$\vec{CD} = -p$

$\vec{DA} = p - q$

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

$\vec{AE} = 3q - 2p$

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

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