10.3 Position Vectors
Given any point
\[
\begin{align}
p &: \Point_{(x, y)}
\\
p &= (x_p, y_p),
\end{align}
\]
the position vector for \( p \) is the vector
\[
\begin{align}
\position p &: \Vector_{(x, y)}
\\
\position p &= x_p \mult \vec{\standard}_{x} + y_p \mult \vec{\standard}_{y}
\end{align}
\]
that travels from the origin to \( p \).
Position vectors give us a way to convert any point to a vector. By changing types, we get access to vector arithmetic.
Formula.
The vector \( v : \Vector_{(x, y)} \) that starts at a point \( p_1 : \Point_{(x, y)} \) and ends at a point \( p_2 : \Point_{(x, y)} \) can be calculated as the difference of position vectors.
\[
\begin{align}
v &= \position p_2 - \position p_1
\end{align}
\]
Whenever we compute a change, we'll use this pattern:
\[
\begin{align}
&\text{to} - \text{from}.
\end{align}
\]
Let's see how this works in an example.
Example.
Let's find the vector \( v \) that starts at the point \( p_1 = (1, 3) \) and ends at the point \( p_2 = (5, 2) \). We'll start by converting our points to position vectors.
\[
\begin{align}
\position p_1 &= \vec{\standard}_{x} + 3 \mult \vec{\standard}_{y}
\\
\position p_2 &= 5 \mult \vec{\standard}_{x} + 2 \mult \vec{\standard}_{y}
\end{align}
\]
And now let's subtract to find the vector \( v \).
\[
\begin{align}
v &= \position p_2 - \position p_1
\\
&= \bigl(5 \mult \vec{\standard}_{x} + 2 \mult \vec{\standard}_{y}\bigr) - \bigl(\vec{\standard}_{x} + 3 \mult \vec{\standard}_{y}\bigr)
\\
&= 4 \mult \vec{\standard}_{x} - \vec{\standard}_{y}
\end{align}
\]
To move from \( p_1 \) to \( p_2 \), we travel four East and one South. We can check our work by drawing our vector \( v \).
Sometimes it can be helpful to rearrange the above formula.
Example.
Suppose we locate the vector \( v = \vec{\standard}_{x} - 3 \mult \vec{\standard}_{y} \) so that it starts at the point \( p_1 = (1, 2) \). At what point \( p_2 \) does the vector end?
To answer this question, let's find the position vector for \( p_2 \).
\[
\begin{align}
\position p_2 &= \position p_1 + v
\\
&= \bigl(\vec{\standard}_{x} + 2 \mult \vec{\standard}_{y}\bigr) + \bigl(\vec{\standard}_{x} - 3 \mult \vec{\standard}_{y}\bigr)
\\
&= 2 \mult \vec{\standard}_{x} - \vec{\standard}_{y}
\end{align}
\]
What point \( p_2 \) has this position vector? Easy.
\[
\begin{align}
p_2 &= (2, -1)
\end{align}
\]
To sum up, if we move one East and three South from \( p_1 = (1, 2) \), we'll end up at \( p_2 = (2, -1) \).
