10.4 Magnitude
The magnitude of a vector \( v : \Vector_{(x, y)} \) is the number
\[
\begin{align}
\magnitude v &: \Number
\\
\magnitude v &= \sqrt[2]{a^2 + b^2}
\end{align}
\]
when we've written
\[
\begin{align}
v &= a \mult \vec{\standard}_{x} + b \mult \vec{\standard}_{y}
\end{align}
\]
in standard form.
Example.
Let's find the magnitude of the vector
\[
\begin{align}
v &= 2 \mult \vec{\standard}_{x} - \vec{\standard}_{y}.
\end{align}
\]
Our vector has \( x \)-part two and \( y \)-part negative one. We calculate.
\[
\begin{align}
\magnitude v &= \sqrt[2]{2^2 + (-1)^2} = \sqrt[2]{4 + 1} = \sqrt[2]{5}
\end{align}
\]
The magnitude of a vector is a distance. We'll sometimes also refer to a vector's magnitude as its length.
Important.
Suppose a vector \( v : \Vector_{(x, y)} \) travels from a point \( p_1 : \Point_{(x, y)} \) to a point \( p_2 : \Point_{(x, y)} \).
Then the magnitude, \( \magnitude v \), is equal to the distance between \( p_1 \) and \( p_2 \).
Why?
By writing our vector in standard form
\[
\begin{align}
v &= a \mult \vec{\standard}_{x} + b \mult \vec{\standard}_{y},
\end{align}
\]
we can picture its parts as the legs of right triangle.
Let \( c : \Number \) be the distance between \( p_1 \) and \( p_2 \) so that we can apply the Pythagorean Theorem.
\[
\begin{align}
a^2 + b^2 &= c^2
\end{align}
\]
Taking square roots,
\[
\begin{align}
c &= \sqrt[2]{c^2} = \sqrt[2]{a^2 + b^2} = \magnitude v
\end{align}
\]
we see that \( c \) is equal to the magnitude of \( v \).
We can calculate the distance between two points by finding the magnitude of a vector between them.
Example.
Let's find the distance between the points
\[
\begin{align}
p_1 &= (-1, 1)
&
&\andSpaced
&
p_2 &= (4, -2).
\end{align}
\]
We'll start by finding the vector from \( p_1 \) to \( p_2 \).
\[
\begin{align}
v &= \position p_2 - \position p_1
\\
&= \bigl(4 \mult \vec{\standard}_{x} - 2 \mult \vec{\standard}_{y}\bigr) - \bigl(- \vec{\standard}_{x} + \vec{\standard}_{y}\bigr)
\\
&= 5 \mult \vec{\standard}_{x} - 3 \mult \vec{\standard}_{y}
\end{align}
\]
Now that we have our vector \( v \), let's find its magnitude.
\[
\begin{align}
\magnitude v &= \sqrt[2]{5^2 + (-3)^2} = \sqrt[2]{25 + 9} = \sqrt[2]{34}
\end{align}
\]
The distance between \( p_1 \) and \( p_2 \) is \( \sqrt[2]{34} \).
Because we've defined magnitude as a square root, magnitudes are always positive.
\[
\begin{align}
\magnitude v &\ge 0
\end{align}
\]
In fact, the only vector to have magnitude zero is the zero vector.
\[
\begin{align}
\magnitude 0 &= \sqrt[2]{0^2 + 0^2} = 0
\end{align}
\]
Every other vector actually travels some distance.