10.5 Scaling
Scaling a vector allows us to change the vector's length without changing its direction. We've seen, for example, that doubling a vector \( v \) produces a vector
\[
\begin{align}
&2 \mult v
\end{align}
\]
that is twice as long. Let's look at a theorem that makes this idea precise by comparing magnitudes.
Theorem.
Suppose \( v : \Vector_{} \) is a vector. Then
\[
\begin{align}
\magnitude \bigl(n \mult v\bigr) &= n \mult \magnitude v
\end{align}
\]
where \( n : \Number \) is any positive number, \( n \ge 0 \).
Why?
Writing \( v \) in standard form
\[
\begin{align}
v &= a \mult \vec{\standard}_{x} + b \mult \vec{\standard}_{y}
\end{align}
\]
allows us to calculate:
\[
\begin{align}
\magnitude \bigl(n \mult v\bigr) &= \magnitude \bigl( n a \mult \vec{\standard}_{x} + n b \mult \vec{\standard}_{y} \bigr)
\\
&= \sqrt[2]{(n a)^2 + (n b)^2}
\\
&= \sqrt[2]{n^2 \mult \bigl(a^2 + b^2\bigr)}
\\
&= \sqrt[2]{n^2} \mult \sqrt[2]{a^2 + b^2}
\\
&= n \mult \magnitude v.
\end{align}
\]
A unit vector is a vector that has magnitude one. Just as magnitude captures the intuitive idea of length, unit vectors capture the intuitive idea of direction. Every vector is determined by a length and a direction.
Important.
We may write any vector \( v : \Vector_{} \) as the product
\[
\begin{align}
v &= \magnitude v \mult v^{\prime}
\end{align}
\]
where \( v^{\prime} : \Vector_{} \) is a unit vector.
In fact, we can find the unit vector \( v^{\prime} \) by scaling
\[
\begin{align}
v^{\prime} &= \frac{v}{\magnitude v}
\end{align}
\]
as long as \( v \) is not the zero vector. This formula for the unit vector \( v^{\prime} \) almost looks like it breaks our rule, "no vectors as denominators." Looking carefully at types, however, we see that we are dividing \( v \), a vector, by \( \magnitude v \), a number.
Example.
Consider the vector
\[
\begin{align}
v &: \Vector_{(x, y)}
\\
v &= \vec{\standard}_{x} + 2 \mult \vec{\standard}_{y}.
\end{align}
\]
Let's find the magnitude of \( v \).
\[
\begin{align}
\magnitude v &= \sqrt[2]{1^2 + 2^2} = \sqrt[2]{5}
\end{align}
\]
We'll scale \( v \) to produce a unit vector \( v^{\prime} \).
\[
\begin{align}
v^{\prime} &= \frac{v}{\magnitude v} = \frac{\vec{\standard}_{x} + 2 \mult \vec{\standard}_{y}}{\sqrt[2]{5}} = \frac{1}{\sqrt[2]{5}} \mult \vec{\standard}_{x} + \frac{2}{\sqrt[2]{5}} \mult \vec{\standard}_{y}
\end{align}
\]
Since \( v^{\prime} \) is a unit vector, it fits as a ray on the unit circle.
