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Section 12.4 Examples
Subsection 12.4.1 The norm of a vector
Example 12.4.1 . Basic computation examples.
Here are a few examples of computing the norm of a vector, in various dimensions.
Consider
\(\uvec{u} = (1,2)\) in
\(\R^2\text{.}\) Then,
\begin{equation*}
\unorm{u} = \sqrt{1^2 + 2^2} = \sqrt{5} \text{.}
\end{equation*}
Consider
\(\uvec{v} = (1,2,-1)\) in
\(\R^3\text{.}\) Then,
\begin{equation*}
\unorm{v} = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \text{.}
\end{equation*}
Consider
\(\uvec{w} = (1,2,-1,5)\) in
\(\R^4\text{.}\) Then,
\begin{equation*}
\unorm{w} = \sqrt{1^2 + 2^2 + (-1)^2 + 5^2} = \sqrt{31} \text{.}
\end{equation*}
Example 12.4.2 . Norms of the standard basis vectors.
The standard basis vectors in \(\R^n\) are always unit vectors :
\begin{align*}
\norm{\uvec{e}_1} \amp= \sqrt{1^2 + 0^2 + \dotsb + 0^2} = \sqrt{1} = 1,\\
\norm{\uvec{e}_2} \amp= \sqrt{0^2 + 1^2 + 0^2 + \dotsb + 0^2} = \sqrt{1} = 1,\\
\amp\vdots\\
\norm{\uvec{e}_n} \amp= \sqrt{0^2 + \dotsb + 0^2 + 1^2} = \sqrt{1} = 1.
\end{align*}
Example 12.4.3 . Normalizing vectors.
We can scale any nonzero vector to a unit vector by dividing by its norm, and this normalized version of the vector will always be parallel to the original.
We computed the norm of
\(\uvec{u} = (1,2)\) to be
\(\unorm{u} = \sqrt{5}\text{.}\) Therefore, the scaled vector
\begin{equation*}
\uvec{u}' = \frac{1}{\sqrt{5}} \uvec{u} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)
\end{equation*}
is a unit vector (i.e.
\(\norm{\uvec{u}'} = 1\) ).
We computed the norm of
\(\uvec{v} = (1,2,-1)\) to be
\(\unorm{v} = \sqrt{6}\text{.}\) Therefore, the scaled vector
\begin{equation*}
\frac{1}{\sqrt{6}} \uvec{v} = \left( \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right)
\end{equation*}
is a unit vector.
We computed the norm of
\(\uvec{w} = (1,2,-1,5)\) to be
\(\unorm{w} = \sqrt{31}\text{.}\) Therefore, the scaled vector
\begin{equation*}
\frac{1}{\sqrt{31}} \uvec{v}
= \left(
\frac{1}{\sqrt{31}},
\frac{2}{\sqrt{31}},
-\frac{1}{\sqrt{31}},
\frac{5}{\sqrt{31}},
\right)
\end{equation*}
is a unit vector.
Subsection 12.4.2 Dot product and the angle between vectors
Here is an example of using the dot product to determine the angle between vectors.
Example 12.4.4 . Computing angle from dot product.
What is the angle between vectors \(\uvec{u} = (1,2)\) and \(\uvec{v} = (-1,3)\) in \(\R^2\text{?}\)
From
Discovery 12.7 , we know that the angle
\(\theta\) between
\(\uvec{u}\) and
\(\uvec{v}\) satisfies
\begin{equation*}
\cos\theta = \frac{\udotprod{u}{v}}{\unorm{u}\unorm{v}} \text{.}
\end{equation*}
So compute
\begin{align*}
\udotprod{u}{v} \amp= 1 \cdot (-1) + 2 \cdot 3 = 5, \amp
\unorm{u} \amp= \sqrt{1^2 + 2^2} = \sqrt{5}, \amp
\unorm{v} \amp= \sqrt{(-1)^2 + 3^2} = \sqrt{10}.
\end{align*}
Therefore,
\begin{equation*}
\cos \theta = \frac{5}{\sqrt{5}\sqrt{10}} = \frac{1}{\sqrt{2}} \text{.}
\end{equation*}
The only angle in the domain \(0 \le \theta \le \pi\) with this cosine value is \(\theta = \pi/4\text{.}\)
