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Section 13.4 Examples

Subsection 13.4.1 The norm of a vector

Example 13.4.1. Basic computation examples.

Here are a few examples of computing the norm of a vector, in various dimensions.

  1. Consider \(\uvec{u} = (1,2)\) in \(\R^2\text{.}\) Then,
    \begin{equation*} \unorm{u} = \sqrt{1^2 + 2^2} = \sqrt{5} \text{.} \end{equation*}
  2. 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*}
  3. 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 13.4.2. 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 13.4.3. 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.

Let's carry this out for the vectors from Example 13.4.1 above.

  1. 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. \(\norm{\uvec{u}'} = 1\)).
  2. 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.
  3. 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 13.4.2 Dot product and the angle between vectors

Here is an example of using the dot product to determine the angle between vectors.

Example 13.4.4. 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 13.7, 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 \(\theta = \pi/4\text{.}\)