Introduction to Vectors

Subsection 3.1 Technical Definition of the Dot Product

Subsection 3.2 The Dot Product and the Angle between Vectors

In this section we will start to look at how we can use the dot product to find the angle between two vectors. This will also give us some context for what the dot product is as well as some of what it can tell us. To start we should quickly review the unit circle and some of the important angles that we will be needing here.

When we are thinking about vectors one of the ways that we can evaluate them in terms of one another is the angle between them. There are three main destinctions that we are going to consider; between \(0^{\circ}\) and \(90^{\circ} \text{,}\) exactly \(90^{\circ} \) and between \(90^{\circ} \) and \(180^{\circ} \text{.}\) To encode these different categories we are going to need to take a look at cosine.

Looking at the unit circle above, recall that the cosine of an angle gives us the \(x \) value of the point on the unit circle associated with that angle. So \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2} \) and \(\cos(120^{\circ}) = \frac{-1}{2} \) and finally \(\cos(90^{\circ}) = 0 \text{.}\) Try a few of these to get the hang of reading the unit circle.

Now what we want to reflect on from the last few questions is the sign on each of the answers that you gave. When the angle that was being evaluated was between \(0 \) and \(90^{\circ} \text{,}\) the cosine reported a positive value. When it was between \(90^{\circ} \) and \(180^{\circ} \) the cosine reported a negative number. And finally, \(\cos(90^{\circ}) = 0 \text{.}\) It is this observation that we are going to use to destinguish some information about the angle between vectors.

We are going to need one additional tool to build the "Angle Between Vectors" equation.

Subsubsection 3.2.1 The Math of the Angle Between Vectors

In this subsection we are going to look at how we can use the law of cosines to build an equation that relates the dot product and angle between vectors. Let's start at the end and take a look at the equation.

Angle Between Vectors.

Let \(\mathbf{v} \) and \(\mathbf{u} \) be vectors and let \(\theta \) be the angle between the vectors.

\begin{equation*} \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}||\mathbf{u}|} \end{equation*}

Recall that \(|\mathbf{u}| \) is the magnitude of \(\mathbf{u} \text{.}\)

Let's work through the math of where this comes from.

Proof.

Let \(\mathbf{u} = \langle u_1, u_2 \rangle \) and \(\mathbf{v} = \langle v_1, v_2 \rangle \) be vectors, and let \(\theta \) be the angle between the vectors. Consider, the figure below. For this proof, we are going to think about the vectors as sides of a triangle, the the angle, \(\theta \) , between them.

We will start by writing the law of cosines using the vectors that we have. So,

\begin{equation*} (|\mathbf{u} - \mathbf{v}|)^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2|\mathbf{u}||\mathbf{v}|\cos(\theta) \end{equation*}

Now we will rewrite this using the components,

\begin{equation*} (|\langle u_1 , u_2 \rangle - \langle v_1, v_2 \rangle |)^2 = |\langle u_1, u_2 \rangle|^2 + |\langle v_1, v_2 \rangle|^2 - 2|\mathbf{u}||\mathbf{v}|\cos(\theta) \end{equation*}

\begin{equation*} (|\langle u_1 - v_1, u_2 - v_2 \rangle |)^2 = u_1^2 + u_2^2 + v_1^2 + v_2^2 - 2|\mathbf{u}||\mathbf{v}|\cos(\theta) \end{equation*}

\begin{equation*} (u_1 - v_1)^2 + (u_2 - v_2)^2 = u_1^2 + u_2^2 + v_1^2 + v_2^2 - 2|\mathbf{u}||\mathbf{v}|\cos(\theta) \end{equation*}

\begin{equation*} u_1^2 -2u_1v_1 + v_1^2 + u_2^2 -2u_2v_2 + v_2^2 = u_1^2 + u_2^2 + v_1^2 + v_2^2 - 2|\mathbf{u}||\mathbf{v}|\cos(\theta) \end{equation*}

\begin{equation*} u_1v_1 + u_2v_2 = |\mathbf{u}||\mathbf{v}|\cos(\theta) \end{equation*}

\begin{equation*} \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos(\theta) \end{equation*}

If we rearrange this last line we get,

\begin{equation*} \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|} = \cos(\theta) \end{equation*}

Let's try a few of these. The steps for the following will be,

1. Find the dot product of the vectors.
2. Find the length of the two vectors.
3. Divide the dot product from (1) by the product of the two lengths.
4. To find \(\theta \) apply inverse cosine to the result from (3).

Use the above steps to find the angle between the given vectors in the next few problems.

Checkpoint 3.13.

Checkpoint 3.14.

Checkpoint 3.15.