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Section 2 Introduction to Vectors

Subsection 2.1 What is a Vector?

Let's start by establishing some vocabulary that we will use on our journey to torque.

Definition 2.1.
A vector is a directed line segment. This means that it will have a finite length, and a direction.

In the last definition we start to see that we are going to need to address the concepts that are foundational in a vector; the length (magnitude) and the direction.

Definition 2.2.
The magnitude of a vector is the length of the vector.

In the definition of a vector, we can see the phrase "line segment". This means that the vector will be "bookended" by points. Meaning that we can use tools that measure the distance between points to find the magnitude of vectors. So, let's note that now:

Definition 2.3. The Distance Between Two Points.
Given two points \(P(a,b) \) and \(Q(c,d) \text{,}\) the distance, \(d \text{,}\) between the points can be measured by,
\begin{equation*} d = \sqrt{(c-a)^2 + (d-b)^2} \end{equation*}

Subsection 2.2 Vector Algebra

In this section we will be looking at vector algebra and the different operations associated with vectors. Start by taking a look at the image below. Note that the vectors look like arrows starting at the origin (initial point) and ending at the arrow (terminal point). We usually use the latters \(\mathbf{u} \text{,}\) \(\mathbf{v} \) and \(\mathbf{w} \text{,}\) but this is just a convention, not a rule.

Figure 2.4. An image of two vectors in the plane.

Subsection 2.3 Vector Notation

In this next section we are going to look at the different ways that we can write vectors. This is a good place to spend some time working through some problems to get familiar with working with vector notation.

Let's start with some important vectors. The vector that will indicate one unit in the positive \(x \) direction is going to be notated \(\mathbf{i} \text{,}\) and the vector that will indicate one unit in the positive \(y \) direction is going to be denoted \(\mathbf{j} \text{.}\) These two vectors will be building blocks for how we will think about all vectors in the plane. Think of these two vectors as a base that we can build other vectors from.

Figure 2.5. A figure generated with TikZ in

As indicated above, one way that we can notate a general vector is by describing it in terms of each of the two base vectors above. The thinking is as follows. If we need to get to \((2,1) \) we would need to go "over" 2 and "up" 1. This would mean that we need 2 of our base vector that gets us "over" and 1 of our base vector that gets us "up". Since these vectors are independent of on another, and the space we are working in only has two components in its point (i.e. (2,1)), we can "stretch" (math word here is scale) \(\mathbf{i} \) to be a new vector 2 times longer, and "stretch" (math word here is scale) \(\mathbf{j} \) to be a new vector 1 times "longer". Finally the sum of these two vectors will give us a new vector that has initial point at \((0,0) \) and terminal point at \((2,1) \text{.}\) The notation for this will be,

\begin{equation*} \mathbf{v} = 2\mathbf{i} + \mathbf{j} \end{equation*}
Figure 2.6. A figure generated with TikZ in

Another way that we will notate a vector is component form. This is a way to encode the "over" and "up" ideas that we started addressing above. The notation for component form is,

\begin{equation*} \mathbf{u} = \langle a , b \rangle \end{equation*}

where \(a \) and \(b \) are called the components of \(\mathbf{u} \text{.}\) One of the utilities of this form is that every vector in this form has the origin as its base point. This allows us to focus on the magnitude and the direction, based around a central point.

One of the things that we want to be able to do is turn two points into a vector in component form. To do this we will use subtraction to translate the direction and magnitude information so that (while still retaining the original values for both) the resulting vector is now based at the origin.

Given the points \(P(1,3) \) and \(Q(2,3) \) find the representative of the vector \(\overrightarrow{PQ} \) that is in component form.

To do this we will subtract the components of the terminal point (in this case \(Q \)) from the initial point (in this case \(P \)). That is,

\begin{equation*} \overrightarrow{PQ} = \langle 2 - 1 , 3 - 3 \rangle = \langle 1 , 0 \rangle \end{equation*}

Bonus: What would the above vector look like using the \(\mathbf{i} \) and \(\mathbf{j} \) notation we used above?

Solution
\(\langle 1 , 0 \rangle = 1\mathbf{i} + 0\mathbf{j} = \mathbf{i} \)

Before we start looking at the next thoughts, try a few of these problems involving the different types of ways that we can represent vectors.

Subsubsection 2.3.1 Vector Notation at a Glance

Glossary Glossary

\(\mathbf{v} = a \mathbf{i} + b \mathbf{j} \) linear combination form.

\(\mathbf{v} = \langle a, b \rangle \) component form

\(\mathbf{i} = \langle 1 , 0 \rangle \) and \(\mathbf{j} = \langle 0,1 \rangle \)

\(|\mathbf{v}| = \sqrt{a^2 + b^2} \) the magnitude of the vector \(\mathbf{v} \)

A vector is a unit vector if the length of the vector is 1.

The zero vector is denoted \(\mathbf{0} \) and is \(\langle 0, 0 \rangle \text{.}\)

Subsection 2.4 Finding Unit Vectors

In this next subsection we are going to take a look at how to make a unit vector. The language that we will use is; "A unit vector in the direction of \(\mathbf{v} \)". What this means is that we are going to make a new vector (we will give it a new name as needed) that has the same direction as \(\mathbf{v} \) but has length one. To do this we will "divide out" the magnitude, just like if we were to divide \(2 \) by \(2 \) to get 1. One note that we should make before we move on is that \(\mathbf{v} \) will need to be a non-zero vector.

Let's take a look at an example. Pay close attention to the way that the question is worded.

Find a unit vector in the direction of \(\mathbf{v} = \langle 3,4 \rangle \text{.}\)

To find this unit vector, we need to divide \(\mathbf{v} \) by its magnitude.

First the magnitude of \(\mathbf{v} \text{,}\)

\begin{equation*} ||\mathbf{v} || = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \end{equation*}

then we need to divide the original vector by the magintude, 5,

\begin{equation*} \mathbf{u} = \frac{1}{5} \mathbf{v} = \frac{1}{5} \langle 3,4 \rangle = \langle \frac{3}{5} , \frac{4}{5} \rangle \end{equation*}

Note that we can report the vector as either, \(\frac{1}{5} \langle 3,4 \rangle \) or \(\langle \frac{3}{5} , \frac{4}{5} \rangle \)

Let's try a few problems,

Now that we have a feel for how to make a unit vector in the direction of a given vector, what we should look at next is a way that we can denote a general unit vector. One way to think about a unit vector is as a tool to indcate direction. Since the length is length 1, we can focus on the direction. To encode this idea of direction, we can use the tools of rotation. Glancing ahead to 3.2 we can take a look at the unit circle, but our plan here is to use \(\sin(\theta) \) and \(\cos(\theta) \) to capture the rotation to a particular direction and use that to get a general equation for a unit vector.

The above thoughts brought together would look like the following,

\begin{equation*} \mathbf{u} = \langle \cos(\theta), \sin(\theta) \rangle \end{equation*}

where \(\theta \) is measured from positive \(x \text{.}\)

Now what we should do next is show that this is infact a unit vector. To do that we will compute the lenght and show that, no matter what \(\theta \) is, the lenght of \(\mathbf{u} \) will still be 1.

\begin{equation*} |\mathbf{u}| = | \langle \cos(\theta) , \sin(\theta) \rangle | = \sqrt{\cos^2(\theta) + \sin^2(\theta)} = \sqrt{1} = 1 \end{equation*}

This shows is that \(\langle \cos(\theta), \sin(\theta) \rangle \) can be used as a general equation for a unit vector.