Chapter 3: Vector Components

3.1 Introduction

It is often necessary to break a vector into its components. In the cartesian coordinate system, this would mean breaking the vector into an x-component and a y-component.

Figure 1 below demonstrates that there are generally two ways to do this as represented by the two different right triangles. It is important to recognize that both triangles are completely equivalent, and neither option is preferable over the other.

The components of a vector are the parts of the vector that correspond to specific axes or directions.

3.2 Solving for the Components

Let’s consider a vector that represents the velocity of a car. This vector might have a magnitude of 60 miles per hour (mph) and a direction of northeast. To break this vector down into its components, we would need to choose specific axes or directions along which to analyze it. Here, one of the components would be along the North direction (y-axis), and the other component would be along the East direction (x-axis).

As a more quantitative example, suppose the car’s velocity vector is represented by in Figure 2. Let (about 60 mph), and let . We want to determine and .

Method 1:

We must first state the following trigonometric identities. Memorize these and never forget them!

In our specific problem, we can see that this means:

Consequently,

Similarly,

Method 2:

Instead of using the right triangle drawn in figure 2, we could use the one with the angle . By simple trigonometry, we can see that . Thus, in our case, .

Now, the y-component is the adjacent side of the triangle, and the x-component is the opposite side. This will yield the same final result as shown below.

A Word of Caution:

Care must be taken to understand the difference between Method 1 and Method 2. There are two different triangles that one could draw, but both are equivalent. Be wise in choosing the correct angle for the triangle you decide to work with.

As shown above, the x-component is not always found through the cosine function, and the y-component is, likewise, not always found through the sine function (though many beginners tend to think that this is the case). It all depends on how you draw your triangle.

3.3 Determining the Direction

Sometimes we are given the components of a vector and asked to determine the magnitude and direction. The magnitude is easy. It is given simply as the result of Pythagorean’s Theorem in two or three dimesions:

The bars represent the absolute value of the vector because we have not yet determined the direction. In a moment, we will be able to remove the bars, but we’re not quite there yet.

In order to determine the direction, we must look to a third trigonometric identity:

Choose your such that the y component is the opposite side and the x-component is the adjacent side (as shown in Figure 2).

However, this formula doesn’t give you the direction right off the bat because we generally measure direction from the positive x-axis. could be anywhere; it’s not always attached to the positive x-axis as we would like. Use Table 1 below to determine which quadrant your vector is in. Then use the rules that follow to determine your angle.

	1	2	3	4
x	positive	negative	negative	positive
y	positive	positive	negative	negative

Rules for Calculating :

• Quadrant 1: Plug in your x and y components to the formula above. You will get an answer that measures (counterclockwise) from the positive x-axis. This is the simplest case. No further work is required.
• Quadrant 2: Plug in your x and y components to the formula above. Your calculator will give you an answer in Quadrant 4. Add 180 degrees. You will now be in Quadrant 2. You will be measuring (counterclockwise) from the positive x-axis.
• Quadrant 3: Plug in your x and y components to the formula above. Your calculator will give you an answer in quadrant 1. Add 180 dregrees. You will now be in Quadrant 3. You will be measuring (counterclockwise) from the positive x-axis.
• Quadrant 4: Plug in your x and y components to the formula above. Your calculator will give you an answer in Quadrant 4. However, the angle will be negative. This is acceptable but not conventional. Add 360 degrees. You will now be measuring (counterclockwise) from the positive x-axis.

Always express your final answer in terms of both magnitude and direction as this is the proper definition of a vector. For this example, we might conclude by stating that:

3.4 Vector Addition Revisited—The Method of Components

Now that we’ve discussed vector components, let’s formalize the discussion on vector addition from the previous chapter. Recall that adding two vectors gives us a resultant vector which we will call The components of this resultant vector are easily calculated then as:

This dramatically simplifies the process of vector addition compared to the tip-to-tail method as no diagram is necessary.

3.5 The Scalar (Dot) Product of Two Vectors

We looked at vector addition and subtraction in the previous chapter, but what if you want to multiply two vectors together? This section is starting to get a bit math heavy, but bear with me. These formulas are extremely simple and useful. Given two vectors:

and

the scalar product can be expressed as a simple sum:

There is another formula for calculating the scalar product when you have the angle between and I will state it without proving it. However, please note that the real power of this equation is in rearranging the variable to instead calculate the angle between the two vectors.

Rearranging then, we see that the angle between two vectors is given as:

It is also worth noting that the scalar product does follow the commutative property such that:

3.5 The Vector (Cross) Product

The vector product is always a tricky one when you’re just getting started with it. However, it doesn’t have to be. Some students will spend 5 to 10 minutes trying to solve a cross product, but truthfully these problems can be completed in less than 60 seconds. There are many tricks to speed up the process, but my personal favorite involves the determinant of a matrix. If you’re unfamiliar with matrices and/or matrix operations, that’s okay. This one is fairly easy to learn and you can do so at my index page here. Let’s write out the equation for a cross product so that you can see why a shortcut trick is useful. Given two vectors, and with components:

and

the cross product can be written as:

The resulting vector is orthogonal to both and , and its magnitude equals the area of the parallelogram spanned by and .

The cross product can also be written as a matrix determinant as follows:

This representation is identical the the earlier equation. What both of these equations tells us is that the components of can be written as:

It is important to keep in mind that the cross product does not follow the commutative property. In other words:

It is also worth noting that the cross product of two parallel or anti-parallel vectors is zero. Go ahead and try that one out for yourself, but I won’t be proving it here.

Finally, to calculate the magnitude of a cross product, you can simply plug in to the following equation:

where is the angle between and

3.6 Unit Vectors

Now that you have an idea for vector components, it’s time to introduce unit vectors. Unit vectors are just vectors with a magnitude of 1. And what happens when you multiply something by 1? Nothing changes. Well, not quite. With scalars, that’s completely true. Nothing changes when you multiply by 1. However, when you multiply a vector by a unit vecotr with magnitude 1, you are taking the component of the vector that is pointing in the same direction as the unit vector. Everything pointing perpendicular to that direction gets zeroed out. This will all make a lot more sense when we get around to looking at an example. But first, know that unit vectors use a “hat” symbol instead of the normal vector arrow symbol. When pronouncing unit vectors, say the “hat” part out loud. Something like would be pronounced as “i-hat.”

Actually, is a special unit vector as it points exactly in the direction of the x-axis. Similarly, points along the y-axis, and points along the z-axis. Some textbooks refer to these unit vectors as and I don’t generally use this notation because it can be confusing when you also have those variables in the problem you are working on. However, this is totally up to your preference.

On to the example then. Suppose you have a vector \vec{T} with components in the x, y, and z directions. So, If you multiply by the unit vector (using the scalar product), you get:

Thus, by multiplying by the unit vector in the x-direction, we have retrieved the x-component of the vector

At this point, we should introduce another formal way of writing a vector, now by using unit vector notation.

Basically, the vector is just some component in the x-direction plus some component in the y-direction plus some component in the z-direction.

Chapter Summary

This chapter introduced the concept of vector components, emphasizing how a vector can be broken down into its x and y components within the cartesian coordinate system. Two methods of solving for these components were described. The first method employs trigonometric identities of sine and cosine with reference to an angle between the vector and the x-axis. The second method uses the same identities but references an angle between the vector and the y-axis. Both methods yield the same results.

We also looked at the process of determining the magnitude and direction of a vector given its components. The magnitude is calculated using Pythagoras’ Theorem, while the direction requires the use of trigonometric identities, specifically the tangent function.

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Continue to Chapter 4: Position, Velocity, and Acceleration

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