Video: Finding the Components of a Vector

Video Transcript

In this lesson, we’re going to learn how to find the components of a vector. We will start with how to find the components of a vector that is drawn graphically and how to represent those components in unit-vector notation. We will also go over how to find the components of a vector using trigonometry.

Before we learn how to find components, let’s refresh our memory on what a vector is and how to represent a vector in graphical form, as well as how to use unit-vector notation. We need to remember that a vector is a quantity that has both magnitude and direction. We also need to recall that a unit vector is a vector of length one. Let’s draw an example of two vectors, 𝐀 and 𝐁. Let’s make vector 𝐀 have a magnitude of three units and be oriented along the horizontal axis to the right of our screen.

When we label vector 𝐀, note that we put a half arrow over the top of the letter. This is a common convention to show that it is a vector. In text form, it is common for the variable to be bolded. We will see this form when we do example problems at the end of our lesson. If we were to write the expression for vector 𝐀 using unit vector notation, it would read 𝐀 is equal to three 𝐢. The 𝐢 is the unit vector that points in the horizontal direction. The hat over the 𝐢 represents that it’s a unit vector. In text form, we could bold the 𝐢 to show that it is a unit vector.

We can draw vector 𝐁 to have the same magnitude of three units, but this time we can orient the vector vertically to the top of the screen. Once again, we label our vector with a half arrow over the top of the letter. The expression for vector 𝐁 using unit-vector notation would be vector 𝐁 is equal to three 𝐣. 𝐣 with a little hat over it is the unit vector that points in the vertical direction. Now that we have done a quick recap on what vectors are, how to draw them graphically, and how to write an expression using unit-vector notation, let’s dive into breaking down a vector into its components.

A vector can be broken down into two parts called components. The magnitude of the 𝑥-component is represented as 𝑎 subscript 𝑥, and the magnitude of the 𝑦-component is represented as 𝑎 subscript 𝑦. The 𝑥-component is how much of the vector points in the horizontal direction and the 𝑦-component is how much of the vector points in the vertical direction. We have drawn vector 𝐀 in on our coordinate system as shown on the grid. From looking at the grid, let’s determine what the 𝑥-component and 𝑦-component are for our vector.

From the grid, we can see that the 𝑥-component of vector 𝐀 is one, two, three, four, five units. Therefore, the magnitude of 𝑎 subscript 𝑥 is equal to five. The 𝑦-component of vector 𝐀 is equal to one, two units. Therefore, we can say 𝑎 subscript 𝑦 has a magnitude of two. In unit vector notation, the magnitude of the 𝑥-component of the vector is what we multiply 𝐢 hat by and the magnitude of the 𝑦-component is what we multiply 𝐣 hat by. If we were to write our vector 𝐀 in unit-vector notation, we would say that it was equal to five 𝐢 plus two 𝐣, where the five 𝐢 is the 𝑥-component of vector 𝐀 and the two 𝐣 is the 𝑦-component of vector 𝐀.

Let’s look at one more vector, vector 𝐁. We’ll break down vector 𝐁 into both its 𝑥- and 𝑦-component and then put it into unit-vector notation. Vector 𝐁 has a magnitude of two units to the left of the screen. Therefore, the magnitude, the 𝑥-component for vector 𝐁, 𝑏 𝑥 would be equal to negative two. Vector 𝐁 also has a magnitude of three units along the vertical to the bottom of the screen. So, the magnitude of vector 𝐁’s 𝑦-component, 𝑏 𝑦, is equal to negative three. Writing out the unit-vector notation for vector 𝐁, we get that vector 𝐁 is equal to negative two 𝐢 minus three 𝐣, where the negative two 𝐢 is the 𝑥-component of vector 𝐁 and negative three 𝐣 is the 𝑦-component of vector 𝐁.

Another way to break down a vector into its components is to use trigonometry. Before we apply trigonometry to our vectors, let’s refresh our memory on how do you use sine, cosine, and tangent with a right triangle. When we have a right angle, we can use trigonometry to solve for an unknown side as long as we know the measure of one angle and the length of one side. If we label this angle 𝜃, then the side that touches the angle or that’s next to it is called the adjacent side. The side that’s opposite to the angle is called the opposite side, and the side that’s opposite to the 90-degree angle is called the hypotenuse.

The sin of the angle 𝜃, the cos of the angle 𝜃, and the tan of the angle 𝜃 are simply the ratios of the sides of similar triangles. The sin of 𝜃 is equal to the ratio of the opposite side of the triangle to the hypotenuse of the triangle. The cos of 𝜃 is equal to the ratio of the adjacent side of the triangle to the hypotenuse of the triangle. And the tan of 𝜃 is the ratio of the opposite side of the triangle to the adjacent side of the triangle. A common acronym to remember the ratios is SOH-CAH-TOA. SOH for sin is equal to opposite over hypotenuse, CAH for cos is equal to adjacent over hypotenuse, and TOA for tangent is equal to opposite over adjacent.

Let’s apply this bit of trigonometry to our vectors. If we need to break down our vector which has a length of eight units and an angle of 35 degrees above the horizontal into its two components, where do we start? We can start by drawing in the components so that the 𝑥-component is along the horizontal axis and the 𝑦-component is along the vertical axis to form a right triangle. We have labeled the 𝑥-component 𝐕 subscript 𝑥 and the 𝑦-component 𝐕 subscript 𝑦.

To find the 𝑥-component of our vector 𝐕, we need to identify where 𝐕 𝑥 is with respect to the angle. We can see that 𝐕 𝑥 is next to the angle of 35 degrees or is the adjacent side. We have been given the hypotenuse of the triangle to be eight, a labeled angle of 35 degrees, and we now know that we’re solving for the 𝑥-component or adjacent side of the triangle. Which of our three trigonometric functions sine, cosine, or tangent should we use? We should use cosine because the cos of 𝜃 is equal to the adjacent side divided by the hypotenuse.

We need to isolate the adjacent side of the triangle, since that is what we’re solving for. We multiply both sides of the equation by the hypotenuse. The hypotenuse cancels out on the right side of the equation. The left side of the equation becomes the hypotenuse times the cos of 𝜃. When we plug in the values of our variables, we put in eight for the hypotenuse, cos of 35 degrees for 𝜃, and 𝐕 𝑥 for the adjacent side of the triangle as we’re solving for the 𝑥-component of our vector. When we multiply our eight by cos of 35 degrees, we get an 𝑥-component of 6.55.

We need to make sure that our calculators in degrees and not radians as the cos of 35 radians is not the same thing as the cos of 35 degrees. We will round our answer to two significant figures because our angle was given to two significant figures. This gives us an 𝑥-component for our vector of 6.5.

Moving on to the 𝑦-component of the vector, which is the opposite side of our triangle, which trigonometric function we will use if we were also given the hypotenuse and the angle? We would use the sin of 𝜃 is equal to the opposite side of the triangle divided by the hypotenuse. To isolate the opposite side, we once again multiply both sides by the hypotenuse. Canceling out the hypotenuse on the right side of the equation and leaving the left side equation to be the hypotenuse times the sin of 𝜃. Substituting the values into our equation, we would have a hypotenuse of eight, a sin of 35 degrees, and our 𝑦-component, 𝐕 𝑦, would represent the opposite side of the triangle.

Multiplying eight by sin of 35 degrees, we get 4.59 as the magnitude of our 𝑦-component. Again, we need to round our component to two significant figures in order to match our problem. 4.59 becomes 4.6. So, the 𝑦-component of our vector is 4.6. Problems can get tricky by which axis the angle is drawn to, so we need to pay careful attention to this. If our angle was 35 degrees to the vertical instead of to the horizontal, what would happen to our components?

In this scenario, the 𝑥-component of our vector is the opposite side of the triangle. So, we do 𝐕 subscript 𝑥 is equal to the sin of 𝜃 times the hypotenuse, just as we did for the 𝑦-component up above. We would once again multiply the sin of 35 degrees by eight. And our 𝑥-component to two significant figures would be 4.6, just as the 𝑦-component was up above. The 𝑦-component of our vector is the adjacent side of the angle. Therefore, we’d use the cos of the angle and multiply it by the hypotenuse to find the 𝑦-component, just as we did for the 𝑥-component up above. We would multiply cos of 35 degrees by eight, which would give us a 𝑦-component of our vector of 6.5 when we round to two significant figures, just as it did up above for our 𝑥-component.

Let’s try two examples of finding the components of vectors, one graphically with unit-vector notation and one with trigonometry.

Write 𝐀 in component form.

The diagram shows vector 𝐀 to be drawn onto a grid. We are asked to write vector 𝐀 in component form. The component form of a vector is typically written as 𝑎 sub 𝑥 𝐢 plus 𝑎 sub 𝑦 𝐣, where 𝑎 sub 𝑥 is the magnitude of the 𝑥-component of the vector and 𝑎 sub 𝑦 is the magnitude of the 𝑦-component of the vector. 𝐢 and 𝐣 represent unit vectors, with 𝐢 being in the direction of the horizontal axis and 𝐣 being in the direction of the vertical axis. Because we’re handwriting our answers, we use a half arrow over the letter 𝐀 to represent that it’s a vector. In text form, we bolded it. In the 𝐢 and 𝐣, we see little hats to represent that they’re unit vectors. In text form, these can also be bolded.

To find the component form of vector 𝐀, we need to determine how much of vector 𝐀 is in the horizontal direction and how much of vector 𝐀 is in the vertical direction. Looking at the grid, we can see that vector 𝐀 is one, two, three units to the left of our screen. This is considered negative three. We could then say that the 𝐢-component of our vector has a magnitude of negative three. Looking back at the grid, we can see that the 𝑦-component or the vertical direction has a magnitude of one, two units pointing towards the top of the screen, which would be a positive two. We could, therefore, say that the 𝐣-component of our vector has a magnitude of two. Based on the diagram, the component form of vector 𝐀 is equal to negative three 𝐢 plus two 𝐣.

In our next example, we will apply trigonometry to solve for the components of a vector.

The diagram shows a vector 𝐀 that has a magnitude of 22. The angle between the vector and the 𝑥-axis is 36 degrees. Work out the horizontal component of the vector. Give your answer to two significant figures.

In the problem, we’re being asked to solve for the 𝑥-component of our vector, or 𝑎 sub 𝑥. With the components drawn in on our diagram, we can see that the triangle is a right triangle. And therefore, we can use trigonometry to solve for our unknown side. Let’s recall that we can use SOH-CAH-TOA. SOH, the sin of the angle is equal to the opposite side of the triangle divided by the hypotenuse. CAH, the cos of the angle is equal to the adjacent side of the triangle divided by the hypotenuse. And TOA, the tan of the angle is equal to the opposite side of the triangle divided by the adjacent side of the triangle.

Looking back at our diagram, we are solving for the 𝑥-component of our vector, we’re given the angle 𝜃, and we’re given the length of the hypotenuse. Which one of our trigonometric functions would we choose, sine, cosine, or tangent? The best one to choose is cosine because our vector component is adjacent to the angle. Now, we need to isolate the adjacent side of the triangle, as that’s what we’re solving for. To do this, we multiply both sides by the hypotenuse. This will cancel out the hypotenuse on the right side of the equation and leave the left side of the equation as the hypotenuse times the cos of 𝜃.

We can now use the values from the problem to substitute in for our variables. The hypotenuse had a magnitude of 22, 𝜃 was 36 degrees, and 𝑎 sub 𝑥 represents the horizontal component of our vector. When we multiply 22 by the cos of 36 degrees, we get 17.8. In the problem, we were instructed to give our answers to two significant figures. Looking at our answer right now, it’s given to three significant figures. Therefore, we can round up 17.8 to 18. The horizontal component of a vector that has a magnitude of 22 and makes an angle with the 𝑥-axis of 36 degrees is 18.

Key Points of the Lesson

To find the 𝑥- and 𝑦-components of a vector represented graphically on a grid, count how many units along the 𝑥-axis and 𝑦-axis the vector goes. In unit-vector notation, a vector can be represented as 𝐀 equals 𝑎 sub 𝑥 𝐢 plus 𝑎 sub 𝑦 𝐣. Trigonometry can be used to determine an unknown component of a vector.