Vector Composition & Decomposition: Study with Video Lessons, Practice Problems & Examples

Vectors are defined by their magnitude, direction (angle θ_x), and components (legs of a triangle). Vector composition involves calculating the resultant vector from given components using the Pythagorean theorem: $\sqrt{a^2+b^2}$. Conversely, vector decomposition finds components from magnitude and angle using $a=A\cos \left(\theta \right)$ and $b=A\sin \left(\theta \right)$.

Vector Composition & Decomposition

Hey, guys. So up until now, we've been able to use vector math on a lot of our vectors without using grids or squares. We could break this up into a triangle. We can count up the boxes to figure out the legs like 3 and 4, stuff like that. Unfortunately, a lot of the vectors that we're going to see are not going to be on diagrams that don't have grids or squares, and we're going to have to do vector math without using them. So, in this video, I'm going to show you that there are really 4 equations that we need in order to describe everything that we need to know about a vector. These four equations are more triangle math equations, and they're called vector composition and decomposition equations. Let's check it out. So, guys, there are 3 things that we need to describe a vector when we have a diagram that doesn't have grids or squares. Vectors have magnitude, the length of this arrow right here. Vectors also have a direction. And without a grid, we specify this direction using an angle relative to the x-axis called theta x. Now the last thing is that vectors are also triangles, which means that they have legs. These legs just get a fancy name; they're called the components. Basically, this vector here, this triangle, can be broken down into its legs. These legs are just called components. So, I mentioned that there are 4 equations, and it really just breaks down into 2 different situations called vector composition and decomposition. And, really, the difference is what you're given and what you're trying to find. Vector composition is when you're given the one-dimensional components or the legs of a triangle, and you're trying to figure out what the magnitude and direction are of the hypotenuse. Decomposition is the opposite. It is when you're already given the magnitude and direction, and now you're trying to figure out what the legs of the vector are. Let's check it out in more detail. Now, guys, when we did vector composition, this is just exactly like what we did when we added perpendicular vectors like 3 and 4. So when we added 3 and 4, we just basically combine them tip to tail, and then we have the resultant or the shortest path from start to finish. To calculate this vector, which I'll call a here, we just use the Pythagorean theorem because this made a triangle. So 3 and 4, the hypotenuse is just the square root of 3 squared and 4 squared, and that's 5 meters. So, these components here or these legs, 3 and 4, just get a fancy name; they're called components. And so, my 3 is my ax component. It's how much this vector a is in the x-axis, and this 4 here is my y component, and that's how much of the vector lies in the y direction, and they combine to form a vector of magnitude a. So, I walked in this direction. It's kind of as if I walked 3 and then 4 in this direction. Now, the one thing we haven't talked about yet is that this vector points at some direction called theta x. And the way we calculate this theta x here is just by using another set of triangle math equations called SOHCAHTOA. So, basically, to figure out this angle here, all I have to use is just an equation that your textbooks and your professors will derive. This theta x is the inverse tangent or the arctangent of my y components over my x components. So in this example, my y component's 4, x is 3. So that means that my theta x here is just going to be the arctangent. So you can find this in your calculators, and this is going to be 4 over 3 and this theta x is just 53 degrees. Make sure your calculator is in degrees mode. And really, this is just an angle that is above the positive x-axis. So it's an angle that goes like this. Alright guys. That's really all there is to it. Just these two equations here and it's when you have the legs of a triangle and you're trying to figure out the magnitude and direction. The other thing is vector decomposition. It's basically the reverse process. So here I have the magnitude and the angle. Now, I want to figure out the legs. So we want to figure out what my ax and I want to figure out what my ay components are. And really, it just comes down to these two equations here that we come, that come from SOHCAHTOA. So now, we want to decompose a into its components ax and ay. So here's the deal. As long as this angle, theta x, is drawn to the nearest x-axis, then we're going to use these equations here. My ax is going to be a times the cosine of the angle. So it's the magnitude times the cosine of the angle, and ay is going to be a times the sine of the angle here. So for example, if I were given an angle like this, like this theta here, this is a theta that's relative to the y direction. This is bad. I can't use this. I won't use that in my equations. This angle over on the other hand, theta x, is good. So this is the one that I'll plug into my equations. So, for example, my ax is going to be the magnitude times the cosine. So, the magnitude is 5, the cosine of the angle is 53 degrees, and so, I'm just going to get 3 and then if you plug in ay, you're going to get the magnitude which is 5 times the, sorry not the cosine, the sine, the sine of 53 and if you plug this in, you're going to get 4. Notice how we've basically just come up with the exact same numbers, 3 and 4 as we had over here, and that should be no surprise because we've basically just formed the exact same triangle as we did over here. It's just we started with the legs of the triangle over here and then here we started with the magnitude and the direction. So, it's no surprise that we got the exact same numbers. Alright, guys. That's really all there is to it. You're just going to use these equations based on what you're trying to find and what you're given. So let's go ahead and get some practice. For each of the following, we're going to draw the vector and then solve for the missing variables. So my ax is 8, my ay is 6. I'm trying to figure out the magnitude and the direction. So I'm given the components, so let me just draw out this little vector here. I've got the legs of the triangle, so I know this is 8 and this is 6. This is my ax. This is my ay. And now, I'm actually trying to find the magnitude and the direction. So that means I'm going to use my vector composition equations. So to figure out my a, I just have to use the Pythagorean theorem. I just use 8 squared plus 6 squared, and if I do that, I get 10. So this is my magnitude. And now for the angle, remember the angle is drawn relative to the x-axis. This is my theta x, and this theta x here is going to be the inverse tangent, the arctangent of my y components, which is 6 over my x component, which is 8. And if you go ahead and work this out, you should get 37 degrees. So this is my angle now. That's 37 degrees. Alright. Let's move on to part b, which is now I have an angle or now I have a magnitude of 13, and I have an angle of 67.4, and now I want to calculate the components. So let's draw this vector here. This is not a component. This is actually just the vector. So I'm just going to draw this like, like this, so I know this B here is 13. Now, the angle relative to the x-axis is 67 degrees. So, you draw this little x-axis like this, and I know this angle here is 67.4 degrees. So now, I actually want to figure out my ax and my ay. So my ax is going to be the well, which equations am I going to use? I know the magnitude and the direction, and I want to figure out the components. I'm going to use my vector decomposition equations. So, I'm sorry, this is actually b. So this is going to be I'm sorry. Whoops. This is by and this is bx. Okay. So, my bx is just going to be 13 times the cosine of 67.4, and then if you work this out, you're going to get 5. And then, if you do by, you're going to do 13 times the sine of 67.4, and you're going to get 12. So these are the components here. So here are my components. Alright, guys. That's it for this one. Let me know if you have any questions.

Calculating Magnitude & Components of a Vector

Hey, guys. Let's work out this example together. We've got a vector and we're told what the y component is. We're also told what the angle is. And in this first part, we're going to find out what the magnitude of this vector is. Let's check it out. Now, I always like to draw out the vector first. I feel like it's a great way to visualize what's going on, making everything make sense. So, just draw out your little coordinate system here. We've got positive y and positive x. Now, we're told this vector has a y component of 12 and it makes an angle of 67.4 degrees. So, start from the origin, 67.4 looks something like this. It doesn't have to be perfect. And we're trying to figure out what the magnitude of this vector is. But if we break it down into its legs of the triangle, we know that the angle that it makes with the positive x axis is 67.4 and I know the y component is going to be 12. Now in part b, I'm actually going to be figuring out what the x component is, so that's unknown. So let's just get to our equations. If I'm trying to figure out the magnitude of a, how am I going to do that? Well, I've got my vector composition and my decomposition equations over here. Remember, there are 4 equations that we know. And the general rule is when we're going from components to vectors, we're going to use our Pythagorean theorem and our tangent inverse. When we're going from vectors down to components, we're going to use our acosine theta and a sine theta equations. So let's just try to figure out the magnitude using that Pythagorean theorem. What you're going to find is that when you try to plug this in, the magnitude, you're going to end up with 2 unknowns because we know what our a y component is, but I don't know what my a x component is. In fact, that's actually what I'm going to be solving for in part b. And I'm also looking for the magnitude. So I've got 2 unknowns in this situation. So that means this equation here is not going to help us. So it might look like we're stuck, but there are actually other equations that we can use, our vector decomposition or component equations, that will also relate the components with the magnitude and the angle. So we're going to figure out which one of these equations we can use. Remember, there are 4 equations. So as long as we have 2 variables, like a component and an angle or something like that, we can always use these equations to figure out the other 2. So let's check it out. So I've got these equations here. My ax component is unknown. My magnitude is unknown, but my angle is known. So I've got 2 unknowns in here. I can't use that. Here, though, I've got my y component. I'm missing my magnitude, but I have the angle theta. So that means that this equation, I can use to solve for the missing variable. So my a y is going to be the magnitude times the sine of the theta x. So that means that I can just move this over here. My y component is 12. I'm looking for magnitude right here. So that means that then I've got to multiply by the sine of 67.4. And now if I just divide this over to the other side, side, I get 12 divided by sine of 67.4 is equal to a. And if you work this out, you're going to get 13 meters as the magnitude of this angle or vector over here. It's going to be 13 meters. So now let's move on to part b. Part B, now we have to figure out the x component of this triangle. So now we actually have 3 out of the 4 variables for our vector. So we can use actually any one of these equations to figure out this x component. So if our x component is missing, we can use any one of these equations. I always like to stick with the easiest one. Now that we actually have the magnitude of this vector, the easiest equation to use is going to be our a cosine of theta. But if you want, if you want to figure this out using the other equations, you are going to get the same answer. So let's just use this equation. Our ax component is going to be 13 times the cosine of 67.4 degrees. And when you plug this in, you're going to get 5. So one way you can also double-check this is that you have now both components, the 5, and 12. And if you use the Pythagorean theorem, that should add up to 13. So that's one way you can kind of double-check it. Anyway, guys, that's it for this one. Let me know if you have any questions.

Here’s what students ask on this topic:

What is vector composition and how is it used in physics?

Vector composition is the process of determining the resultant vector from its components. In physics, this is used to find the magnitude and direction of a vector when its components along the x and y axes are known. The magnitude is calculated using the Pythagorean theorem: $\sqrt{a^2+b^2}$. The direction is found using the inverse tangent function: $\theta =tan^-1\frac{y}{x}$. This is essential for understanding the overall effect of multiple forces or movements in different directions.

How do you decompose a vector into its components?

Vector decomposition involves breaking down a vector into its x and y components. Given a vector with magnitude $A$ and direction $\theta$, the components are found using trigonometric functions. The x-component is calculated as $A\cos \left(\theta \right)$, and the y-component is $A\sin \left(\theta \right)$. This process is crucial for analyzing forces, velocities, and other vector quantities in physics.

What are the key equations for vector composition and decomposition?

The key equations for vector composition and decomposition are based on trigonometry and the Pythagorean theorem. For vector composition, the magnitude is found using $\sqrt{a^2+b^2}$, and the direction is $\theta =tan^-1\frac{y}{x}$. For vector decomposition, the x-component is $A\cos \left(\theta \right)$ and the y-component is $A\sin \left(\theta \right)$. These equations are fundamental for solving problems involving vectors in physics.

How do you calculate the magnitude and direction of a vector from its components?

To calculate the magnitude and direction of a vector from its components, use the Pythagorean theorem and trigonometric functions. The magnitude is given by $\sqrt{a^2+b^2}$, where $a$ and $b$ are the x and y components, respectively. The direction is found using the inverse tangent function: $\theta =tan^-1\frac{y}{x}$. This angle $\theta$ is measured relative to the x-axis.

What is the difference between vector composition and vector decomposition?

Vector composition and vector decomposition are inverse processes. Vector composition involves finding the resultant vector's magnitude and direction from its components. This is done using the Pythagorean theorem and trigonometric functions. Conversely, vector decomposition involves breaking down a vector into its x and y components given its magnitude and direction. This is achieved using the equations $A\cos \left(\theta \right)$ for the x-component and $A\sin \left(\theta \right)$ for the y-component. Both processes are essential for analyzing vector quantities in physics.