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Vector Composition & Decomposition

Patrick Ford
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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 so we could break this up into a triangle. We can count up the boxes to figure out the legs, like three and four stuff like that. Unfortunately, a lot of vectors that we're gonna see are not are gonna be on diagrams that don't have grids or squares. And we're gonna have to do vector math without using them. So in this video, I'm gonna show you that there's really four equations that we need in order to describe everything that we need to know about a vector. These four equations arm or triangle math equations. They're called vector composition and decomposition equations. Let's check it out. So, guys, there's three things that we need to describe a vector. When we have a grid that doesn't have, you know when we have ah, diagram doesn't grids of squares. So 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. The last thing is that vectors also are triangles, which means that they have legs. These legs just get a fancy name. They're called the components. So basically, this vector here, this triangle can be broken down into its legs. These legs are just called components. So I mentioned that there's four equations and it really just breaks down into two 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 is off the high pot news decomposition is the opposite. It's our 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 we did what we did when we added perpendicular vectors like three and four. So when we added three and four, we just basically combine them. Tip to tail on Ben, we had 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 theory because this made a triangle. So three and four, the high pot news is just square into three squared four squared and that's m. So these components here are these legs three and four just get a fancy name. They're called components. And so my three is my ex component. It's how much of this vector isn't. The A is in the X axis, and this four here is in the Why is my wife components? That's how much of the vector lies in the Y direction. And they combined to form a vector of magnitude. A. So a walked in this direction. It's kind of a Ziff. I walked three and then four 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 Data X here is just by using another set of triangle math equations called Sakata. So basically, to figure out this angle here, um, all I have to use is just a new equation that your textbooks and your professors will derive this Data X is the inverse tangent or the arc tangent of my white components over my ex components. So in this example, my wife components for X is three. So that means that my theta X here is just gonna be the arc tangent. So you can find this in your calculators and this is gonna be 4/3. And this theta X is just 53 degrees. Make sure your calculator is in degrees mode. And so 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 the direction. The other thing is vector decomposition. It's basically the reverse process. So here I have the magnitude of the angle. Now I want to figure out the legs, so we want to figure out what my ex and I want to figure out what my A Y components are. And really, it just comes down to these two equations here that we come that come from so Kitona. So now we want to decompose a into its components a X and A y. So here's the deal. As long as this angle fatal X is drawn to the nearest x axis, then we're gonna use these equations here. My ex is gonna be a times the co sine of the angle. So it's the magnitude Times CO sign of angle and A Y is gonna be a times the sign of the angle here. So, for example, if I were given an angle like this, like this data here, this is a data that's relative to the Y direction. This is bad. I can't use this. Well, I won't use that in my equations. This this angle over. On the other hand, Data X is good. So this is the one that I plug into my equations. So, for example, by X is gonna be the magnitude times the cosine. So the magnitude is five. The cosine of the angle is 53 degrees, and so I'm just gonna get three. And then if you plug in a y, you're gonna get the magnitude, which is five times the sorry not the coastline of sign the sign of 53. And if you plug this in, you're gonna get four. Notice how we basically just come up with the exact same numbers three and four as we had over here. And that should be no surprise because we 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 start with the magnitude and the direction. So it's no surprise that we got the same exact numbers. Alright, guys, that's really all there is to it. You're just gonna 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. You gonna draw the vector and then solve for the missing variables. So my exes ate my A y six 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 eights and this is six. This is my ex. This is my A y. And now I'm actually trying to find the magnitude and the direction. So that means I'm gonna use my vector composition equations. So to figure out my A, I just have to use the Pythagorean theorem. I just use eight squared plus six square. And if I do that, I get 10. So this is my magnitude. And now for the angle. Remember the angles drawn relative to the X axis? This is my tha X and this fatal X here is gonna be the inverse tangent, the arc tangent of my white components, which is six over my ex component, which is eight. And if you go ahead and work this out, you should get 37 degrees. So this is my angle now that's 37 degrees. All right, let's move on to the 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. Eso I'm just gonna draw this like like this So I know that this be here is 13 now, the angle relative to the X axis, it's 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 ex and my a y So my ex is gonna be the well, it's what equations, um, I'm gonna use I know the magnitude and the direction. And I want to figure out the components. I'm gonna use my vector decomposition equations. So I'm sorry. This is actually be so this is gonna be I'm sorry. What's this is B y. And this is B X. Okay, so my BX is just gonna be 13 times the cosine of 67.4. And then if you work this out, you're gonna get five. And then if you do be, why you're gonna do 13 times the sign of 67. and you're gonna get 12. So these the components here. So here my components. Alright, guys, that's it for this one. Let me know if you have any questions.