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3. Vectors

# Calculating Cross Product Using Components

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concept

## Calculating the Vector (Cross) Product Using Components 8m
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Hey guys. So in earlier videos I showed you how to calculate the vector product, like a cross B. By using this equation, A B. Sine theta. So the magnitudes of the two vectors times the sine of theta, the angle between them. But in some problems, you'll need to calculate this vector product in terms of unit vector components. And you can't use this equation. For example, the example, we're gonna work out down here, we're gonna write a cross B. But we're not giving angles and directions were not even given a diagram. All we're given is these vectors in terms of their unit vectors, Eyes, Js and Ks. So we definitely can't use a B sine theta. So what I'm gonna do in this video, I'm gonna show you how to calculate the cross product by using components. Now, right off the bat, I'm just going to mention that different textbooks and professors have their own method of doing this if your professor has a really strong preference and wants you to learn it their way, do it that way. If not, I'm gonna show you what I think is the easiest way to do this and I'm gonna show you a list of steps. You get the right answer every single time. So let's go ahead and jump right in. Alright, so remember the whole idea here is that this vector product, a cross B just generates a new vector C. So we need to do is we need to find a way to calculate its components. Remember each vector, like whatever vector can be written in terms of its X, Y and Z components, I J and K. Now, when we did this for scalar products and we did this, we didn't really have to calculate the components because all you had to do was just multiply the like components straight down and then add everything up together. But what you got was just number you didn't have to actually solve for any components. But see this new vector here actually does have components. So in other words, it can be written as C X ni I hats plus C YJ hats plus C Z K hats. So we would need to do is we need to figure out a way to actually solve for C X, C Y and so easy. So let's just jump right into our problem here because that's exactly what we're gonna do. We're gonna write the A cross B in terms of its components. So, we're basically gonna get something that looks like this. All right. So let's just jump right into the first step here. The first thing you want to do is you're gonna want to build a table of all of the x and y and Z components for the two vectors that you have. But what you're gonna do is you're going to repeat the X and Y columns twice. What I mean by this is you're gonna build a little table that looks like this and what I have to do is sort of extract the numbers for a X, Y, Z and so on and so forth. If you look this vector, this is I hat plus two J. Hats. In other words this is a one, right? It's kind of implied here and this is a two in the y directions. We have one and two does a have a Z component. Well that would mean it has a K hat direction but we have no K hat here. So we're just gonna write a zero, there's no Z components. And then you're just going to repeat this, X and y twice. You're going to have one and two. Now we're gonna do the exact same thing for the B. So this bee has negative two in the X And then it has three in the UAE and then it has four in the K. And then you just repeat the other the first two columns. So it's negative two and three. So that's the first step, you just build out the table of components. Now, the next step here is we're going to write a b minus B. A. For each component? Remember we're really just trying to figure out what R, C, X, y and Z. So we can write our vector in this form. So what happens here is I'm gonna write a b minus B. A. For each of them. So I'm gonna write a B minus B. A. And I'm always gonna leave a little space and you'll see why in just a second here. So maybe minus B. A. Um It's kind of one way I remember that. Right? So A B minus B. A. All right so that's done. The next thing you want to do is you want to always multiply the unlike components diagonally. What do I mean by this? Well if you're trying to figure out C. X. Then what I'm gonna do is I'm gonna multiply the components that are not X. So what I always like to do is like like to just draw a strike through that column. So two multiplier to get the X. Components. I'm gonna have to multiply the unlike components Y. And Z. And the way I'm going to do this is I'm going to multiply these numbers diagonally. So one way to think about this is that remember this is called the cross product? And the reason for that is you're gonna take this table here and you're gonna multiply diagonally like this. So you're gonna go this way multiply two and four and then three and zero. So you're always going to go down diagonally and then up diagonally. So this is why it's called the cross product. So in other words what I'm really doing is I'm multiplying a. Y. Times BZ. That's the first thing. So I'm doing a Y. B. Z. And then minus and then I'm gonna do a. Sorry B. Y. Times Z. So this is gonna be B. Y. Times A Z. Now I'm just gonna plug and chug. Right, so this is gonna be two times four. So we always cross going down first And then -3 times zero. Like this what happens here is this last term is gonna go away and you're just gonna end up with eight. So that's my ex components. Alright? So you just cancel out the component that you don't want to that you are the light component. You multiply the unlike components and you just plug and chug. Now we're gonna do the same exact thing for the Y. Component. So you basically just repeat the step over and over again for the Y. Component. See why I'm gonna have to look at the Y column and I want to multiply the unlike components. So basically I'm gonna cross this out because I don't need it anymore. And I'm gonna multiply the Z. And the X. Components. And I'm gonna go like this down and then up. So what you end up with here is that you have a Z be x minus bc AX. So in other words you have zero times negative two -4 times one. So this zero goes away, right? Because anything times 00 and you just end up with -4. And now I want you to go ahead and pause the video and see if you can do this last one yourself. Alright, so we're gonna do the same thing. All right? So we've got uh we want to figure out the Z components. I'm going to cross out the Z. Column. I don't need it anymore to figure out Z. You're gonna have to multiply A. Or sorry, X. And Y. And you're gonna cross, you're gonna go down like this and then up like that. So you're gonna have a X. B, Y minus bx A. Y. So what you end up with here is one times three minus negative, two times two. Alright. So what you end up with here is three minus negative four. And we work this out, you're gonna get seven. Just be careful here because you have a negative, you have two negatives there. So you have to, you know, don't get tripped up with the minus signs, but that's basically it. So, I want you to sort of notice that there's a pattern here. It's always a B. B. A. Always a B B A A B B. A. And what happens is that the two letters here are always gonna be the ones that are next in the table. So for example, you have X. If you're calculating X, it's always going to be Y. Z, Y. Z. Because that's what happens in the table. You have X. Y and Z. For why you're gonna have the next two columns on the table, which is Z and X. So you're gonna have Z. X. Z. X. And then finally for Z. You're gonna have xy xy. Because that's what's next on the table. So that's why I think this is the easiest way to learn this. Alright, so if you put all of this stuff together here, all these components, basically what you're gonna get is that this c vector can be written in terms of these components here. What you have is you have C. X. Which is eight in the I. Hat And you have negative force, you're going to have -4 in the Jihad and then you have seven in the K. Hat. And that's how you write the C vector in terms of its unit vector notation. Alright, So if you put all this stuff together here, basically these equations, I'm just gonna write out sort of a long generic equation. This is something you might need to know, you might see in a textbook but basically I'm just gonna repeat these over again. So this is gonna be a Y, b, z minus B, y, A Z plus a Z, B x minus B, Z X. And then finally A X B y -7, So that's basically sort of the generic way to write any unit vector in terms of the cross product. Alright, so that's it for this one. Guys, let me know if you have any questions
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example

## Finding Vector Product Using 2 Methods 9m
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