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3. Vectors

Calculating Cross Product Using Components


Calculating the Vector (Cross) Product Using Components

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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

Finding Vector Product Using 2 Methods

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Hey guys, so let's get started with our problem here. So we've got these two vectors A and B. And what I want to do this problem is I want to calculate find the magnitude and direction of the cross product. So a cross B. But I'm gonna use two different approaches to get it. The first, in the first part, I'm gonna use a B. Sine theta and the right hand rule. And then what happens in part B. I'm going to write these this cross product in terms of the unit vector components. Okay, so let's get started. I'm gonna go ahead and start off with part eight but before I start, I actually want to go ahead and just draw out the vectors. So I've got four, I and three J. So this is basically going to look something like this. I've got four. I've got three and this vector over here is going to be my A. Alright, we've seen this kind of thing before, so this be vector here is gonna be negative two plus three. Some of the words the components are gonna be, So this is gonna be let's see is like negative two. And then I've got three over here, this is my components and the actual vector is going to be this guy. So this is my B. Over here and I've got a over here. Alright, so let's go ahead and start with part. I want to find the magnitude in the direction of the cross product. So the magnitude of this magnitude of C. Remember is going to be a B times the sine of theta. But we're gonna have to use the absolute values magnitude of a magnitude of B times the sine of the angle between them. If you look through your problem or you look through this question, we actually have none of those things. We don't have the magnitudes and we don't have the angle between them. So we have some work to do. The first thing we have to do is calculate the magnitude of each of these vectors, but we can do that because we're told what their X and Y components are, right? So we basically can figure out what this magnitude is. Just by using, you know, some uh some Pythagorean theorem, this is pretty easy, this is a 345 triangle. So with this a vector we know is five. That's the hypotenuse, so that's one that's pretty, you know, pretty quick. The other one here is not really a special triangle. So we're going to have to calculate this really quickly here. So this magnitude of B. I'm just gonna go over here real quick. The magnitude of B is remember just the pythagorean theorem. So you just do negative too squared plus three squared. And what you're gonna get here is you're gonna get this is Let's see, I've got 3.6. Okay, so that's the other magnitude. So basically this is gonna be let's this is gonna be five times 3.6. And now we just need to find the angle between these two vectors. In other words we need to figure out what this value is. This is my theta. Alright so how do we do that? Well um I don't know what this data is but what I can do is if I know the components of each of the triangles, what I can do is I can find each of these angles. I'm gonna call this one theta A. And I'm gonna call this one day to be because we have the components we can just use inverse tangent to find those. So basically what happens here because I'm gonna I'm gonna um sort of write this sign, I don't know what I'm gonna stick in here just yet but I'm gonna have to figure out this data. So this data here is going to be well what I can do is this whole entire angle is 180 minus theta A minus theta B. Basically what I'm gonna do here is I'm gonna subtract from 1 80. I'm gonna subtract these two angles which I can find and that will just give me whatever is left over and that's the angle between the two vectors. Okay so theta A. Is just equal to inverse tangents Of this is gonna be three divided by four, right? So that's why over X. If you work this out you're gonna get 36.9°. We've seen that angle before. This is the special angle when you have a 345 triangle. Alright. Day to be is gonna be the inverse tangents. And we're gonna use a positive values. Otherwise you're gonna get like a weird negative number. We're gonna use three and two. So this is going to be Y. Component of the X. Components. And what you'll get here is you'll get 56.3 degrees. All right. So basically this is the 36.9 and this is the 56.3. So what I'm gonna do here is my data value, This data is going to be 36.9 -56.3. And what you're gonna get here is you're gonna get that this angle here is 86.8°. So that's what we plug into our equation. So this is gonna be the sign of 86.8. All right. All of this stuff is stuff that we've seen before when you work this out, What you're gonna get here is you're gonna get um You are going to get 18. So in other words, the magnitude of this vector here, C. Is 18. That's the first answer. So, we know that this vector here is 18. Now interesting to figure out which direction it points and to do that we're gonna use the right hand rule. So remember we're gonna point our fingers along A. So we're gonna go like this right? And then you're going to curl towards the B vector. So you're going to curl your fingers this way. And what happens is when you do that, I want you to do this. Your thumb is going to be pointing right at you. So in other words, this C vector here is actually gonna be pointing out of the page. So direction, it's going to be our circle with a dot in it. In other words, Z. We're going to call this the Z direction. In other words, that's really just K. Hat, Right? So k hat, that's the Z component that goes out towards you. So basically that's it for part A. Alright, so that's the magnitude, we've got 18 indicate hot components. So here's what's going on now we're gonna do is we're gonna do the same exact thing, We're gonna write the we're gonna figure out what this C vector is. But now we're just gonna use the other method to do it, which is building out the table of components and then doing the A B. B. A thing and we're gonna calculate the magnitude of C. And hopefully we should come up with is the right is the same answer that we just got Hopefully we should just end up with a magnitude of 18 and it should point in the K. Hat direction after all, it's the same thing. We're just using two different methods to get it. So let's get started. The first thing we have to do is just build our components of our a and b vectors. So I'm just gonna draw this real quick here. So I've got let's see more space. So this is A. And be and then I need 1234. So this is going to be X. Y. Z X. Y. Right? You're gonna repeat them and then we're just gonna extract what each one of these numbers is from these vectors. So what I've got here is that four, I. N. Three J. So in other words this is gonna be four And three, there's no z components and then four and 3 and then the other one is going to be negative two and three. Right? So we've got negative two and three. So this is just going to be negative two and 30 negative two and three. Just repeat it. Alright, so now the next step here is we're going to write a B B. A for each components. So in other words, I'm going to do C x, c, Y and c. Z. So this is going to be a B b minus B. A a b minus b A a b minus oops B A. Okay, so now we're gonna do here is we're gonna go ahead and start crossing things out, right? So remember the C X components? We're gonna cross out that one first and we're gonna do down and then up we're gonna cross the unlike multiply the unlike components, some of the words we're gonna get Y. Z. And then Y. Z. So what you get here when you multiply this out is you're gonna get three times zero minus three times zero. So what happens here if both of these things will go away and your X component will just be zero? Alright, so that's the first one, So now the next one, remember we're gonna do this, the Y component, we're gonna cross that out and we're gonna do down minus up. So now where this is going to be a Z. X Z X. In other words this is going to be zero times 4 zero times negative two. What happens is both of these things again go away and you're just gonna get zero for the Y components and now finally you're just going to do the Z components, you cross that out and then you're gonna do down, you're gonna cross down and then minus up. So this is gonna be a X b y minus B b X A Y. So in other words what you're gonna get here is you're gonna get four times three minus and this is gonna be negative two times three. Alright, when you work this out you're gonna get 12 over here and this is gonna be negative six, you're gonna have 12 minus negative six. And what you end up with here is 18. So in other words, what we just found here is that using the unit vector component method, we can actually write the c vector here as 18 and this is going to be in the k hat direction. And that's exactly what we got for the first one. For the first method, you know, using a B sine theta, figuring all the angles of the magnitudes and things like that, we actually just got a magnitude of 18. And using the right hand rule, we figured out that points directly out towards you out of the page. And that's exactly what we got here by unit by using the unit vector component method. Alright, so that's it for this one guys, let me know if you have any questions.