Intro to Cross Product (Vector Product) - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Vector (Cross) Product and the Right-Hand-Rule

Video duration:

10m

Play a video:

Hey guys. So in a previous video I covered that. There was two ways of multiplying vectors. One was called the scalar product or sometimes called the dot product. We've already covered that. And in this video, I want to show you the second way to multiply vectors, which is called the vector product or sometimes called the cross product. I'm also going to cover a really important rule for cross products that you'll need to know which is called the right hand rule. And you're gonna be using this a lot sometimes in physics. So let's go ahead and check this out here and we're gonna do a bunch of examples. Now, before we begin, we begin, I just want to briefly recap dot products. So imagine I have these two vectors here, A and B. They're offset by some angle of 60 degrees. Remember the dot product or the scalar product is just one way I can multiply this three and this four vector. But what I get out of it is just a number, right? I don't get a vector out of it. So the equation that we use for this is this a B. Cosine of theta. Right? So if we just do this here, A times B or a dot B was a B. Cosine of theta. So in other words, it was just the magnitude of a which is three magnitude of B, which is four times the cosine of the angle between them, which is just 60 degrees. What you get out of this is you just get six right? You don't get a vector, you just get the number six. And what the six really just represents here is it just represents the multiplication of parallel components of A. And B. So for example this be vector here has a component that lies in the same direction parallel to A. And if you work this out you're gonna get the B. X. Component is too. So if you multiply these two components here that are parallel, you end up just getting six out of it. Alright, so that's the scalar products. Now we're gonna do the same thing, We're gonna multiply these using the vector products and the whole idea here guys, the thing that's different about the vector product is that you get a third vector out of it. You're gonna get a new vector. That's why we call it the vector product. And this vector here is gonna be called C. Alright, so with the scalar product, you multiply them, you just get a number with the vector product. We're gonna multiply these two vectors and we're gonna get a third one out of it called C. That's what you need to know about the C vector. Is that it is perpendicular to both A and B. Alright. The equation that we're gonna use to calculate the magnitude is going to be very similar to the dot product. We're going to use a times B. Times the sine of theta. So before we use the cosine. But now we're going to use the sign. So it's the magnitude of a magnitude of B. Times the sine of the angle. And this angle here, Theta represents the smallest of the angles between A. And B. So in other words there's two ways to represent this angle. You can go all the way around, but that's not the smallest angle. So we're just gonna use the 60. Alright? So if we do the vector product of these two vectors, what you're gonna get is a third vector, which is C. And the magnitude of the C vector is just gonna be a B times sine of theta. So we're gonna do the same thing we did over here, this is gonna be three and 43 times four. Except now we're gonna do sign of 30. Now, if you I'm sorry, this is a sign of 60. And if you work this out, what you're gonna get here is 10.4. The magnitude of this vector is not gonna be six, it's going to be 10.4 because now we're using this sign. Okay, but this is a vector. Remember magnetic vectors have both magnitude and direction. Where does this 10.4 vector point. And to do that, we're gonna use something called the right hand rule. So to find the direction of the vector products, we're going to use something called the right hand rule. Now different people have different rules for doing this. If your professor has a preference, I would say stick with that. But I'm gonna show you the way I think is the easiest way to do this. And I highly suggest that you do this with me every single time, because the more you do it, the more you'll get familiar with it. All right, So here's how the right hand rule works. What you're gonna do is you're gonna take your hand and you're gonna point your fingers along the first vector always you're always gonna point along the first vector, which is in our case is a. So we're gonna point our fingers along a like this. Okay, now what you wanna do is you want to curl your fingers towards the second vector. So you're gonna curl them towards B. So in other words, I'm gonna curl my fingers up like this in the direction where B. Is and what you do that, what you're going to see is that your thumb points in the direction of C. So your thumb is going to give you the direction of that new vector. So you're gonna take your hand pointed a, curl it towards B. And now my thumb is going to be pointing out towards me. So the way that we represent this on a two dimensional graph, like a you know xy xy plane is we just draw a little circle with a dot going through this this symbol here, The circle of the dot means it's going out of the page towards you, right? So you can kind of imagine like an arrow that's coming straight towards you. All you see is just the tip. Okay, what's really important about this? Is that your curling first? You're pointing your fingers along the first and then your curling towards the second one. So let me show you, for example what happens if you do it the other way around. So in this diagram I have the same exact numbers, the same exact vectors. Except now they're flipped. Right now I have a that's pointing up like this and be that's going down. Alright, so what happens here Is that the magnitude of this is still gonna be 10.4, right? And I'm still going to have another vector. But now what happens is I'm going to point my fingers along a and in order to curl towards B, I have to flip my hand around, right, you're going to do this so you're gonna have to sort of contort your hand like this. My fingers point along a curled towards B. And your thumb is now gonna be pointing away from you. My thumb is pointing away from me towards you, right? So it's pointing away from me. And so the symbol that we use for that is a circle but now we've drawn X through it like this, that indicates that it's going out of the sort of into the page away from you. Okay, so you can kind of imagine that if you were to sort of fire an arrow and look behind it, you would see a little feathers or something going away from you. Okay, so that's basically all there is to it. So you're gonna use the magnitude, you're gonna use this equation to calculate the magnitude and then you're gonna use the right hand rule to calculate the direction. Alright, so let's go ahead and just do a bunch of examples together. Alright, so now we want to do is we wanna sort of calculate the magnitude and direction of these vector products here. Now what I want to mention off the bat is that sometimes sometimes in some books you may be given sort of a three dimensional view like this. Um And some books will even do X. And y like this and then Z. And y like this. There's actually no right way to do it. It really just comes down to preference. I'm just gonna show you both of them just in case you run into them. Okay, so here's what happens. So we've got these two vectors, I've got six and five. So what I wanna do is calculate the magnitude and direction of a cross B. So in other words, when I do this we're gonna we're gonna get is that the magnitude of C. Which is gonna be a. B. Times the sine of theta. Okay, so the magnitude of a is just six. The magnitude of B is five And the sine of the angle between them. Well, what happens here is we have an an era that goes along the X axis and one that's still kind of like in the XZ plane. So what I want you to do is kind of imagine that you had two pens on like a tabletop and they're both flat but they're just sort of pointing offset. Right? So one of the pens is sort of like facing away from you But they're basically both still flats. Okay, so the angle between them is going to be the 30°. So we're gonna have six times five times the sine of 30, which is 15. Alright, so that's the magnitude, where does it points? We're gonna have to use the right hand rule. So again take your fingers, point them along A So point them sort of flat on your tabletop like this. And I want you to do is curl them away from you towards be so curl your fingers like this away from you. And what you'll see is that your thumb is going to be pointing up. So in other words, this vector here is going to be pointing along the y axis in this problem. So we've got C. Is equal to 15 and it points along the plus y axis. Okay, so the one way, one way you can kind of see this is that if you are looking at this from the top down, right? If you are looking down from the tabletop then this vector C. Would just be coming out towards you right? It's basically just going up like this, you're looking straight up at it. Okay let's do the next one here. So now we've got here is we've got this Y and Z. So again we have a different sort of orientation but it works the same. So we want to calculate the cross products. So we've got see the magnitude of C. Is a B. Sine Theta. Okay so the magnitude here is three magnitude here is eight. And then what about the angle that's between them? What do I plug into my sign here? Do I plug in 23? Do I plug in 67? Remember? It's the smallest angle between these two vectors here between them. So what you can do here is if they sort of both lie on the Y. Z. Plane like this. Remember that a straight line Is 180. So if you subtract 67 and 23, what you're gonna get out of this? Is that this is actually a perfect right angle. This is 90°. So we've got 90 degrees like this. So you're gonna stick in sine of 90. And what you're gonna get here is that this vector here is 24? Okay, so the sine of 90 is just one? Alright so where's the point? Well along the three D. View. What happens is we're gonna take your finger curl it, sorry point along a curl towards B. So if you do that, what happens is that your thumb is going to be pointing away from you? So the way this looks on a three dimensional sort of plane is that it be going backwards like this? Remember this is sort of like the front to back dimension. So it's gonna be pointing out like this. So this c vector here is gonna be 24 points sort of away from you like that and we would represent that again. It's just an X. A circle with an X. Like that. Okay now the last one here is we're going to do two and 4. So we got the magnitude which is just a be sine theta. Okay so now we've got the magnitude which is four magnitude which is two. And what about the sine of the angle between them? Well what happens is here, is that on a two dimensional sort of grid like this these these vectors are parallel. So what happens here is that the angle between them is zero? And if you plug this in, what happens is that the sine of zero is just zero. The whole thing just goes away. So in other words what happens is that the cross product is 0? If the vectors are parallel and this can happen one of two ways either the angle between them is zero or you can also have 1 80. If you were to plug in sign of 1 80 you're also gonna get zero. That happens if they're pointing away from each other. Alright, so that's it for this one guys, let me know if you have any questions.

2

Problem

Problem

Find the magnitude and direction of the vector C =2B × A.

A

|C| = 27; along the –x direction

B

|C| = 27; along the +x direction

C

|C| = 54; along the +x direction

D

|C| = 54; along the +y direction

E

|C| = 19; along the +x direction

3

example

Find the Vector Product

Video duration:

4m

Play a video:

Hey guys, so hopefully got a chance to check this one out on your own. So we have these two vectors on our diagram here and we want to find the magnitude and direction of the cross product. So we're gonna have this new vector C. Which is a cross B. So let's go ahead and get started. Now I can't draw this vector on this diagram because I don't know what the direction is, but I can go ahead and find out the magnitude by using my equation for the cross product. The magnitude is just a. B. Times the sine of the angle between them, right? So stated here is between A and B. The magnitude of A is just giving us 12, the magnitude of B. Is eight. Now we just need the sine of the angle between them. So it's gonna be this 30 degrees, It is gonna be something else. Well this is where sort of these three D diagrams can be kind of tricky because because you sort of can mess you up on the perspective of things. So this X axis here is kind of like back and forth. Right? So this xy plane is kind of like the flat plane and the Z plane or the z axis goes vertical. So what happens is this a vector? It might look like it's kind of raised off the ground, like it's kind of like doing this but it's actually not. So what happens here is these dash lines are supposed to indicate that it's along the X. Y plane. So imagine that you have sort of a tabletop like this and imagine you have this vector that's pointing to the side except it's kind of tilt away from you so it's kind of like tilting away like that so that's what this A vector is, and then B. Is just pointing straight up. So what is the angle between something that points horizontal and vertical? It's 90 degrees. So it might not look like it's 90 degrees. But this actually is because sort of this is like vertical and flat. Alright, so this is actually 90 degrees here, so that's the sign of 90. And so therefore the magnitude of our vector is going to be 96. Alright, so that's the magnitude have about the direction. Well for the direction we're gonna need to use the right hand rule. Remember the rule is for a cross B. You always point your fingers towards A. And you curl towards B. So here's what I want you to do, don't you take your right hand and point your fingers in the direction of A. So what happens is it's kind of flat like this. But remember you're gonna have to turn your fingers away from you, it looks like I'm moving towards you but I want you to do it yourself, your finger's gonna be pointing away from you and then you're gonna curl your fingers sort of up vertically towards B. So when you do that, what happens is your thumb is going to be pointing not straight at you, but kind of off to the right like this. So it's gonna be kind of looking like this all right now, what we want to do is this is our c vector. It's a it has a magnitude 96 it points here along this axis. Now this thumb, my c vector is still flats, right? It's still flat, it's just still on the xy plane, just pointing in a different direction. So I'm gonna use the same sort of dash lines that I used for the other one to kind of indicate that it's on this plane. What I want to do is essentially figure out what's the angle of the positive angle um from the X axis, remember? So that's this one over here. So basically want to figure out what is this data now? So, which kind of equation can we use remember this data? Is is the angle between A and B. Tells us nothing about this angle over here. So we can't use that equation. Well, it turns out we're gonna have to use the property of the of the vector product. Remember that? The definition of this vector is that it's perpendicular to both of the vectors that make it up, in other words, C is perpendicular to the vertical, that's 90 degrees and it's also perpendicular to the A vector that's also 90 degrees. So what happens is I'm gonna look at these two vectors over here because I'm told some information about this 30 degrees. And this angle. So what I'm gonna do is I'm gonna try to try to draw sort of like a top down view of this diagram and this is what I end up with. Right? So you end up with a little diagram looks like this. This is your plus Y and your plus X. Right? So basically I'm just imagining that I'm kind of just looking down at the table top from above. So I've got my vector a. That points over here, this is a equals 12. We know this is 30 degrees and this is my C vector, this is my new cross products, right? So it looks like this actually when we look it's gonna be like that. Right? So this see here is 96. Alright, So this angle here, Theta is actually just this angle. The angle between this X axis over here and see. All right. So how do we find that? Well, we know that The difference between X&Y. This angle here is 90. And if we know that this here is 30° and we know that this is also a 90° angle because the vector product is perpendicular. Then what happens here is if this is 90, then this angle also has to be 30°. So we have 30 and then we have 90 over here. And that means that this vector points off at 30 as well. So, that's actually your answer. So the answer is that data here is equal to 30 degrees as a positive angle from the X axis. Alright, so hopefully that makes sense guys, let me know if you have any questions.

4

example

Finding Angle Using Scalar & Vector Products

Video duration:

4m

Play a video:

Hey guys, So let's get started with our problem here. So we've got these two vectors A. And B. And we're told some information about their scalar and vector products there dot and cross products and we want to calculate what their angle between them is. So in other words, what we're asked to find here is theta. But if you notice, I actually know nothing about these vectors, I don't not given a graph, I don't know what their magnitudes and directions or anything like that. So how do we figure this out? Well, if you're looking for theta, the one equation that pops up here with the theta term is going to be that the magnitude of the vector products, remember this really just produces another vector. Just going to call it C. Is equal to a B. Sine theta, the two magnitudes and the angle between them. So really this is our target variable and we actually know what this magnitude is, right? Usually we're asked to find it, but we actually know it's already 12. So if I try to rearrange this equation, what I'm gonna end up with is that the sine of theta is equal to 12 over a B. And I can't really do anything else with this, right? So I can't even if I were to try to do this inverse sine, I still don't know what this A and B are. So, because I don't know what the magnets of those vectors, I can't really finish off this equation, I'm going to need something else. I'm gonna need another equation. Well, the one thing that we haven't seen that we haven't used yet, is that the scalar product is negative eight. Remember the scalar product is A dot B. So what I'm gonna do here is I'm gonna start out another equation and I'm gonna say that a dot B is equal to remember that the definition of a dot B. It doesn't have a magnitude, it's just a number and it was just a B. Cosine theta. Except we already know what the scalar product is. We know that A dot B. Whatever. Whenever you do A B. Cosine, theta, you're gonna get negative eight. Alright, so here's what I'm gonna do, I'm gonna rewrite another equation. I'm gonna get cosine of theta, right? Because this is my other, this is my target variable, just in another equation. And this is going to be -8 divided by a B. So here's what I'm gonna do, right? I've got these two equations and they actually both have three unknowns. I've got the theta that's unknown and I've got my A and B. That's unknown. So what I can do to get rid of them is I can actually divide the two equations. So what I'm gonna do here is I'm gonna do sine of theta divided by cosine of theta. And what you're going to get here is that 12 divided by A B divided by negative eight Over A B. And what happens here is that the A B. S are going to cancel when you divide these two equations, the abs cancel out. And then basically what you end up with here is just a tangent. Theta. Right? Sine over cosine equals tangent. So you've got the tangent, theta is equal to and this is just 12 over negative eight. So this is just gonna work out to negative 1.5. So we've got these two equations that had three unknowns and by dividing them we were actually able to get down to only one equation with one unknown. So now what I have to do is just I just take the inverse tangent of this and just make sure that your calculator is in degrees. You're gonna take the inverse tangent of negative 1.5. What you're gonna get here is you're gonna get negative 56 points. So you're gonna get like negative 60 56.3 degrees. Alright. So is that our final answer? Well remember that this angle here, this theta is the angle between these two vectors A and B. But it has to be the smallest positive value. So what we can do here is if this angle here is negative 56 then we're gonna have to add 100 and 80 degrees to it. one way you can kind of think about this Um is that these two vectors have a scalar product of -8. So what happens here is I'm just gonna draw like a like a sketch real quickly, what these vectors might look like. So this is a sketch. So this is my X and Y axis. So, if this vector is like, let's say this is a then in order for them to have a scalar product that's negative, It means that they have to be pointing in opposite directions. You only get scalar products that are negative whenever you have some components that point anti parallel. So basically what happens is we know that the angle between A and B has to be greater than 90 degrees. So this data here must be Greater than 90° because a dot B is negative is less than zero. Alright, So now that we've added 182 this what you're gonna get is 100 and 23.7 degrees and that is the right answer. All right. So now we know that these angle this angle here has to be 123.7. That's the only way you can get A vector product that has a magnitude of 12. But a scalar product that has a magnitude Or that's a scalar product, that's negative eight. Alright, so hopefully that makes sense. Let me know, guys, if you guys have any questions and I'll see in the next one

Do you want more practice?

We have more practice problems on Intro to Cross Product (Vector Product)