Calculating Cross Product Using Components - Online Tutor, Practice Problems & Exam Prep

To calculate the vector product \( \mathbf{A} \times \mathbf{B} \) using unit vector components, first create a table of the x, y, and z components for both vectors. Then, apply the formula \( C_x = A_y B_z - A_z B_y \), \( C_y = A_z B_x - A_x B_z \), and \( C_z = A_x B_y - A_y B_x \) to find each component. This method emphasizes the importance of understanding vector components and their relationships in physics, particularly in contexts like angular momentum and electromagnetic interactions.

Calculating the Vector (Cross) Product Using Components

Hey, guys. So in earlier videos, I showed you how to calculate the vector product like a cross b by using this equation absinθ. 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 going to work out down here, we're going to write A cross B, but we're not given angles and directions. We're not even given a diagram. All we're given is these vectors in terms of their unit vectors, i's, j's, and k's. So we definitely can't use absinθ. So what I'm going to do in this video is I'm going to 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 going to show you what I think is the easiest way to do this, and I'm going to show you a list of steps so 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 what 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 a number. You didn't have to actually solve for any components. But C, this new vector here, actually does have components. So in other words, it can be written as cx⋅I+cy⋅j+cz⋅k. So we need to figure out a way to actually solve for cx, cy, and cz. So let's just jump right into our problem here because that's exactly what we're going to do. We're going to write the A cross B in terms of its components.

So let's just jump right into the first step here. The first thing you want to do is you're going to 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 going to do is you're going to repeat the X and Y columns twice. What I mean by this is you're going to build a little table that looks like this, and you have to extract the numbers for ax, ay, az and so on and so forth. If you look at the vector, this is i hat plus 2 j hats. In other words, this is a one. Right? It's kind of implied here, and this is a 2 in the y direction. So we have 12. 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 going to write a 0. There's no z component. And then you're just going to repeat this x and y twice. You're going to have 1 and 2. Now we're going to do the exact same thing for B. So this B has negative 2 in the x, and then it has 3 in the Y, and then it has 4 in the K. Then you just repeat the first two columns. So this is negative 2 and 3. 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 AB−BA for each component. Remember, we're really just trying to figure out what are cx, cy, and cz so we can write our vector in this form. So what happens here is I'm going to write AB−BA. I'm always going to leave a little space and you'll see why in just a second here. So AB−BA. Alright. So that's done.

Finding Vector Product Using 2 Methods

Hey, guys. So let's get started with our problem here. We've got these two vectors, a and b. What I want to do in this problem is calculate and find the magnitude and direction of the cross product, so a cross b. But I'm going to use two different approaches to get it. The first, in the first part, I'm going to use absinθ and the right-hand rule, and then what happens in part b, I'm going to write this cross product in terms of the unit vector components. Okay? So let's get started. I'm going to go ahead and start off with part a, but before I start, I actually want to go ahead and just draw out the vectors. I've got 4i and 3j. This is basically going to look something like this: I've got 4, I've got 3, and this vector over here is going to be my a. Alright. We've seen this kind of thing before. So, this b vector here is going to be negative 2 and 3.

So, in other words, the components are going to be negative 2 and then I've got 3 over here; those are 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 a. I want to find the magnitude and the direction of the cross product. So the magnitude, this magnitude of c, remember is going to be a b times the sine of theta. But we're going to 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 magnitudes of each of these vectors, but we can do that because we're told what their x and y components are. Right? So, we can basically figure out what this magnitude is just by using some Pythagorean theorem. This is pretty easy. This is a 3, 4, 5 triangle. So with this a vector, we know is 5. That's the hypotenuse. So that's one. That's 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 going to go over here real quick. The magnitude of b is, remember, just the Pythagorean theorem. So you just do negative 2 squared plus 3 squared, and what you're going to get here is, let's see. I've got it's 3.6. It's 3.6. Okay. So that's the other magnitude.

So, basically, this is going to be let's this is going to be 5 times 3.6. And now we just need to find the angle between these two vectors. So in other words, we need to figure out what this value is. This is my theta. Alright? So, how do we do that? Well, I don't know what this theta 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 going to call this one theta a, and I'm going to call this one theta b. Because we have the components, we can just use inverse tangent to find those. So basically, what happens here, I'm so I'm going to, I'm going to, sort of, write this sign. Don't know what I'm going to stick in here just yet, but I'm going to have to figure out this theta. So this theta 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 going to do here is I'm going to subtract from 180. I'm going to subtract these two angles, which I can find, and that will just give me whatever's left over and that's the angle between the two vectors. Okay? So theta a is just equal to inverse tangent of, this is going to be 3 divided by 4, right? So that's y over x. If we work this out, you're going to get 36.9 degrees. We've seen that angle before. This is the special angle when you have a 3, 4, 5 triangle.

Theta b is going to be the inverse tangent, and we're going to use positive values. Otherwise, you're going to get a weird negative number. We're going to use 3 and 2. So this is going to be the y component over the x component, and what you'll get here is you'll get 56.3 degrees. Alright? So basically, this is 36.9 and this is 56.3. So what I'm going to do here is my theta value, this theta is going to be 180 minus 36.9 minus 56.3, and what you're going to get here is you're going to get that this angle here is 86.8 degrees. So that's what we plug into our equation. So this is going to be the sin of 86.8. Alright. All of this is stuff is stuff that we've seen before. When you work this out, what you're going to get here, is you're going to get, 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 we just need to figure out which direction it points, and to do that, we're going to use the right-hand rule. So remember, we're going to point our fingers along A, so we're going to go like this, right, and then you're going to curl towards the B vector. So in other words, 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 going to be pointing out of the Page. So the direction is 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 that's it for part a. Alright. So that's the magnitude. We've got 18 in the k hat components. So here's what's going on. Now, what we're going to do is we're going to use the same technique. We're going to figure out what this c vector is, but now we're just going to use the other method to do it, which is building out the table of components and then doing the ab ba thing. And we're going to calculate the magnitude of c. And hopefully, what 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 build our components of our a and b vectors. So I'm just going to draw this real quick here. So I've got, let's see. It's a little more space. So this is a and b and then I need 1, 2, 3, 4. So this is going to be xyzxy, right? You're going to repeat them, and then we're just going to extract what each one of these numbers is from these vectors. So what I've got here is that 4i and 3j. So in other words, this is going to be 4 and 3. There's no z components, and then 4 and 3. And then the other one's going to be negative 23. Right? So we've got negative 23. So this is just going to be negative 23, 0, negative 23. Just repeat it. Alright.

So now the next step here is we're going to write ABBA for each component. So in other words, I'm going to do CX, CY, and CZ. So this is going to be ab minus ba, ab minus ba, ab minus ba. Okay. So now what we're going to do here is we're going to go ahead and start crossing things out. Right? So remember the CX components, we're going to cross out that one first, and we're going to do down and then up. We're going to cross the unlike multiply the unlike components. In other words, are we going to get YZ and then YZ. So what you get here when you multiply this out is you're going to get 3 times 0 minus 3 times 0. So what happens here, both of these things will go away and your x component will just be 0. Alright. So that's the first one. So now the next one, remember, we're going to do this the y component. We're going to cross that out and we're going to do down minus up. So now this is going to be zx, zx. In other words, this is going to be 0 times 4 minus 0 times negative 2. What happens is both of these things again go away and you're just going to get 0 for the y components. And now finally, you're just going to do the z components. You cross that out and then you're going to do down. You're going to cross down and then minus up. So this is going to be axby minus baxay. So in other words, what you're going to get here is you're going to get 4 times 3 minus, and this is going to be negative 2 times 3. Alright. When you work this out, you're going to get 12 over here, and this is going to be negative six. You're going to 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 ab sin theta, figuring all the angles, all the magnitudes, and things like that, we actually just got a magnitude of 18, and using the right-hand rule, we figured out it points directly out towards you, out of the page. And that's exactly what we got here by using the unit vector component method.

So that's it for this one, guys. Let me know if you have any questions.

Here’s what students ask on this topic:

How do you calculate the cross product of two vectors using their components?

To calculate the cross product of two vectors $A$ and $B$ using their components, follow these steps:

1. Create a table of the x, y, and z components for both vectors.

2. Use the following formulas to find each component of the resulting vector $C$:

$C_x=A_y{B}_z-A_z{B}_y$

$C_y=A_z{B}_x-A_x{B}_z$

$C_z=A_x{B}_y-A_y{B}_x$

3. Combine these components to form the resulting vector $C$ in unit vector notation.

What is the formula for the cross product of two vectors in component form?

The formula for the cross product of two vectors $A$ and $B$ in component form is:

$A\times B=(A_y{B}_z-A_z{B}_y)i+(A_z{B}_x-A_x{B}_z)j+(A_x{B}_y-A_y{B}_x)k$

Why can't you use the formula $ABsin\theta$ to calculate the cross product in terms of components?

The formula $ABsin\theta$ calculates the magnitude of the cross product but does not provide the direction or the components of the resulting vector. When vectors are given in component form (i.e., in terms of $i$, $j$, and $k$), you need to use the component-wise method to find the exact vector result, including its direction and magnitude.

What are the steps to find the cross product of two vectors using the component method?

To find the cross product of two vectors using the component method, follow these steps:

1. List the x, y, and z components of both vectors.

2. Use the formulas:

$C_x=A_y{B}_z-A_z{B}_y$

$C_y=A_z{B}_x-A_x{B}_z$

$C_z=A_x{B}_y-A_y{B}_x$

3. Combine these components to form the resulting vector $C$ in unit vector notation.

What is the significance of the cross product in physics?

The cross product is significant in physics because it is used to determine a vector that is perpendicular to two given vectors. This is crucial in various applications such as calculating torque, angular momentum, and the magnetic force on a moving charge. The direction of the resulting vector follows the right-hand rule, and its magnitude is given by the product of the magnitudes of the original vectors and the sine of the angle between them.