Intro to Cross Product (Vector Product) - Online Tutor, Practice Problems & Exam Prep

Vectors can be multiplied using two methods: the scalar (dot) product and the vector (cross) product. The scalar product yields a number, calculated as $a^b\cos \left(\theta \right)$, while the vector product results in a new vector, calculated as $a^b\sin (\mathrm{\theta <mo>)</mo>}$. The direction of the vector product is determined using the right-hand rule, indicating its perpendicular orientation to the original vectors.

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

Hey, guys. So in a previous video, I covered that there were 2 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 going to be using this a lot sometimes in physics. So let's go ahead and check this out here, and we're going to do a bunch of examples.

Now before we begin, I just want to briefly recap dot products. So imagine I had these 2 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 3 and this 4 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 abcosθ. So, if we just do this here, a dot b was a b cos. So in other words, it was just the magnitude of A, which is 3, magnitude of B, which is 4 times the cosine of the angle between them, which is 60 degrees. What you get out of this is you just get the number 6. And what the 6 really just represents here is it just represents the multiplication of parallel components of A and B. So, for example, this B vector here has a component that lies in the same direction parallel to A, and if you work this out, you're going to get the B x component is 2. So if you multiply these two components here that are parallel, you end up just getting 6 out of it.

Alright. So that's the scalar product. Now we're going to do the same thing. We're going to multiply these using the vector product 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 going to get a new vector. That's why we call it the vector product, and this vector here is going to be called C. Alright? So with the scalar product, you multiply them, you just get a number. With the vector product, we're going to multiply these 2 vectors and we're going to get a third one out of it called C. Now 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 going to use to calculate the magnitude is going to be very similar to the dot product. We're going to use a×bsinθ. So before we used the cosine, but now we're going to use the sine. So it's the magnitude of A, magnitude of B times the sine of the angle, and this angle here θ represents the smallest of the angles between A and B. So in other words, there are 2 ways to represent this angle. You can go all the way around, but that's not the smallest angle, so we're just going to use the 60. Alright? So if we do the vector product of these two vectors, what you're going to get is a third vector, which is C, and the magnitude of the C vector is just going to be a b times sine of θ. So we're going to do the same thing we did over here. This is going to be 3 and 4, 3 times 4, except now we're going to do sine of 60. And if you work this out, what you're going to get here is 10.4. The magnitude of this vector is not going to be 6. It's going to be 10.4 because now we're using the sine. Okay? But this is a vector. So remember the vectors have both magnitude and direction. Where does this 10.4 vector point? And to do that, we're going to 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 going to 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. Alright? So here's how the right-hand rule works. What you're going to do is you're going to take your hand and you're going to point your fingers along the first vector, always. You're always going to point along the first vector which in our case is A. So we're going to point our fingers along A like this. Okay? Now what you want to do is you want to curl your fingers towards the second vector. So you're going to curl them towards B. So in other words, I'm going to curl my fingers up like this in the direction where B is. And once 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 going to take your hand, point at 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 an xy-plane is we just draw a little circle with a dot going through this. This symbol here, the circle with 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 you're curling, first to you're pointing your fingers along the first and then you're curling towards the second one.

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

Find the Vector Product

Hey, guys. So hopefully you got a chance to check this one out on your own. We have these two vectors on our diagram here, and we want to find the magnitude and direction of the cross product. We're going to have this new vector \( \mathbf{c} \), which is \( \mathbf{a} \times \mathbf{b} \). 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 \( AB \times \sin(\text{angle}) \). Right? So the angle here is between \( \mathbf{a} \) and \( \mathbf{b} \). The magnitude of \( \mathbf{a} \) is given as 12. The magnitude of \( \mathbf{b} \) is 8. Now we just need the sine of the angle between them. So, is it going to be this 30 degrees, or is it going to be something else? Well, this is where these three-dimensional diagrams can be tricky because they can mess up your perspective. So, this X-axis here is kind of back and forth. Right? So this XY plane is like the flat plane, and the Z-axis goes vertical. What happens is this \( \mathbf{a} \) vector might look like it's raised off the ground, but it's actually not. These dashed lines indicate that it's along the XY plane. So imagine that you have a tabletop like this, and imagine you have a vector pointing to the side except it's tilted away from you. So, that's what this \( \mathbf{a} \) vector is, and \( \mathbf{b} \) is pointing straight up. So, what is the angle between something that points horizontally and vertically? It's 90 degrees. It might not look like it, but this actually is, because this represents vertical and flat. The sine of 90 results in 1, and therefore the magnitude of our vector is going to be 96.

Now, how about the direction? For the direction, we're going to need to use the right-hand rule. The rule is for \( \mathbf{a} \times \mathbf{b} \), you always point your fingers towards \( \mathbf{a} \) and curl towards \( \mathbf{b} \). Take your right hand and point your fingers in the direction of \( \mathbf{a} \). It looks like I'm moving towards you, but you're going to do it yourself. Your fingers are pointing away from you and then you curl your fingers up vertically towards \( \mathbf{b} \). When you do that, your thumb will be pointing not straight at you but off to the right. This is our \( \mathbf{c} \) vector. It has a magnitude of 96 and it points here along this axis. This thumb, my \( \mathbf{c} \) vector, is still flat. It's still on the XY plane, just pointing in a different direction. We're going to use the same sort of dashed lines that I used for the other one to indicate that it's on this plane. We want to figure out what is the angle, or rather the positive angle, from the X-axis. So, what is this theta? The definition of this vector product is that it's perpendicular to both vectors that make it up. In other words, \( \mathbf{c} \) is perpendicular to the vertical, that's 90 degrees, and it's also perpendicular to the \( \mathbf{a} \) vector, that's also 90 degrees. If this is 30 degrees and this is 90 degrees, then the \( \mathbf{c} \) vector points off at 30 degrees as well. So, the answer is that \( \theta \) here is equal to 30 degrees as a positive angle from the X-axis. Let me know if you have any questions.

Finding Angle Using Scalar & Vector Products

Hey, guys. So let's get started with our problem here. So we've got these 2 vectors a and b, and we're told some information about their scalar and vector products, their 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 know 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. I'm just going to call it c. Is equal to absinθ, 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 going to end up with is that sinθ is equal to 12/ab. 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 magnitude of those vectors, I can’t really finish off this equation. I'm going to need something else. I'm going to need another equation. Well, the one thing that we haven't seen or that we haven't used yet is that the scalar product is negative 8. Remember the scalar product is a dot b. So what I'm going to do here is I'm going to start out another equation, and I'm going to say that a·b is equal to remember that the definition of a dot b, it doesn't have a magnitude. It's just a number. And remember, it was just abcosθ, except we already know what the scalar product is. We know that a·b, whenever you do abcosθ, you're going to get negative 8. Alright? So here's what I'm going to do. I'm going to rewrite another equation. I'm going to get cosθ. Right? Because this is my other this is my target variable just in another equation, and this is going to be negative 8 divided by AB. So here's what I'm going to do. Right? I've got these two equations and they actually both have 3 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 going to do here is I'm going to do sinθ divided by cosθ. And what you're going to get here is that 12 divided by AB divided by negative 8 over AB. And what happens here is that the ABs 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 8, so this is just going to work out to negative 1.5. So we've got these two equations that had 3 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. So you're going to take the inverse tangent of negative 1.5. What you're going to get here is you're going to get negative 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 going to have to add 180 degrees to it. One way you can kind of think about this, is that these two vectors have a scalar product of negative 8. So what happens here is I'm just going to draw like a sketch real quickly of 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 antiparallel. So, basically, what happens is we know that the angle between A and B has to be greater than 90 degrees. So this theta here must be greater than 90 degrees because a dot b is negative. It’s less than 0. Alright? So now that we've added a 180 to this, what you're going to get is 123.7 degrees, and that is the right answer. Alright? So now we know that this 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 or that's a scalar product that's negative 8. So hopefully, that makes sense. Let me know if you guys if you guys have any questions, and I'll see you in the next one.

Here’s what students ask on this topic:

What is the difference between the dot product and the cross product of vectors?

The dot product (scalar product) of two vectors results in a scalar (a single number), calculated using the formula $ab\cos \theta$. The cross product (vector product), on the other hand, results in a new vector, calculated using the formula $ab\sin \theta$. The direction of the resulting vector from the cross product is perpendicular to the plane formed by the original vectors, determined using the right-hand rule.

How do you use the right-hand rule to determine the direction of the cross product?

To use the right-hand rule, follow these steps: 1) Point your fingers in the direction of the first vector (A). 2) Curl your fingers towards the direction of the second vector (B). 3) Your thumb will point in the direction of the resulting vector (C). If your thumb points out of the page, it is represented by a dot inside a circle. If it points into the page, it is represented by an 'X' inside a circle.

What is the formula for calculating the magnitude of the cross product of two vectors?

The magnitude of the cross product of two vectors A and B is given by the formula $\mathrm{|C|}=\mathrm{|A|}\mathrm{|B|}\sin \theta$, where $\mathrm{|A|}$ and $\mathrm{|B|}$ are the magnitudes of vectors A and B, and $\theta$ is the angle between them.

What happens to the cross product if the two vectors are parallel?

If two vectors are parallel, the angle between them is either 0 or 180 degrees. In both cases, the sine of the angle is zero ($\sin 0=0$ and $\sin 180=0$), making the magnitude of the cross product zero. Therefore, the cross product of two parallel vectors is a zero vector.

Can the cross product be used in three-dimensional space, and how is it represented?

Yes, the cross product can be used in three-dimensional space. It is represented by a new vector that is perpendicular to the plane formed by the original vectors. The direction of this vector is determined using the right-hand rule. In a 3D coordinate system, the cross product of vectors A and B can be calculated using the determinant of a matrix formed by the unit vectors i, j, k and the components of A and B.