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Video transcript

We've learned a good bit about the dot product. But when I first introduced it, I mentioned that this was only one type of vector multiplication, and the other type is the cross product, which you're probably familiar with from your vector calculus course or from your physics course. But the cross product is actually much more limited than the dot product. It's useful, but it's much more limited. The dot product is defined in any dimension. So this is defined for any two vectors that are in Rn. You could take the dot product of vectors that have two components. You could take the dot product of vectors that have a million components. The cross product is only defined in R3. And the other, I guess, major difference is the dot produc, and we're going to see this in a second when I define the dot product for you, I haven't defined it yet. The dot product results in a scalar. You take the dot product of two vectors, you just get a number. But in the cross product you're going to see that we're going to get another vector. And the vector we're going to get is actually going to be a vector that's orthogonal to the two vectors that we're taking the cross product of. So now that I have you excited with anticipation, let me define it for you. And you probably already have seen this once or twice in your mathematical careers. Let's say I have the vector a. It has to be in R3, so it only has three components: a1, a2 and a3. And I'm going to cross that with the vector b and it has three components: b1, b2 and b3. a cross b is defined as a third vector. And now this is going to seem a little bit bizarre and hard to essentially memorize because this is a definition. But I'll show you how I think about it when I have my vectors written in this column form. If you watch the physics playlist, I have a bunch of videos on the cross product and I show you how I think about the cross product when I have it in the i, j, k form. But when I have it like this, the way you think about this first term up here, this is going to be another three vector or another vector in R3, so it's going to have 1, 2, 3 terms. For the first term, what you do is you ignore these top two terms of this vector and then you look at the bottom two and you say, a2 times b3 minus a3 times b2. And I've made a few videos on determinants, although I haven't formally done them in kind of this linear algebra playlist yet. But if you remember kind of co-factor-- finding out the co-factor terms for when you're determining the determinant or if you're just taking the determinant for a 2x2 matrix, this might seem very familiar. So this first term right here is essentially the determinant of-- if you get rid of this first row out of both of these guys right here, you take a2 times b3 minus a3 times b2. So it's a2 times b3 minus a3 times b2. That was hopefully pretty straightforward. Now, not to make your life any more complicated, when you do the second, when you do the middle row, when you do this one right here, so you cross that out. And you might want to do a1 times b3 minus a3 times b1. And that would be natural because that's what we did up there. But the middle row you do the opposite. You do a3 times b1 minus a1 times b3. Or you can kind of view it as the negative of what you would have done naturally. So you would have done a1 b3 minus a3 b1. Now we're going to do a3 b1 minus a1 b3. And that was only for that middle row. And then, for the bottom row, we cross that out again or ignore it. And we do a1 times b2, just like we do with the first row. Times a2 b1. Or minus a2 b1. This seems all hard to-- and it is hard to remember. That's why I kind of have to get that system in place like I just talked to you about. But this might seem pretty bizarre and hairy. So let me do a couple of examples with you, just so you get the hang of our definition of the dot product in R3. So let's say that I have the vector-- let's say I'm crossing the vector. I have the vector 1, minus 7, and 1. And I'm going to cross that with the vector 5, 2, 4. So this is going to be equal to a third vector. Let me get some space to do my mathematics. So for the first element in this vector, the first component, we just ignore the first components of these vectors and we say minus 7 times 4 minus 1 times 2. And these are just regular multiplication. I'm not taking the dot product. These are just regular numbers. Then for the middle term, we ignore the middle terms here and then we do the opposite. We do 1 times 5 minus 1 times 4. Remember, you might have been tempted to do 1 times 4 minus 1 times 5 because that's how we essentially did it in the first term. But the middle term is the opposite. And then finally, the third term you ignore the third terms here and then you do it just like the first term. You start in the top left. 1 times 2 minus 7. Put that in parentheses. Minus minus 7 times 5. And so that is equal to-- let me see. What do we get? Minus 7 times 4. I don't want to make a careless mistake here. That's minus 28. Minus 2. So this is minus 30 for that first term. This one is 5 minus 4. 5 minus 4 is just 1. And then 2 minus minus 35. So 2 minus minus 35. That's 2 plus 35. That's 37. So there you go. Hopefully you understand at least the mechanics of the cross product. So the next thing you're saying well, OK. I can find the cross product of two things. But what is this good for? What does this do for me? And the answer is, is that this third vector right here, and depending on whether I stay in the abstract case or whether this case with numbers, this is orthogonal to the two vectors that we took the cross product of. So this vector right here is orthogonal to a and b. Which is pretty neat. If you just go think about the last video when we were talking about normal vectors to a plane, we can define a vector by-- we can define a plane by two vectors. If we define a plane-- let's say that I have vector a right there, and then I have vector b. Let me do vector b like this. Those define a plane in R3. Let me define your plane. So all the linear combinations of those two guys, that's a plane in R3. You can kind of view it as they might form a subspace in R3. That forms a plane. If you take a cross b, you get a third vector that's orthogonal to those two. And so a cross b will pop out like this. It'll be orthogonal to both of them and look like that. And so this vector right there is a cross b. And you might say, Sal, how did you know-- I mean, there's multiple vectors that are orthogonal. Obviously, the length of the vector, and I didn't specify that there, but it could pop straight up like that or why didn't it-- you know, you just as easily could have popped straight down like that. That also would be orthogonal to a and b. And the way that a cross b is defined, you can essentially figure out the direction visually by using what's called the right hand rule. And the way I think about it is you take your right hand and let me see if I can draw a suitable right hand. Point your index finger in the direction of a. So if your index finger is in the direction of a and then I point my middle finger in the direction of b. So my middle finger, in this case, is going to go something like that. My middle finger is going to do something like that. And then my other fingers do nothing. Then my thumb will go in the direction of a cross b. You could see that there. My thumb is in the direction of a cross b. And assuming that you are anatomically similar to me, then you still get the same result. Let me draw it all. So this is vector a. Vector b goes in that direction. Hopefully you don't have a thumb hanging down here. You know that a cross b in this example will point up and it's orthogonal to both. To kind of satisfy you a little bit, that the vector's definitely orthogonal or that this thing is definitely orthogonal to both of these, let's just play with it and see that that definitely is the case. And what is orthogonal? What is in our context, the definition of orthogonal? Orthogonal vectors. If a and b are orthogonal, that means that a dot b is equal to 0. Remember, the difference between orthogonal and perpendicular is that orthogonal also applies to 0 vectors. So these could also be 0 vectors. Notice that I didn't say that any of these guys up here had to be nonzero. Well, in a little bit, we'll talk about the angle between vectors and then you have to assume nonzero. But if you're just taking a cross product, nothing to stop you from taking-- no reason why any of these numbers can't be 0. But let me show you that a cross b is definitely orthogonal to both a and b. I think that might be somewhat satisfying to you. So let me copy a cross b here. I don't feel like rewriting it. OK. Let me paste it. OK, bundle up little other stuff with it. Let me take the dot product of that with just my vector a, which was just a1, a2, a3. So what does the dot product look like? That term times that, so it's a1-- let me get some space. It's a1 times a2 b3 minus a1 times that. Minus a1 times a3 b2. And then you have plus this times this. So plus a2 times a3. So plus a2 times a3 times b1. And then minus a2 a1 b3. And then finally, plus-- I'll just continue it down here. Plus a3 times a1 b2 minus a3 times a2 b1. All I did I just took the cross-- the dot product of these two things. I just took each of this. This guy times that was equal to those two terms. That guy times that was equal to the next two terms, equal to those two terms. And then this guy times that was equal to those two terms. And if these guys are really orthogonal, then this should be equal to 0. So let's see if that's the case. So I have an a1 a2 b3, a positive here, and then I'm subtracting the same thing here. This is the same thing as a1 a2 b3, but it's just a minus. So that will cancel out with that. Let's see, what else do we have? We have a minus a1 a3 b2. We have a plus a1 a3 b2 there, so these two are going to cancel out. And I think you see where this is going. You have a positive a2 a3 b1 and then you have a negative a2 a3 b1 there. So these will also cancel out. Now, I just showed you that it's orthogonal to a. Let me show you that it's orthogonal to b. Let me get another version of my-- the cross product of the two vectors. Probably scroll down a little bit. And let me go back. And let me multiply that times the vector b. b1, b2 and b3. I'll do it here just so I have some space. So b1 times this whole thing right here is b1 a2 b3 minus b1 times this. Minus b1 a3 b2. Let me switch colors. And then b2 times this thing here is going to be b2-- so it's going to be plus. This is all really one expression, I'm just writing it on multiple lines. This isn't a vector. Remember, when you take the dot product of two things, you get a scalar quantity. So plus b2 times this thing. So b2 a3 b1 minus b2 a1 b3. And then finally, b3 times this. So plus b3 a1 b2 minus b3 a2 b1. So if these guys are definitely orthogonal, then this thing needs to equal 0. And let's see if that's the case. We have a b1, a2, b3. So b1 and a b3. b1 a2 b3, that's a positive one and this is a negative one. You have a b3, an a2 and a b1 so that and that cancel out. Here you have a minus b1 a3 b2. So you have a b1 and a b2. It's a minus b1 a3 b2. This is a plus the same thing. Just switching the order of the multiplication. But these two are the same term. They're just opposites of each other, so they cancel out. And then finally, you have a b2, an a1 and a b3. It's a negative. And then you have a positive version of the same thing. So these two guys cancel out. So you see that this is also equal to 0. So hopefully you're satisfied that this vector right here is definitely orthogonal to both a and b. And that's because that's how it was designed. This is a definition. You could do a little bit of algebra and you could have without me explaining this definition to you, you could have actually come up with this definition on your own. But obviously this was kind of designed to have other interesting properties to it. And I'll cover those in the next few videos, so hopefully you found that helpful.