Learn about The dot product

Khan Academy Presents: Introduction to the vector dot product.

Learn about The dot product Let’s learn a little bit about the Dot product. The Dot product frankly out of the two ways of multiplying vectors, I think it's the easier one. So, what is the Dot product do? One I’ll give you definition and then I’ll give you the intuition. So, if I have two vectors. Let say vector A and vector B. That’s how I draw my arrows and I can draw my arrows like that. That is equal to the magnitude of vector A times the magnitude of vector B times the cosine of the angle between them. Now where this is cosine come from? This might seem a little arbitrarily but I think with the visual explanation, I’ll make a little bit more sense. So, let me draw arbitrarily these two vectors. So, that is my vector A, nice big and fat vector. It’s good for showing the point. And let me draw of vector b like that and then let me draw the cosine or let me start angle between them and this is θ. So, there is two way of viewing this. You can view it and let me label them. This is vector A. I’m trying to be color consistent and this is vector B. So, there is two ways of viewing this product and you can view it as vector A. Because multiplication associative, you can twist the order. So, this could also be written as the magnitude to the vector A times cosine θ times and now I’ll do it in color appropriate times vector B. And these times, this is the Dot product. I mean you almost have to write it. This is regular multiplication because these are all scalar quantities. When you see the dot between the vectors you are talking about the vector of that product. So, if we are just rearrange this expression this way and what is it mean? What is a cosine θ? Let me ask you a question, if I were to draw A right angle right here perpendicular to B. So, I just drop right angle there cosine θ, SOCATOA. SOCA is equal to adjacent of a hypotenuse, right. Well let see adjacent is equal to this and the hypotenuse is equal to the magnitude of A, right. So, let me rewrite that. So, cosine θ and all of this applies to the A vector and cosine θ of this angle is equal to adjacent which is, I don’t what you can call this. Let’s call this the projection of A on to B, right. It’s like if you where to shine a light perpendicular to B. If there was a light show what where straight down, it would be the shadow of A on to B. And you could almost think of it is the part of the A that goes in the same direction of B. So, this projection they call it and at least the way I get intuition of what a projection is I can if this shadow, if you had a light source that came up perpendicular and what would be the shadow of that factor on to this one. So, if you think about it this shadow right here. We can call that the projection of A on to B or I don’t know, let just call it A sub B and it's magnitude of it, right. It’s how much of vector a goes on vector B over adjacent side over the hypotenuse. The hypotenuse is just the magnitude of vector A, right. This is start on basic calculus. Or another way that you can view it, just multiply both sides by the magnitude of vector A. And you get the projection of a on to which is just a fancy way of saying that this side, the part of a that goes in the same direction as B is another way to say it is equal to, just multiplying both sides times the magnitude of A. Is equal to the magnitude of A cosine θ which is exactly of what we have appeared in the definition of the Dot product. So, another way of visualizing the Dot product is you can replace this term with the magnitude of the projection of A on to B which is just this times the magnitude of B. That’s interesting. So, what are the Dot product of two vectors says? It says, “Well, just take one vector.” Let’s figure out how much of that vector. What component of its magnitude goes in the same direction as the other vector and let’s just multiply them. And where is that useful? Well, think about it. What about work when we learn in physics. Work is force times the distance but it’s not just the total force times the total distance. It's’ the force in the same direction as the distance. So, if I have a, you should review the physics play list if you’re watching this within the calculus play list. But let say I have a, it’s a 10 Newton object that sitting on ice. So, there is no friction and we don’t worry about friction right now. And let say I pull on it. Let say it my force vector, this is my force vector. Let say my force vector is 100 Newton’s. I’m making a numbers up. And now as I pull it, I slide it to the right. So, my distance vector is 10 meters. My distance vector is 10 meters fell to the ground. And the angle between them is equal to 60 degrees which is the same thing as pie over three. Well, let see the degrees is little even more in too depth. It is 60 degrees. The distance right here is 10 meter. So, my question is by pulling on this rope or whatever at this angle, at the 60 degree angle with a force of 100 Newton’s in pulling this block to the right for 10 meters. How much work am I doing? Well, work is force times the distance but not just the total force, the magnitude to the force in the direction of the distance. So, what is the magnitude of the force to the direction of the distance? It would be the component. It would be the horizontal component of this force vector, right. So, it would be 100 Newton’s times the cosine of 60 degrees. And I tell how much of that 100 Newton’s goes to the right or in other way, you can view it as this is the force vector and this down here is the distance vector. And you could say that the total work you performed is equal to the force vector that the distance vector. So, time using the Dot product and the Dot product to the force and the distance vector. And we know that the definition is the magnitude of the force vector which is 100 Newton’s times the magnitude of the distance vector which is 10 meters times the cosine of the angle between them. The cosine of the angle is 60 degrees. So, that’s equal to 1000 Newton-meters times cosine of 60 degrees is what? Its square root of three over two, right. The cosine of square root of three over two if I remember it correctly, so times the square root of three over two. So, the two is becomes 500. So, it becomes 500 square roots of three—. Whatever that is minus 700 something, I’m guessing, it maybe 800 something. I’m not quite sure but the important thing to realize is that the Dot product is useful. It applies to work. It actually calculates what component of what vector goes into the other direction. And you can interpret it the other. And you can say this is the magnitude of AB cosine θ. And that’s completely valid. And what is B cosine θ? Well, if you took B cosine θ and you could work this out as an exercise for yourself. That’s the magnitude of the B vector that is going in the A direction. So, it doesn’t matter what order you go. So, it contradict you know, and then when we take the Cross product it matters what do you do? A cross B or B cross A. When you doing the Dot product, it doesn’t matter on what order so B cosine θ would be the magnitude of vector B that goes in the direction of A. So, if you want to draw a perpendicular line here, B cosine θ would this vector. The magnitude of b is cosine θ so that you could say how much of vector B goes at the same direction as A and then multiply the two magnitudes. Or you could say how much of vector a goes in the same direction as vector B and then multiply the two magnitudes. And now this is I think a good time to just make sure you understand the difference between the Dot product and the Cross product. The Dot product ends up with just a number. You multiply two vectors and all you have is a number. You don’t end up with a scalar quantity. And why is that interesting? What it tells you? It just tell how much do this, you can almost said this vectors reinforce each other because you’re taking their magnitudes. The parts of their magnitudes are going on the same direction and multiplying them. The Cross product is actually almost the opposite. You’re taking the orthogonal components, right. The difference was this was a sine θ. And I don’t want to mess you this picture too much but you should review the Cross product videos. And I’ll do another video where you compared and contrast them. But the Cross product as your saying, let’s multiply the magnitude of the vectors that are perpendicular to each other. That isn’t going in the same direction. That they are actually orthogonal to each other. And then you have to pick a direction since you are not saying the same direction that they are going in. So, you are picking the direction that is orthogonal to both vectors. And that’s why the orientation matters and you have to take the right hand rule because there are actually two vectors that go, that are perpendicular to any other two vectors in three dimensions. Anyway I’m all out of time. I will continue this hopefully you are not too confusing discussion. In the next video I’ll compare and contrast the Cross product and that the Dot product. See you in the next video.