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Video: Vector Dot Product

Tim Burnham

Understand how to perform dot product vector calculations and some examples of how to use dot products to find the measure of the angle between two vectors.

14:59

Video Transcript

There are three ways to multiply a vector. You can perform a scalar multiplication in which you multiply each component of the vector by a real number or a scalar. So for example three 𝑣, there we would just multiply each component in vector 𝑣 by the number three. You can also multiply vectors by vectors and there’s two different flavours of that: dot products and cross products. In this video, we are only gonna look at dot products, but we are gonna see some nice way to use them, and we’re also gonna try out a few examples.

Right! So we’ve got vector 𝑢, which is seven two, and vector 𝑣 which is three six, and we’ve been asked to find the dot product of vectors 𝑢 and 𝑣. So, this here is the notation that we use: the 𝑢 with a dot and then the 𝑣. The dot isn’t right down on the ground (on the line). It is sort of half way up, if you see what I mean, the space between 𝑢 and 𝑣, but that’s the notation then for dot product.

Now, when we calculate the dot product, we get an answer which is a scalar or just a real number. So let’s have a look at that. So, we’re just gonna write the vectors out in their component form. So it’s seven comma two dot three comma six. And to work this out, we can sum together the product of the horizontal components and the product of the vertical components. So first of all, we’re gonna do seven times three, and then we’re gonna add two times six.

Now that looks a little bit weird, seven point three plus two point six. Well, it doesn’t mean seven point three plus two point six. It means seven times three two times three, and the times is the dot product times. We don’t use the normal cross multiplication sign because that would confuse this sort of multiplication with vector cross multiplication. So, sorry for the confusion here but that’s just a piece of notation that you’re gonna have to get used to cause that’s what we use for dot products. So when you see this dot, and raised up off the ground a little bit, it means we’re multiplying those two things together. So seven times three is twenty-one two times six is twelve. When we add those together, we get thirty-three. So as we said, the dot product of two vectors, we just multiply each corresponding component together and then add up all those results, and the answer that we get is just a number, a scalar. So let’s summarise that then. The dot product of two vectors is the sum of the products of the corresponding components, so we multiply the 𝑥-components together and then we added the product of the 𝑦-components in this case.

So we were working with a two-dimensional example, but that can also be scaled up to three- or four- or any dimensional that we like. Now obviously vectors 𝑢 and 𝑣 have to have the same number of dimensions as each other but, you know, we can have one two three four as many as we like here up to 𝑛 dimensions in each, and all we’re gonna do is multiply together the corresponding components. So that’s 𝑢 one times 𝑣 one, and that’s just a scalar, a real number result. Then, 𝑢 two times 𝑣 two. Then, we’re gonna add 𝑢 three times 𝑣 three and all of the others 𝑢 four 𝑣 four 𝑢, five 𝑣 five and so on all the way up to 𝑢 𝑛 times 𝑣 𝑛. And so, we’re just multiplying a number by a number, a number by a number, a number by number, and so on and adding them all together. The answer we’re gonna get is just a real number or a scalar. So, that’s the process expanded up to 𝑛 dimensions, as many dimensions as we like.

Now, while we’re here, let’s just quickly revisit magnitudes of vectors and just bring something else into place. So, here, we’ve got vector 𝐴𝐵 which has got an 𝑥-component of five and a 𝑦-component of twelve. Now, you should remember that if we wanted to work out the magnitude of vector 𝐴𝐵, what we need to do is use a bit of Pythagorean theorem and work out the square root of five squared plus twelve squared, which, for those of you who are curious about it, the answer comes out to be thirteen. But that’s kind of irrelevant for what I’m gonna say. I just wanna take a quick look inside this square root here, inside this radical here, and look at that expression that we’ve got. Now, just imagine that we did vector 𝐴𝐵 dot vector 𝐴𝐵. That will be five twelve dot five twelve. So, we’re gonna just multiply the component — the corresponding components together and add those results. So, that’s gonna be five times five add twelve times twelve, which gives us five squared plus twelve squared. So, actually what we’ve got here is the same as the thing that we’ve got here, so what’s inside that square root there is 𝐴𝐵 dot 𝐴𝐵. So the magnitude of vector 𝐴𝐵 is just the square root of 𝐴𝐵 dot 𝐴𝐵, and we’ve done that on a two-dimensional example, but that also works on a three-dimensional, four-dimensional, as many dimensions as we like. So that’s quite a nice simple useful result that we can use. So, all it means is we’re taking the square of each component and then adding them all together and then putting that inside the square root.

Okay, so now we’ve met dot products and we’ve tried one out, and — don’t worry we’re gonna practise some little bit more in a moment. Here is another useful result that will help us to find the angle between two vectors. If we’ve got two vectors 𝑢 and 𝑣, they could be two-dimensional, could be three dimensional, could be any dimensional, as we said before, cos of the angle between them is the vector 𝑢 divided by the magnitude of 𝑢. So that’s the unit vector in the direction of vector 𝑢 dot the unit vector of 𝑣 in the direction of vector 𝑣. And so, 𝜃, obviously as we said, is the angle between those two vectors, so this is a great result; we can rearrange that. So obviously if cos 𝜃 is equal to all that stuff, then 𝜃 is equal to cos to the minus one of all of that stuff. So that’s just another slight rearrangement of the formula. So if we find the dot product of the unit vectors in the directions of 𝑢 and 𝑣, it gives us the cosine of the angle between them, and from there we can find the angle between them. So let’s take a look at a couple of examples of that in action.

Right! so we’re given a vector 𝑢 is four one and vector 𝑣 and its component form is two five. We’re just gonna do two things: 1) find the dot product to those two vectors, and 2) find the angle between them. So first, let’s just quickly do a sketch of that situation. So, I’ve got vector 𝑢 is four one vector 𝑣 is two five, looks roughly like that, and what we’re also trying to find is this angle 𝜃, which is the angle between the two vectors. Okay! So let’s go ahead and do that. So, to find the dot product of the two vectors, remember we’re just gonna multiply the 𝑥-components together and add that to the product of the 𝑦-components. So, that’s four times two and one times five, which is eight plus five, which is equal to thirteen. So, that part of the question was pretty quick. The dot product of 𝑢 and 𝑣 is thirteen.

Now to find the angle between them, we know that the cos of the angle between them is the dot product of the unit vectors in the direction 𝑢 and 𝑣. And to find the unit vectors, we take vector 𝑢 and we divide it by the magnitude of itself, and we take the vector 𝑣 and we divide it by the magnitude of itself. So let’s write some numbers in. So, for vector 𝑢, which was four one, to work out the magnitude of that, I’m gonna do the square root of four squared plus one squared. So take each of those components, square them, add the result together, take the square root of that. So it’s basically our Pythagorean theorem thing that we were doing, and let’s do the same again for 𝑣. And because the components of 𝑣 were two and five, it’s just one over two squared plus five squared lots of vector 𝑣. So all this means is we’ve got each component of 𝑢 divided by the magnitude, and then we’ll have each component of 𝑣 divided by its magnitude as well. So four squared plus one squared; that’s sixteen plus one. So we got root seventeen for that one. And then we’re just dividing the 𝑥-component by root seventeen and the 𝑦-component by root seventeen. So let’s do the same for 𝑣, two squared is four, five squared is twenty-five, so we got one over root twenty-nine. And dividing each of the 𝑥- and the 𝑦-components by root twenty-nine, that vector now becomes two over root twenty-nine, five over root twenty-nine. So doing the dot product, we’re gonna be multiplying this term by this term, and we’re gonna be multiplying this term by this term and adding the results together. So, cos 𝜃 then is equal to four over root seventeen times two over root twenty-nine plus one over root seventeen times five over root twenty-nine. And, four times two is eight, and seventeen times twenty-nine is four hundred and ninety-three. So we got eight over root four nine three. One lot of five is five, so that’s five over root four nine three. Eight and five makes thirteen, so cos 𝜃 is equal to thirteen over root four nine three. So we now just need to do the cos inverse of that. So, cos to the minus one of thirteen over root four nine three which is fifty-four point two to one decimal place, so that’s our answer. Well, that’s our answer in degrees to one decimal place so this angle here is fifty four point two degrees to one decimal place.

So, that whole kinda part b there hinges on the fact that we the cosine of the angle between the vectors is equal to the dot product of the unit vectors in the directions of 𝑢 and 𝑣. And to work out a unit vector, you take your vector and divide it by the length of itself so it has a length of one. We followed that through, did the inverse cos, and we got our final answer.

Okay, one more example then. Find the angle between vectors 𝑢 which is three negative two and 𝑣 which is negative five negative three. So let’s just do a quick sketch to see what this looks like.

So, 𝑢 is three negative two and 𝑣 is negative five negative three, and the angle between them is gonna be this angle here hopefully. Imma call that 𝜃, okay? So, we just — we don’t need to work out the dot product of 𝑢 and 𝑣 to start off with. That was just in the previous question. All this is asking for is for us to use the result that cos 𝜃 is equal to the dot product of the unit vectors in the directions of 𝑢 and 𝑣. So in fact, I’m gonna rearrange that and do 𝜃 equals cos to the minus one of all that, so let’s just write that down. It’s always a good idea to write down the result that you’re starting off with so the thing is gonna drive your calculation. Right! Let’s do a bit of substituting in now. So we know how to work out the magnitude of 𝑢; it’s just gonna be the square root of three squared times negative two squared, and we can let’s just write that bit out now. And the same again for vector 𝑣. So you can already see this starts to look a little bit clumsy because I’ve rearranged this into the 𝜃 equals cos to the minus one. I’m writing this — all these big brackets out every time, so you can kinda make your decision about whether you use this form straight away or whether you use the previous form that I just used and then do your conversion at the end take cos to the minus one. But let’s plow on with the exercise anyway, so let’s evaluate each of those terms. Okay, so three squared plus negative two squared is thirteen, so the 𝑢 vector becomes three over root thirteen negative two over root thirteen. And then moving on to the 𝑣 vector, five squared plus three squared gives us thirty-four. So the 𝑣 components are minus five over root thirty-four and minus three over root thirty-four. So we’ve just gotta do the dot product of those. So that’s gonna be this component times this component plus this component times this component, which gives us this. And now, I just got to — I’ve got three over root thirteen times negative five over root thirty-four, and I’ve got negative two times negative three over root thirteen times root thirty-four. So, that gives us negative fifteen over root four four two plus six over root four four two. Add those two together, and I’ve got minus nine over root four four two. So if I get my calculator out and do the inverse cos of nine over root four four two, that gives me an angle of a hundred and fifteen point three degrees to one decimal place. So there it is! That’s my answer. Now, let’s just quickly check back to the diagram just to make sure that it all makes sense. And it has given me an angle of a hundred and fifteen degrees, which fits in with this angle. It hasn’t given me this angle running outside that. Sometimes you just need to check exactly which angle it’s calculated for you, so hopefully that’s nice and clear on that.

Okay, so let’s just do a quick summary of what we’ve looked at in this video. The dot product of two vectors: we just take each of the corresponding components of the vectors and multiply them together, so 𝑢 dot 𝑣. We got 𝑢 one 𝑣 one plus 𝑢 two 𝑣 two plus 𝑢 three 𝑣 three all the way up to 𝑢 𝑛 𝑣 𝑛, so that’s a simple case as we’ve seen in our example if we got a two-dimensional example. But it can be scaled up to any dimensions that we want. And the result is just a number, a real number, a scalar. And also the dot product of unit vectors in the directions of the two vectors gives us the cosine of the angle between them, which enables us to work out what the angle between two vectors is. Very very useful. Okay, that’s it for now.