### Video Transcript

We already know that a vector is a set of numbers that can be represented in a suitable space by a line segment with a specific length and direction. Weβve also seen that a line segment has a magnitude and direction, which basically means that we can describe it by saying how long it is and in which direction itβs pointing. In this video, weβre gonna talk about horizontal and vertical components of two-dimensional vectors and introduce the π and π unit vector notation.

Any two-dimensional vector has two components. The first represents the amount of movement in the π₯-direction and the second, the amount of movement in the π¦-direction. Of course, when I say movement, Iβm just talking about differences in π₯- and π¦-coordinates in a graphical representation of the vector by a line segment on a graph. The vector itself might be representing something completely different, force or acceleration, for example. So in this case, the amount of π₯-movement is this distance here. So weβre going from an π₯-coordinate of three to an π₯-coordinate of six, so thatβs a difference of plus three. So our π₯-component of the vector is positive three. The π¦-component is this bit here. The π¦-coordinate of π΄ was two, the π¦-coordinate of π΅ was nine, so thatβs a difference of plus seven. And we can write that as π΄π΅, vector π΄π΅, in this format here: three, seven. So the three is the π₯-component and the seven is the π¦-component.

And if we had a three-dimensional coordinate system with π₯- and π¦- and π§-coordinates, we would just be inserting another number at the end here onto our vector. So we can extend this system to any number of dimensions.

We define two special vectors, π and π, to be positive one in the π₯-direction or positive one in the π¦-direction, respectively. So this is π, itβs just a movement of one in the π₯-direction. And hereβs π, a movement of one in the π¦-direction. So remember that the π and π vectors β are at one, zero and zero, one β can be placed anywhere on the graph. They donβt have to start at the origin. So there we are; I placed them somewhere else. Each vector is just describing a particular journey. In this particular case here, for π, weβre adding one to the π₯-coordinate and weβre leaving the π¦-coordinate as it is. So weβre doing this journey from here to here. In the π case, weβre not adding anything to the π₯-coordinate, but weβre adding one to the π¦-coordinate. It describes this movement from here up to here.

Now remember, we can also use the rules of adding and subtracting vectors to stack up these πs and πs, to create bigger vectors. So, for example, this vector here, π, represents a movement of one in the π₯-direction and none in the π¦-direction. If I increase that to this vector here, so this whole length here, that would be two πs following on from each other. Or, it will be a translation of two in the π₯-direction and zero in the π¦-direction. Now if I added on to that this vector here β which starts at the end here and then moves up not one, not two, but three β that would be three π vectors added on together, making a vector of three π or zero, three. So if I add two π plus three π, that represents this green journey from the start point to the end point up here. So two π, the π₯-component would be two. Three π, the π¦-component would be three. And the direct journey two π plus three π is the green line. So this means the π₯-component was two and the π¦-component was three. So weβve now got two different ways of representing this green vector here. We can use it in the standard vector notation that weβre familiar with. But weβve got this new notation here, in terms of π and π vectors. The number of steps in the π₯-direction is the πs and the number of steps in the π¦-direction is the πs.

So letβs sum that up in the general case. If we start off at point πΆ here with coordinates π₯ one, π¦ one and we end up at point π· here with coordinates π₯ two, π¦ two, then vector πΆπ· is, the π₯-component here is the difference in the π₯-coordinates and the π¦-component here is the difference in the π¦-coordinates. So thatβs our standard vector way of writing it. Or, we can say that that is π₯ two minus π₯ one lots of the π unit vector plus π¦ two minus π¦ one lots of the π unit vector. That make some kind of sense. All weβre saying is that the π₯-component of a vector is the coefficient of π, in this format, and the π¦-component of the vector is the coefficient of π, in this format. So weβve just got this new notation where we have the π vector, which is a step of one in the π₯-direction, the π vector, which is a step of one in the π¦-direction, and we just say how many of those weβre doing in each case to make up our vector.

Right. So now we know this new format. Letβs just have a look at a couple of quick questions that involve using πs and πs in our questions. So weβre going to do this question here. Weβve got π΄π΅; itβs the vector three π plus four π. ππ is the vector negative two π plus three π. And weβve just got to add those two vectors together.

So the first stage is just to take vector π΄π΅ and add vector ππ. And all we have to do is add the πs together first, and then add the πs together second. So three π add negative two π is just one π, so thatβs π. And four π add positive three π is seven π. So thereβs our answer, π plus seven π. Simple as that, add the π-components together, add the π-components together, look out for the negative signs and- when youβre doing those calculations; but otherwise, thatβs a pretty straightforward process.

So letβs just visualise that example. So we had three π plus four π. So weβll be going positive three π and then positive four π, this is the ππ vector. So there we are. Weβve just laid that down at the origin. We couldβve laid it anywhere on the-on the graph. Now adding vectors, we just lay them end-to-end. So what we were adding, π₯π¦, was negative two π plus three π. So weβre effectually starting π₯ from π΅, so π₯ lays on top of π΅ and weβre going negative two π. So weβre going two in the negative π₯-direction and weβre going positive three π, up to here.

So adding vectors is just a matter of laying them end-to-end on the graph. So we laid π΄π΅ down, which started here and ended here, and then we just added ππ to the end of that, laid that onto the end. So that started from where we just finished off and then ended up here. So the resultant vector is this green one here. And to get from the beginning of the green vector to the end of the green vector, we had to go positive one in the π₯-direction. So thatβs one π, or just π. And in the π¦-direction, weβre going up seven all the way up here. So thatβs plus seven π.

So when youβre doing these questions, it really is just a matter of adding the π₯-components together, adding the π¦-components together, and coming up with a simple answer. You donβt need to do all this graphical checking. But Iβm just hoping that thatβs giving you some extra insight into the process and why it works.

Right. Letβs take a look at one final question then. Weβve got vector π΄π΅, is three π take away four π. Vector ππ is negative four π add seven π. And weβve got to find vector π΄π΅ take away vector ππ.

So just writing that out, π΄π΅ is three π take away four π. ππ is negative four π plus seven π, and thatβs what weβre taking away from vector π΄π΅. So we just need to be really careful, when weβre taking these away, about the signs here because weβve got a negative outside the bracket, weβve got negatives and positive inside the bracket.

So letβs just start off with the π-components then. Iβve got three π and Iβm taking away negative four π, so that means Iβm adding four π. So three π add four π is seven π. And then for the π-components, Iβm starting off with negative four π and Iβm taking away positive seven π, so f- negative four take away another seven is negative eleven π. So thereβs our answer, seven π take away eleven π.

And just to summarise then what weβve learnt. π is the unit vector in the π₯-direction one, zero and π is the unit vector in the π¦-direction zero, one. And given any vector β like π΄π΅ is five, three β this here is the π₯-component, this here is the π¦-component. We can rewrite this as five π, because thatβs the number of π₯s, plus three π, because thatβs the number of π¦s. And given two vectors, π΄π΅ and πΆπ·, with their π- and π-components like this, in this format, we can add or subtract them just by adding or subtracting their π-components and their π-components, separately.

So, for example, adding π΄π΅ and πΆπ·, we can add the two and the negative two π and we can add the negative three and the four π. Well, in this case, two plus negative two π, thatβs zero πs, so we donβt need to bother writing zero π. And negative three plus four is just positive one, so we end up with an answer of just π, one π. And if we want to subtract the vectors π΄π΅ minus πΆπ·, weβve got two of the πs here take away negative two πs, and we had negative three take away four of the πs. Then two take away negative two is two add two, so thatβs four of the πs. And negative three take away another four is negative seven πs. So weβre adding negative seven π. So we probably not write plus negative π, we just write four π take away seven π.

So hopefully, youβll be comfortable using the π and π unit vectors just to represent the π₯- and the π¦-component of any vectors that you come across now.