WEBVTT
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In this video, weβre gonna look at what happens when you add or subtract vectors to or from each other.
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We know that vectors can be represented by line segments with a specific length and direction, and using these will really help us to visualise vector addition and subtraction.
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The addition process amounts to laying vectors end-to-end and calculating the components of the resultant vector by examining the difference between the coordinates of the initial point of the first vector and the terminal point of the last vector in the chain.
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So letβs see an example of that.
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Weβve got π which is the vector three, two and π which is the vector four, negative one.
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And we want to add those two vectors together.
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So just quickly sketching those out, here weβve got vector π which has an π₯-component of positive three and a π¦-component of positive two β so weβve gone three in that direction, two in that direction β and π, which has an π₯-component of four and a π¦-component of negative one.
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So if we add these two vectors, itβs like laying them end to end and then working out how we get from the beginning of the first vector to the end of the last vector.
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So to get from here to here, weβve taken a journey of three in the π₯-direction here and another four in the π₯-direction here.
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So to find the π₯-component of the resultant vector, π plus π, we simply add the π₯-component of π to the π₯-component of π.
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And now letβs visit the π¦-components.
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So to get from here to here again, weβve done our positive two to take us up to the top of the diagram but then we came back one.
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So weβve got to add those two components together.
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So thatβs two plus negative one.
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And three and four makes seven, so the resultant π₯-component is seven.
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And two add negative one is one, so the resultant π¦-component is one.
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So addition of vectors is just a case of adding the π₯-components and adding the π¦-components.
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Now letβs add three vectors together.
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So weβve got vector π, which is four minus two; π is one, six; and π is negative five, negative four.
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And weβve got to add all three of those vectors together.
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So first letβs just sketch out a diagram.
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Well hereβs vector π going positive four in the π₯-direction and negative two in the π¦-direction.
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Thereβs π going positive one in the π₯-direction and positive six in the π¦-direction.
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And lastly we lay π onto the end of π, and weβve got negative five in the π₯-direction and negative four in the π¦-direction.
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And we can see that when weβve added π, π, and π, the initial point of the first vector π is in exactly the same place as the terminal point of the last vector π.
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So although weβve gone around the houses, we started off here and weβve ended up in exactly the same place here.
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Okay letβs add those π₯-components together.
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So weβve gone four and then another one, and then weβve taken away five.
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And in the π¦-direction, we went negative two then added six and then finally took away four.
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And four add one take away five is zero, and negative two add six take away four is also zero.
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So the resultant vector by doing the adding up of the components is zero, zero.
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And that tallies with what we knew at the beginning: that we started off and we ended in exactly the same position.
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So the resultant vector, we havenβt moved in the π₯-direction we havenβt moved in the π¦-direction; our initial point and our terminal point are coincident.
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Having looked at some basic addition of vectors, letβs take a quick look at repeated addition of vectors, or as itβs more properly called multiplying vectors by scalars.
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So letβs start up with π is equal to the vector two, three.
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Letβs say we wanted to find out the what the value of two π was.
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Well two π is just π plus π.
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And as weβve just seen, to add two vectors together you just add the components together, so two plus two and three plus three.
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So yep weβre labouring the point here a little bit.
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But that means, weβve got two lots of the π₯-component and two lots of the π¦-component, so this two here.
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Weβve multiplied each of the components by two, which gives us the vector four, six.
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Okay letβs think about three π now, so thatβs π plus π plus π.
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Well Iβm just gonna jump straight to the stage.
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When weβre multiplying the components, because itβs three π weβre multiplying the π₯-component by three and the π¦-component by three, which gives us six, nine.
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And letβs just look at one more example: a fractional scalar, so a quarter.
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So a quarter of π, weβre just gonna multiply each component by a quarter.
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So thatβs a quarter of two and a quarter of three.
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So that gives us a half and three-quarters as our components.
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So just to summarise that, the process of multiplying a vector by a scalar, we take the scalar and we just multiply each component of the vector by that scalar to get our components of the resultant vector.
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So letβs start off with the same vector π.
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And now letβs have a look at negative scalar multiplication, works in exactly the same way.
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So letβs just see a couple of examples.
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Okay weβre trying to find negative π.
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That just means minus one times π, so weβre gonna take the scalar negative one and weβre gonna multiply both components by negative one.
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And that gives us an answer of negative two and negative three.
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And okay, letβs look at negative two π.
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So this time the scalar is negative two and weβre gonna multiply each of our π₯- and π¦-components by negative two.
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And that gives us an answer of negative four, negative six.
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Now itβs just worth taking a moment to notice π and negative π: the components have the same numbers, two- two and two for the π₯-components and three and three for the π¦-components.
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But the in the negative π, the signs are the opposite signs.
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In fact, what weβre doing is weβre going exactly the opposite direction.
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So vector π, weβre going positive two in the π₯-direction and positive three in the π¦-direction; thatβs this journey here.
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Negative π, weβre going negative two in the π₯-direction and negative three in the π¦-direction.
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So weβve done the exact reverse journey in the neg- in the negative direction, in the opposite direction.
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So positive π and negative π are the same length of line, the same angle of vector, but theyβre just going in the opposite direction.
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Okay letβs look at a quick example or two of subtracting some vectors.
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So weβve got vector π is two, three and vector π is four, negative one.
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And we wanna calculate the resultant of vector π take away vector π.
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Well thereβs different ways, you could just take the π₯-components of π and subtract the π₯-components of π and the π¦-components of π subtract the π¦-component of π to get your resultant.
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But I just like to suggest that we look at it in this slightly different way so π take away π is the same as π add the negative of π.
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So when we write that out, weβve got two plus the negative of four and we got three plus the negative of negative one, which looks a bit on odd on the page.
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But there it is.
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So two add negative four is negative two, and three add the negative of negative one, which is positive one, makes four.
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Now this business of converting π minus π into π plus the negative of π seems like weβre overcomplicated things.
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But letβs just take a look at it visually first and see how that pans out.
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So weβve got here vector π, which Iβm going across two and up three.
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So if I was adding π, Iβd be going right four and down one, but Iβm adding the negative of π.
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So adding the negative of π takes me from the end of π left four, so weβre going four in that direction.
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And the negative of negative one, weβre going up one in that direction.
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So this idea of laying vectors end to end, thinking of taking away π as adding negative π, means we can just lay down the negative vector to the end of π and we can then look for this resultant vector, going from the beginning of π to the end of π here.
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And when we do that, we were going negative two in this direction and up here positive four in that direction.
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So when subtracting vectors, we can just subtract the components if you want to or we can add the negative of the components for the second vector, which helps us to visualise the vector diagram a bit more easily.
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Letβs just take a look at one more example.
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So weβre gonna start off with the same π and π vectors.
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But this time weβre gonna calculate π take away π, which is the same as π add the negative of π, which gives us four add negative two and negative one add negative three.
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So thatβs a resultant vector of two, negative four.
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Well letβs look at the diagram again.
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Well first of all, vector π, weβre going positive four in the π₯-direction and negative one in the π¦-direction and then adding the negative of π, thatβs negative two in the π₯-direction, so weβre coming back, negative two.
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And negative of three so weβre gonna go down three β one, two, three β in the π¦-direction.
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So we start off at the initial point of π; we add the negative of π; and we end up at the terminal point of π.
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So our resulting journey is this one here.
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And in completing that journey, we went positive four in the π₯-direction but then we came back two, which gave us an π₯-component of two.
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So this distance here is two.
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And we went down one and then we went down another three, which gave us our π¦-component of negative four, which means weβve come down here four.
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So this vector here is π take away π, π subtract π, or π plus negative π.
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So just putting those two results side by side, π take away π was negative two four, and π take away π was two negative four.
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So they were both in the same position here, this was the position.
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And this was the length of the vector, but one was going in this direction and the other was going in that direction.
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So theyβre the same vector, but in opposite directions; one vector is the negative of the other.
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So vector π minus π is the negative of vector π minus π.
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And that makes sense because negative of π is negative π and the negative of negative π is positive π.
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So algebraically, they mean the same thing; and in vector format, they mean the same thing as well.
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Now letβs look at one last problem.
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Weβve got a regular hexagon π΄π΅πΆπ·πΈπΉ and πΊ is the midpoint of that and we have to express π΄πΈ in terms of vectors π’ and π£.
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So the vector π£ is from πΊ to πΆ and the vector π’ is from π· to πΆ.
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Now because this is a regular hexagon, we know that a number of these things are parallel.
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So π΄π΅ and πΈπ· and πΉπΊ and πΊπΆ are all parallel; π΄πΉ and π΅πΊ and πΊπΈ and πΆπ· are all parallel; and πΈπΉ, π·πΊ, πΊπ΄, and πΆπ΅ are all parallel.
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So we know for example that vector π’ runs from π· to πΆ, or we can put in vector π’ in some various different places in there as well.
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So those distances are parallel, but theyβre also the same length.
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So they- we can pick the vector π’ up and place them in each of those locations.
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And likewise, vector π£, running from πΊ to πΆ, that will also be vector π£; that will also be vector π£; and that will also be vector π£.
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So we got a few gaps on our hexagon here.
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How would I get for example from πΊ to π· along this vector here?
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Well I could go the straight-line route, but that doesnβt tell me anything in terms of π’ and π£.
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So I could also go by this other route; I could go along from πΊ to πΆ, which is the vector π£, and I could go from πΆ to π·, which is the opposite way to π’, so itβs a negative π’.
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So vector πΊπ· as we said is πΊπΆ plus πΆπ·, which is π£ plus the negative of π’.
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In other words, π£ take away π’.
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So letβs draw that in on the diagram then: πΊπ· is π£ take away π’.
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And likewise, πΉπΈ is parallel and the same length, so that is also π£ minus π’; AπΊ is too; and so is π΅πΆ.
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So when youβre trying to summarise the journey from π΄ to πΈ in terms of π’ and π£, so all of these journeys between individual points on our hexagon are already in terms of π’ and π£, so we just need to pick a convenient route.
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So letβs go along here, which is a negative π’ β itβs the opposite direction to a π’ β and then down here, which is π£ minus π’.
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We better tidy those up.
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So when we write that out, weβve got π’ plus π£ minus π’.
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So when we write that out, weβve got a negative π’ plus π£ minus π’.
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And it doesnβt matter what route we took; however convoluted, wouldβve still come up with that same answer for π΄πΈ.