Video: Adding and Subtracting Vectors (More Complicated Questions) | Nagwa Video: Adding and Subtracting Vectors (More Complicated Questions) | Nagwa

# Video: Adding and Subtracting Vectors (More Complicated Questions)

Understand how to add, subtract, and scalar multiply vectors to solve vector problems, including showing that two vectors are parallel and that three points are collinear.

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### Video Transcript

In this video, weβre gonna take a look at a few vector addition and subtraction questions that you might attempt after youβve had a go at the more basic introductory ones. In this question, weβve got triangle ππ΄π΅ and weβre told that the point π is on the side π΄π΅ and that the distance π΄π is a third of the distance from π΄ to π΅.

Now we got to express vector π΄π΅ in terms of π and π, and weβve also got to express the vector ππ in terms of π and π. Now one of these is much easier than the other, but letβs have a look at the question.

Now if we think about that side π΄π΅, if π΄π is a third of the way from π΄ to π΅, then I can split that into three equal sized chunks. Now the first one from π΄ to π, thatβs vector π΄π, thatβs repeated on the other two because if itβs third the distance, then if I do the same thing three times, that will make the distance π΄π΅.

So here weβve laid the same vector in to in three times, and thatβs something that we will come back and use later on in the question. So the f- the first part of the question, express π΄π΅ in terms of π and π.

How am I gonna get from point π΄ to point π΅? Well I could go in a straight line from π΄ to π΅, but that doesnβt give me an answer in terms of vectors π and vector π. So the other way I can do it is go from π΄ to π and then from π to π΅. And in going from π΄ to π, Iβm going on backtracking along an π vector; Iβm going the opposite direction of an π vector; thatβs negative π.

And in going from π to π΅, thatβs just going a positive way along the π vector, so thatβs just π. So the vector π΄π΅ can be expressed as the negative of an π vector plus the positive of a π vector. Well there we go; thatβs the first part done.

Now for part b, Iβve got to find out ππ. So Iβve got- Iβve got this journey here from π to π, and Iβve got to express that in terms of π and π. Now the direct line from π to π doesnβt include any information about vectors π΄ and π΅, so Iβve got to try and think about how can I get from π to π via other routes which do have π and π in them.

So I could use this route from π to π΄ and then from π΄ to π. And we know that π to π΄ is just simply the-the- this vector here π and from π΄ to π is a third of the way from π΄ to π΅, and we just worked out what π΄ to π΅ was.

So letβs start writing that down. So ππ is equal to π plus a third π΄π΅ and we know that π΄π΅ is negative π plus π, so ππ is π plus a third of negative π plus π. And multiplying out the brackets, that means that ππ is equal to π minus a third of π plus a third of π, so our answer is that vector ππ is equal to two-thirds of vector π plus one-third of vector π.

Now in our next question, weβve got a quadrilateral π΄π΅πΆπ· which is a parallelogram. πΈ and πΉ are points on the diagonal from π· to π΅, and the distance from π· to πΈ is the same as the distance from πΈ to πΉ, which is the same as the distance from πΉ to π΅.

Weβre told that π· to πΆ is a vector called π₯ and πΆ to π΅ is a vector called π¦. What weβre gonna do is show that π΄πΉπΆπΈ is also a parallelogram. So looking on that diagonal there, Iβve just shown that those three links are the same, and Iβve drawn in the shape π΄πΉπΆπΈ, and we- basically to prove that thatβs a parallelogram, a parallelogram has got two pairs of parallel sides.

So weβve got to prove that the vector from π΄ to πΉ is parallel to the vector from πΈ to πΆ and the vector from π΄ to πΈ is parallel to the vector from πΉ to πΆ. If we can do that, we know itβs a parallelogram.

Now because we know π΄π΅πΆπ· is a parallelogram, we know that π΄π΅ is parallel to π·πΆ, so vector π΄π΅ is equal to vector π·πΆ, which is equal to vector π₯. So I can mark that in. And likewise, the vector from π· to π΄ must be parallel to the vector from πΆ to π΅. In fact, they must be equal because itβs a parallelogram, and that is equal to π¦. So letβs try and fill in some of the other missing vectors on this diagram.

Well from π· to π΅, there are two ways we can go direct from π· to π΅ or I can go from π· to πΆ and then from πΆ to π΅. And doing that, I can express that in terms of vectors π₯ and π¦. So in going from π· to π΅, we can see that that is the same as π₯ plus π¦. So in going from π· to πΈ, thatβs a third of the way from π· to π΅, and πΈ to πΉ is the same and πΉ to π΅ is the same.

So all of those vectors are equal to a third of π₯ plus π¦, so we can add those to the diagram as well. Now our diagram is getting a little bit cozy, but we can at least work out what some of the other vectors that we want to know π΄πΉ and πΈπΆ and πΉπΆ and π΄πΈ and so on. So to go from π΄ to πΉ, Iβve got two ways. Iβve got lots of ways that I can go. And one of them would be to go from π΄ to π΅ and then from π΅ to πΉ.

And I can see that going from π΄ to π΅ is the same as vector π₯, and from π΅ to πΉ itβs the negative of going π₯ plus π¦, a third of π₯ plus π¦. So I can write that down. Now, I can actually work out what they are. π₯ plus the negative a third of π₯ is two-thirds of π₯, and then Iβve got- take away a third of π¦.

Now I need to work out πΈ to πΆ and letβs hope we get the same vector. So from πΈ to πΆ, I can go back from πΈ down to π· and then from π· across to πΆ. And this time going from πΈ to π· is the negative cause weβre going backwards along that vector of a third π₯ plus π¦. And going from π· to πΆ, weβre just moving along vector π₯. So thatβs minus a third π₯ plus π₯ is two-thirds π₯ and then weβve got minus a third π¦.

So we can see that vector π΄πΉ, this side here, is equal to vector πΈπΆ which is this side here. So those lengths of the same side and theyβre parallel; theyβre the same length and theyβre parallel. Now if theyβre the same length and theyβre parallel, that means weβve taken the same vector and just picked it up and moved it to a different place.

Now letβs look at the other sides, π΄πΈ and πΉπΆ. And to get from π΄ to πΈ, I could go direct but that doesnβt tell me anything about vectors π₯ and π¦, or I could go back the e- from e- back along the π¦ the negative of the π¦ and then up from π· to πΈ; thatβs positive a third π₯ plus π¦. So the negative of π¦ plus a third of π¦ is minus two-thirds π¦, and weβre adding a third of π₯ to that. So vector π΄πΈ is equivalent to a third π₯ minus two-thirds of π¦.

Now letβs do the same for πΉ to πΆ. And I think the easiest way to get from πΉ to πΆ apart from doing directly πΉ to πΆ is to go from πΉ up to π΅, which is a third of π₯ plus π¦, and then come down from π΅ to πΆ, the negative of a π¦. And when I do that, I find that Iβve got the same answer, a third π₯ minus two-thirds π¦.

So this vector here and this vector here are both exactly the same vectors. That means theyβre the same length and theyβre parallel. So just written that down in my solution, π΄πΈ is equal to πΉπΆ so theyβre the same length and theyβre parallel.

So Iβm just gonna write down the definition of a parallelogram; itβs a quadrilateral with two pairs of parallel edges. Now we gather the information weβve got and see if we can show that that is a parallelogram. Now we saw that vector π΄πΉ was equal to vector πΈπΆ, so that must mean that these two sides are parallel. And likewise sides π΄πΈ and πΉπΆ are parallel because vector π΄πΈ is equal to vector πΉπΆ.

We can also show that the shape π΄πΉπΆπΈ has got four sides. There we are, one two three four. Therefore, we meet all of the criteria. And we can just write our conclusion: π΄πΉπΆπΈ is a parallelogram. So for our last question then, weβve got parallelogram π΄π΅πΆπ· and weβve got a point πΈ which lies on the diagonal π΄πΆ, so that the length of π΄πΈ is two-thirds of the length of π΄πΆ.

Weβve also got the vector from π· to πΆ is three π₯ and the vector from π· to π΄ is called three π¦, and πΉ is the midpoint of π΅πΆ, so this distance is the same as this distance. Weβve gotta prove that π·πΈπΉ, so this line going from π· through πΈ to πΉ, is a straight line.

So our approach here is to find an equivalent vector for π·πΈ and to find an equivalent vector for π·πΉ and prove that one is a simple multiple of the other. And if one is a multiple of the other, then we know that theyβre parallel because the π₯ and π¦ are in the same proportions and also weβll know that π· is common to both of them. So not only are they parallel, but theyβve got a point in common, so that will mean that they make a straight line.

So thatβs gonna be our general approach. Letβs have a look at what weβre gonna do first. So since there- since π΄π΅πΆπ· is a parallelogram, we can just fill in those two things. We know that π΄π· and π΅πΆ are parallel, so πΆ to π΅ is three π¦, and we know that π΄π΅ and π·πΆ are parallel, so π΄ to π΅ is the vector three π₯.

Now also because weβve got that line side π΅πΆ as three π¦, we know that πΉ is the midpoint of those, we can fill in the vectors for πΆ to πΉ and for πΉ to π΅. Theyβre simply half of three π¦.

Now the question told us that π΄πΈ is two-thirds of π΄πΆ, so if we can find a vector for π΄ to πΆ in terms of π₯s and π¦s, then we can work out what the vector from π΄ to πΈ is and weβve got piles in every direction through our parallelogram then. Now to get from π΄ to πΆ, we could do the direct route, but we could also do this route here; we could come the wrong way down a three π¦ vector and then we could go forward through a three π₯ vector. So thatβs negative three π¦ plus three π₯.

And as we said, π΄πΈ is two-thirds of π΄πΆ, so itβs gonna be equal to two-thirds of negative three π¦ plus three π₯. And we can add that to our diagram then. π΄πΈ is two π₯ take away two π¦. So that just leaves us with πΈπΆ to fill out and we know that πΈπΆ is the difference between π΄πΆ and π΄πΈ, so thatβs π΄πΆ take away π΄πΈ.

And we saw up here that π΄πΆ was three π₯ take away three π¦, so three π₯ take away three π¦ take away two π₯ take away two π¦ gives us three π₯ take away two π₯ and negative three π¦ add two π¦. So πΈπΆ comes out to be π₯ minus π¦, so we can add that on here as well.

Right with all that information added to our diagram, now we can go about the real business of trying to prove that π·πΈπΉ is a straight line. So how do we get from π· to πΈ? Well we could come along here from π· to πΆ and then up to πΈ from there, so we can go from π· to πΆ and then from πΆ to πΈ. And vector π·πΆ is just three π₯ and vector πΆ to πΈ; well we know that πΈ to πΆ was π₯ minus π¦, so πΆ to πΈ is the opposite of that, the opposite direction is negative of π₯ minus π¦.

We have to be careful with our negative signs here. But when we work all that out, weβve got π·πΈ is equal to two π₯ plus π¦. So letβs work out now a vector for π· to πΉ. And I think the easiest way to get from π· to πΉ is to go from π· to πΆ and then from πΆ up to πΉ. And from π· to πΆ, that was a vector of three π₯, and from πΆ to πΉ thatβs three over two π¦.

So Iβm just gonna factorize that out; take the three out as a common factor. And that shows us that π·πΉ is equal three lots of π₯ plus a half π¦. Now letβs go back to π·πΈ. And I can see that if I factorize the two out there, Iβve got two lots of π₯ plus a half π¦. So one of these is two lots of π₯ plus a half π¦, and the other one is three lots of π₯ plus a half π¦. So one is simply a multiple of the other. So π·πΉ is equal to three over two lots of π·πΈ. If I multiply π·πΈ by three over two, Iβll get exactly π·πΉ.

Now that proves that theyβre parallel. So one vector being a simple scale a multiple of another shows that they are parallel, and point π· is common to both lines π·πΈ and π·πΉ. So not only are they parallel, but they have one point at least one point in the same line as well.

So this situation where weβve got π·πΉ and π·πΈ being parallel and theyβve both got a point in common, we say that those two vectors are collinear. In other words, they make a straight line. So we can say that π·πΈπΉ is a straight line.