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Video: Adding and Subtracting Vectors (More Complicated Questions)

Tim Burnham

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.

14:38

Video Transcript

In this video, we’re gonna take a look at a few vector addition and subtraction questions that you might attempt after you’ve had a go at the more basic introductory ones. In this question, we’ve got triangle 𝑂𝐴𝐡 and we’re told that the point 𝑋 is on the side 𝐴𝐡 and that the distance 𝐴𝑋 is a third of the distance from 𝐴 to 𝐡.

Now we got to express vector 𝐴𝐡 in terms of π‘Ž and 𝑏, and we’ve also got to express the vector 𝑂𝑋 in terms of π‘Ž and 𝑏. Now one of these is much easier than the other, but let’s have a look at the question.

Now if we think about that side 𝐴𝐡, if 𝐴𝑋 is a third of the way from 𝐴 to 𝐡, then I can split that into three equal sized chunks. Now the first one from 𝐴 to 𝑋, that’s vector 𝐴𝑋, that’s repeated on the other two because if it’s third the distance, then if I do the same thing three times, that will make the distance 𝐴𝐡.

So here we’ve laid the same vector in to in three times, and that’s something that we will come back and use later on in the question. So the f- the first part of the question, express 𝐴𝐡 in terms of π‘Ž and 𝑏.

How am I gonna get from point 𝐴 to point 𝐡? Well I could go in a straight line from 𝐴 to 𝐡, but that doesn’t give me an answer in terms of vectors π‘Ž and vector 𝑏. So the other way I can do it is go from 𝐴 to 𝑂 and then from 𝑂 to 𝐡. And in going from 𝐴 to 𝑂, I’m going on backtracking along an π‘Ž vector; I’m going the opposite direction of an π‘Ž vector; that’s negative π‘Ž.

And in going from 𝑂 to 𝐡, that’s just going a positive way along the 𝑏 vector, so that’s just 𝑏. So the vector 𝐴𝐡 can be expressed as the negative of an π‘Ž vector plus the positive of a 𝑏 vector. Well there we go; that’s the first part done.

Now for part b, I’ve got to find out 𝑂𝑋. So I’ve got- I’ve got this journey here from 𝑂 to 𝑋, and I’ve got to express that in terms of π‘Ž and 𝑏. Now the direct line from 𝑂 to 𝑋 doesn’t include any information about vectors 𝐴 and 𝐡, so I’ve got to try and think about how can I get from 𝑂 to 𝑋 via other routes which do have π‘Ž and 𝑏 in them.

So I could use this route from 𝑂 to 𝐴 and then from 𝐴 to 𝑋. And we know that 𝑂 to 𝐴 is just simply the-the- this vector here π‘Ž and from 𝐴 to 𝑋 is a third of the way from 𝐴 to 𝐡, and we just worked out what 𝐴 to 𝐡 was.

So let’s start writing that down. So 𝑂𝑋 is equal to π‘Ž plus a third 𝐴𝐡 and we know that 𝐴𝐡 is negative π‘Ž plus 𝑏, so 𝑂𝑋 is π‘Ž plus a third of negative π‘Ž plus 𝑏. And multiplying out the brackets, that means that 𝑂𝑋 is equal to π‘Ž minus a third of π‘Ž plus a third of 𝑏, so our answer is that vector 𝑂𝑋 is equal to two-thirds of vector π‘Ž plus one-third of vector 𝑏.

Now in our next question, we’ve got a quadrilateral 𝐴𝐡𝐢𝐷 which is a parallelogram. 𝐸 and 𝐹 are points on the diagonal from 𝐷 to 𝐡, and the distance from 𝐷 to 𝐸 is the same as the distance from 𝐸 to 𝐹, which is the same as the distance from 𝐹 to 𝐡.

We’re told that 𝐷 to 𝐢 is a vector called π‘₯ and 𝐢 to 𝐡 is a vector called 𝑦. What we’re gonna do is show that 𝐴𝐹𝐢𝐸 is also a parallelogram. So looking on that diagonal there, I’ve just shown that those three links are the same, and I’ve drawn in the shape 𝐴𝐹𝐢𝐸, and we- basically to prove that that’s a parallelogram, a parallelogram has got two pairs of parallel sides.

So we’ve got to prove that the vector from 𝐴 to 𝐹 is parallel to the vector from 𝐸 to 𝐢 and the vector from 𝐴 to 𝐸 is parallel to the vector from 𝐹 to 𝐢. If we can do that, we know it’s a parallelogram.

Now because we know 𝐴𝐡𝐢𝐷 is a parallelogram, we know that 𝐴𝐡 is parallel to 𝐷𝐢, so vector 𝐴𝐡 is equal to vector 𝐷𝐢, which is equal to vector π‘₯. So I can mark that in. And likewise, the vector from 𝐷 to 𝐴 must be parallel to the vector from 𝐢 to 𝐡. In fact, they must be equal because it’s a parallelogram, and that is equal to 𝑦. So let’s try and fill in some of the other missing vectors on this diagram.

Well from 𝐷 to 𝐡, there are two ways we can go direct from 𝐷 to 𝐡 or I can go from 𝐷 to 𝐢 and then from 𝐢 to 𝐡. And doing that, I can express that in terms of vectors π‘₯ and 𝑦. So in going from 𝐷 to 𝐡, we can see that that is the same as π‘₯ plus 𝑦. So in going from 𝐷 to 𝐸, that’s a third of the way from 𝐷 to 𝐡, and 𝐸 to 𝐹 is the same and 𝐹 to 𝐡 is the same.

So all of those vectors are equal to a third of π‘₯ plus 𝑦, 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 𝐴𝐹 and 𝐸𝐢 and 𝐹𝐢 and 𝐴𝐸 and so on. So to go from 𝐴 to 𝐹, I’ve got two ways. I’ve got lots of ways that I can go. And one of them would be to go from 𝐴 to 𝐡 and then from 𝐡 to 𝐹.

And I can see that going from 𝐴 to 𝐡 is the same as vector π‘₯, and from 𝐡 to 𝐹 it’s the negative of going π‘₯ plus 𝑦, a third of π‘₯ plus 𝑦. 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 I’ve got- take away a third of 𝑦.

Now I need to work out 𝐸 to 𝐢 and let’s hope we get the same vector. So from 𝐸 to 𝐢, I can go back from 𝐸 down to 𝐷 and then from 𝐷 across to 𝐢. And this time going from 𝐸 to 𝐷 is the negative cause we’re going backwards along that vector of a third π‘₯ plus 𝑦. And going from 𝐷 to 𝐢, we’re just moving along vector π‘₯. So that’s minus a third π‘₯ plus π‘₯ is two-thirds π‘₯ and then we’ve got minus a third 𝑦.

So we can see that vector 𝐴𝐹, this side here, is equal to vector 𝐸𝐢 which is this side here. So those lengths of the same side and they’re parallel; they’re the same length and they’re parallel. Now if they’re the same length and they’re parallel, that means we’ve taken the same vector and just picked it up and moved it to a different place.

Now let’s look at the other sides, 𝐴𝐸 and 𝐹𝐢. And to get from 𝐴 to 𝐸, I could go direct but that doesn’t tell me anything about vectors π‘₯ and 𝑦, or I could go back the e- from e- back along the 𝑦 the negative of the 𝑦 and then up from 𝐷 to 𝐸; that’s positive a third π‘₯ plus 𝑦. So the negative of 𝑦 plus a third of 𝑦 is minus two-thirds 𝑦, and we’re adding a third of π‘₯ to that. So vector 𝐴𝐸 is equivalent to a third π‘₯ minus two-thirds of 𝑦.

Now let’s do the same for 𝐹 to 𝐢. And I think the easiest way to get from 𝐹 to 𝐢 apart from doing directly 𝐹 to 𝐢 is to go from 𝐹 up to 𝐡, which is a third of π‘₯ plus 𝑦, and then come down from 𝐡 to 𝐢, the negative of a 𝑦. And when I do that, I find that I’ve got the same answer, a third π‘₯ minus two-thirds 𝑦.

So this vector here and this vector here are both exactly the same vectors. That means they’re the same length and they’re parallel. So just written that down in my solution, 𝐴𝐸 is equal to 𝐹𝐢 so they’re the same length and they’re parallel.

So I’m just gonna write down the definition of a parallelogram; it’s a quadrilateral with two pairs of parallel edges. Now we gather the information we’ve got and see if we can show that that is a parallelogram. Now we saw that vector 𝐴𝐹 was equal to vector 𝐸𝐢, so that must mean that these two sides are parallel. And likewise sides 𝐴𝐸 and 𝐹𝐢 are parallel because vector 𝐴𝐸 is equal to vector 𝐹𝐢.

We can also show that the shape 𝐴𝐹𝐢𝐸 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: 𝐴𝐹𝐢𝐸 is a parallelogram. So for our last question then, we’ve got parallelogram 𝐴𝐡𝐢𝐷 and we’ve got a point 𝐸 which lies on the diagonal 𝐴𝐢, so that the length of 𝐴𝐸 is two-thirds of the length of 𝐴𝐢.

We’ve also got the vector from 𝐷 to 𝐢 is three π‘₯ and the vector from 𝐷 to 𝐴 is called three 𝑦, and 𝐹 is the midpoint of 𝐡𝐢, so this distance is the same as this distance. We’ve gotta prove that 𝐷𝐸𝐹, so this line going from 𝐷 through 𝐸 to 𝐹, is a straight line.

So our approach here is to find an equivalent vector for 𝐷𝐸 and to find an equivalent vector for 𝐷𝐹 and prove that one is a simple multiple of the other. And if one is a multiple of the other, then we know that they’re parallel because the π‘₯ and 𝑦 are in the same proportions and also we’ll know that 𝐷 is common to both of them. So not only are they parallel, but they’ve got a point in common, so that will mean that they make a straight line.

So that’s gonna be our general approach. Let’s have a look at what we’re gonna do first. So since there- since 𝐴𝐡𝐢𝐷 is a parallelogram, we can just fill in those two things. We know that 𝐴𝐷 and 𝐡𝐢 are parallel, so 𝐢 to 𝐡 is three 𝑦, and we know that 𝐴𝐡 and 𝐷𝐢 are parallel, so 𝐴 to 𝐡 is the vector three π‘₯.

Now also because we’ve got that line side 𝐡𝐢 as three 𝑦, we know that 𝐹 is the midpoint of those, we can fill in the vectors for 𝐢 to 𝐹 and for 𝐹 to 𝐡. They’re simply half of three 𝑦.

Now the question told us that 𝐴𝐸 is two-thirds of 𝐴𝐢, so if we can find a vector for 𝐴 to 𝐢 in terms of π‘₯s and 𝑦s, then we can work out what the vector from 𝐴 to 𝐸 is and we’ve got piles in every direction through our parallelogram then. Now to get from 𝐴 to 𝐢, we could do the direct route, but we could also do this route here; we could come the wrong way down a three 𝑦 vector and then we could go forward through a three π‘₯ vector. So that’s negative three 𝑦 plus three π‘₯.

And as we said, 𝐴𝐸 is two-thirds of 𝐴𝐢, so it’s gonna be equal to two-thirds of negative three 𝑦 plus three π‘₯. And we can add that to our diagram then. 𝐴𝐸 is two π‘₯ take away two 𝑦. So that just leaves us with 𝐸𝐢 to fill out and we know that 𝐸𝐢 is the difference between 𝐴𝐢 and 𝐴𝐸, so that’s 𝐴𝐢 take away 𝐴𝐸.

And we saw up here that 𝐴𝐢 was three π‘₯ take away three 𝑦, so three π‘₯ take away three 𝑦 take away two π‘₯ take away two 𝑦 gives us three π‘₯ take away two π‘₯ and negative three 𝑦 add two 𝑦. So 𝐸𝐢 comes out to be π‘₯ minus 𝑦, 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 𝐷𝐸𝐹 is a straight line. So how do we get from 𝐷 to 𝐸? Well we could come along here from 𝐷 to 𝐢 and then up to 𝐸 from there, so we can go from 𝐷 to 𝐢 and then from 𝐢 to 𝐸. And vector 𝐷𝐢 is just three π‘₯ and vector 𝐢 to 𝐸; well we know that 𝐸 to 𝐢 was π‘₯ minus 𝑦, so 𝐢 to 𝐸 is the opposite of that, the opposite direction is negative of π‘₯ minus 𝑦.

We have to be careful with our negative signs here. But when we work all that out, we’ve got 𝐷𝐸 is equal to two π‘₯ plus 𝑦. So let’s work out now a vector for 𝐷 to 𝐹. And I think the easiest way to get from 𝐷 to 𝐹 is to go from 𝐷 to 𝐢 and then from 𝐢 up to 𝐹. And from 𝐷 to 𝐢, that was a vector of three π‘₯, and from 𝐢 to 𝐹 that’s three over two 𝑦.

So I’m just gonna factorize that out; take the three out as a common factor. And that shows us that 𝐷𝐹 is equal three lots of π‘₯ plus a half 𝑦. Now let’s go back to 𝐷𝐸. And I can see that if I factorize the two out there, I’ve got two lots of π‘₯ plus a half 𝑦. So one of these is two lots of π‘₯ plus a half 𝑦, and the other one is three lots of π‘₯ plus a half 𝑦. So one is simply a multiple of the other. So 𝐷𝐹 is equal to three over two lots of 𝐷𝐸. If I multiply 𝐷𝐸 by three over two, I’ll get exactly 𝐷𝐹.

Now that proves that they’re parallel. So one vector being a simple scale a multiple of another shows that they are parallel, and point 𝐷 is common to both lines 𝐷𝐸 and 𝐷𝐹. 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 we’ve got 𝐷𝐹 and 𝐷𝐸 being parallel and they’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 𝐷𝐸𝐹 is a straight line.