# Video: AQA GCSE Mathematics Higher Tier Pack 3 • Paper 1 • Question 25

𝐴𝐵𝐶 is a triangle. 𝐷, 𝐸, and 𝐹 are points on the triangle such that 𝐴𝐷𝐸𝐹 is a parallelogram and 𝐵𝐸 : 𝐵𝐶 = 3 : 5. The vector 𝐵𝐸 = 𝐚 and the vector 𝐶𝐹 = 𝐛. a) Work out the vector 𝐸𝐹 in terms of 𝐚 and 𝐛. b) Work out the vector 𝐴𝐸 in terms of 𝐚 and 𝐛. Simplify your answer as much as possible.

10:43

### Video Transcript

𝐴𝐵𝐶 is a triangle. 𝐷, 𝐸, and 𝐹 are points on the triangle such that 𝐴𝐷𝐸𝐹 is a parallelogram and the ratio of the side lengths 𝐵𝐸 to 𝐵𝐶 is three to five. The vector 𝐵𝐸 is 𝐚 and the vector 𝐶𝐹 is 𝐛. Part a) asks us “work out the vector 𝐸𝐹 in terms of 𝐚 and 𝐛.” And part b) asks us “work out the vector 𝐴𝐸 in terms of 𝐚 and 𝐛. Simplify your answer as much as possible.”

Let’s look at our diagram. We’re told in the question that the vector from 𝐵 to 𝐸 is represented by the lowercase letter 𝐚. And we can identify the vector from 𝐵 to 𝐸 on the diagram; it’s this vector here. We’re also told that the vector from 𝐶 to 𝐹 is represented by the lowercase letter 𝐛. And again, we can identify this vector on our diagram; it’s this vector here.

Each part of the question involves us finding another vector in terms of these two vectors: lowercase 𝐚 and lowercase 𝐛. But to do this, we’re going to have to use some other information that we’re given in the question, for example, that 𝐴𝐷𝐸𝐹 is a parallelogram.

And by definition, the opposite sides of a parallelogram are parallel. So we can mark on the diagram that the side 𝐷𝐸 is parallel to the side 𝐴𝐹 and the side 𝐴𝐷 is parallel to the side 𝐹𝐸. We’ll come back to thinking about the consequences of these parallel sides later. But all we need for part a of the question is that the ratio 𝐵𝐸 to 𝐵𝐶 is three to five. So if 𝐵𝐸 is three units long, then 𝐵𝐶 is five units long.

Our task in part a is to find the vector 𝐸𝐹 in terms of 𝐚 and 𝐛. We can locate this vector 𝐸𝐹 on our diagram; it’s here pointing from 𝐸 to 𝐹. And to find 𝐸𝐹 in terms of 𝐚 and 𝐛, we want to write it as a sum or difference of vectors whose value in terms of 𝐚 and 𝐛 we either know or can easily work out. We see from the diagram that we know what 𝐶𝐹 is. This is 𝐛 and so we write 𝐸𝐹 as 𝐸𝐶 plus 𝐶𝐹.

So we travel from 𝐸 to 𝐹 via the point 𝐶 going along the vector 𝐸𝐶 and then along the vector 𝐶𝐹 which we know to be 𝐛. All we have left to work out then is what the vector 𝐸𝐶 is in terms of 𝐚 and 𝐛. Well, we can see that the vector 𝐸𝐶 points in the same direction as the vector 𝐚. And so, it must be a multiple of this vector. So we can say the 𝐸𝐶 is some number 𝑘 times the vector 𝐚.

We can make it clear that 𝐚 is a vector, whereas 𝑘 is a number by underlining 𝐚. You might like to do this. Now, when we look at the diagram, we said that if the vector 𝐚 had length of three, then the vector 𝐵𝐶 had length of five. And as a result, the vector 𝐸𝐶 will have a length of two using these units.

And as 𝐸𝐶 is two-thirds the length of 𝐚 pointing in the same direction as 𝐚, it must be two-thirds 𝐚. And looking at the diagram, this seems very reasonable. So we can substitute this value into our expression for 𝐸𝐹. 𝐸𝐶 is two-thirds 𝐚. And remember that we know that 𝐶𝐹 is just 𝐛. So 𝐸𝐹 is two-thirds 𝐚 plus 𝐛. So we found the vector 𝐸𝐹 in terms of the vectors 𝐚 and 𝐛 as required.

Just as a quick note before moving under part b, you might be worried that the ratio of 𝐵𝐸 to 𝐵𝐶 being three to five doesn’t mean that 𝐵𝐸 has a length of three and 𝐵𝐶 has a length of five. For example, maybe 𝐵𝐸 has a length of six and 𝐵𝐶 has a length of 10.

But in this question, we’re not told any unit. So we’re not given any length. So we can pick our units and we can pick them to make 𝐵𝐸 be three units long and hence 𝐵𝐶 be five units long. If you prefer, you can make 𝐵𝐸 three 𝑥 units long for some 𝑥 and 𝐵𝐶 five 𝑥 units long for some 𝑥 and then 𝐸𝐶 will be two 𝑥 units long. We still find that 𝐸𝐶 has two-thirds the length of 𝐚 and hence is two-thirds of 𝐚.

Okay, let’s now move on to part b, where we have to find the vector 𝐴𝐸; that’s this vector here in terms of the vectors lowercase 𝐚 and lowercase 𝐛. Again, the approach is to write the vector 𝐴𝐸 as a sum or difference of vectors. But here, there are many parts from 𝐴 to 𝐸 and it’s not clear which one is the best.

For example, we could travel via the point 𝐵. The vector 𝐴𝐸 is the sum of the vectors 𝐴𝐵 and 𝐵𝐸. And this is nice because we know that the vector 𝐵𝐸 is the vector lowercase 𝐛. Unfortunately, it’s not particularly obvious what the vector 𝐴𝐵 is. Alternatively, we could go via the point 𝐷. The vector 𝐴𝐸 is the sum of the vectors 𝐴𝐷 and 𝐷𝐸. But again, it’s not clear how easy it’s going to be to find these two vectors, 𝐴𝐷 and 𝐷𝐸.

I’m going to use the solution where we go via 𝐹, writing the vector 𝐴𝐸 as the sum of the vectors 𝐴𝐹 and 𝐹𝐸. Looking at the diagram, we can see that the vector 𝐴𝐹 appears to be pointing in the opposite direction to the vector 𝐛 and so is a multiple of that vector. And of course, the vector 𝐹𝐸 is just the opposite of the vector 𝐸𝐹 which we worked out in the first part, part a.

I’m going to choose this path because I think it’s going to lead to the least working. But any one of these parts would work. And so rather than looking at all the paths and trying to work out which will lead to the least amount of working, it might be quicker just to pick a path and going through the working, however much there ends up being.

So 𝐴𝐸 is 𝐴𝐹 plus 𝐹𝐸 this vector 𝐹𝐸 is the opposite of the vector 𝐸𝐹. And you worked out that the vector 𝐸𝐹 was two-thirds of 𝐚 plus 𝐛 in part a. So really, we just have to find the vector 𝐴𝐹 in terms of the vectors lowercase 𝐚 and lowercase 𝐛. And as we said before, this vector 𝐴𝐹 points in the opposite direction to the vector 𝐶𝐹, which is 𝐛. And so, it’s a multiple of the vector 𝐛, some number 𝑐 times 𝐛. And as it points in the opposite direction to 𝐛, we know that this number 𝑐 must be negative.

Now, we’d like to say something about the relative lengths of the vectors 𝐴𝐹 and lowercase 𝐛 so that we can find the value of 𝑐 much like we found the value of 𝑘 in the first part a of the question. This is where we use the fact that 𝐴𝐷𝐸𝐹 is a parallelogram. Notice that we haven’t used this fact before.

As we have lots of parallel lines on our diagram, we have lots of equal angles. For example, with the parallel lines 𝐴𝐵 and 𝐹𝐸 and the transversal 𝐴𝐶, we can see that these two angles are corresponding and hence equal. And with the transversal 𝐵𝐶, we can see that these two angles are corresponding angles in parallel lines and hence equal.

And as they have two pairs of equal angles, the triangle 𝐴𝐵𝐶 — that’s the big triangle — and triangle 𝐹𝐸𝐶 — that’s the smallest triangle are similar. And as a result, their sides are proportional. That is, the ratios of corresponding sides are all equal.

Now we know that the length of the side 𝐵𝐶 is five 𝑥 and the length of the corresponding side 𝐸𝐶 is two 𝑥. So the ratio of corresponding sides of these two triangles is five 𝑥 to two 𝑥 or just five to two. And so, if you let the side 𝐴𝐶 have length five 𝑦, then the length of the corresponding side 𝐹𝐶 to have the ratio five to two must be two 𝑦. And this means that the vector 𝐴𝐹 whose length we really want to find must be three 𝑦.

Now we’re ready to find the value of 𝑐 in 𝐴𝐹 equals 𝑐 times 𝐛. The vector 𝐴𝐹 has a length of three 𝑦, whereas the vector lowercase 𝐛 has a length of two 𝑦. So 𝐴𝐹 is three over two times as long as 𝐛. But 𝑐 is negative as 𝐴𝐹 points in the opposite direction to the vector lowercase 𝐛. And so, the vector 𝐴𝐹 is not three over two 𝐛, but negative three over two 𝐛.

Now, we can substitute this for 𝐴𝐹 in our expression for 𝐴𝐸. And of course, we do the same for the vector 𝐹𝐸. We get negative three over two 𝐛 minus two-thirds 𝐚 plus 𝐛. Let’s simplify. We distribute the minus sign over the terms in the brackets, making sure that we get minus 𝐛 here. And we combine negative three over two 𝐛 with the minus 𝐛 to get minus five over two 𝐛. And so our final and fully simplified answer is that the vector 𝐴𝐸 is negative two-thirds 𝐚 minus five over two 𝐛.

To summarize then, in both parts 𝐚 and 𝐛, we found the vectors by writing them as sums of other vectors. Now to find those vectors, we used the fact that if a vector points in the same or the opposite direction to a given vector, then it is a multiple of that vector. In part a, we used a ratio given in the question to work out which multiple of the vector this was. And in part b, we used some triangular similarity.