# Video: AQA GCSE Mathematics Higher Tier Pack 3 β’ Paper 1 β’ Question 25

π΄π΅πΆ is a triangle. π·, πΈ, and πΉ are points on the triangle such that π΄π·πΈπΉ is a parallelogram and π΅πΈ : π΅πΆ = 3 : 5. The vector π΅πΈ = π and the vector πΆπΉ = π. a) Work out the vector πΈπΉ in terms of π and π. b) Work out the vector π΄πΈ in terms of π and π. Simplify your answer as much as possible.

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### Video Transcript

π΄π΅πΆ is a triangle. π·, πΈ, and πΉ are points on the triangle such that π΄π·πΈπΉ is a parallelogram and the ratio of the side lengths π΅πΈ to π΅πΆ is three to five. The vector π΅πΈ is π and the vector πΆπΉ is π. Part a) asks us βwork out the vector πΈπΉ in terms of π and π.β And part b) asks us βwork out the vector π΄πΈ in terms of π and π. Simplify your answer as much as possible.β

Letβs look at our diagram. Weβre told in the question that the vector from π΅ to πΈ is represented by the lowercase letter π. And we can identify the vector from π΅ to πΈ on the diagram; itβs this vector here. Weβre also told that the vector from πΆ to πΉ is represented by the lowercase letter π. And again, we can identify this vector on our diagram; itβs this vector here.

Each part of the question involves us finding another vector in terms of these two vectors: lowercase π and lowercase π. But to do this, weβre going to have to use some other information that weβre given in the question, for example, that π΄π·πΈπΉ is a parallelogram.

And by definition, the opposite sides of a parallelogram are parallel. So we can mark on the diagram that the side π·πΈ is parallel to the side π΄πΉ and the side π΄π· is parallel to the side πΉπΈ. Weβ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 π΅πΈ to π΅πΆ is three to five. So if π΅πΈ is three units long, then π΅πΆ is five units long.

Our task in part a is to find the vector πΈπΉ in terms of π and π. We can locate this vector πΈπΉ on our diagram; itβs here pointing from πΈ to πΉ. And to find πΈπΉ in terms of π and π, we want to write it as a sum or difference of vectors whose value in terms of π and π we either know or can easily work out. We see from the diagram that we know what πΆπΉ is. This is π and so we write πΈπΉ as πΈπΆ plus πΆπΉ.

So we travel from πΈ to πΉ via the point πΆ going along the vector πΈπΆ and then along the vector πΆπΉ which we know to be π. All we have left to work out then is what the vector πΈπΆ is in terms of π and π. Well, we can see that the vector πΈπΆ points in the same direction as the vector π. And so, it must be a multiple of this vector. So we can say the πΈπΆ is some number π times the vector π.

We can make it clear that π is a vector, whereas π is a number by underlining π. You might like to do this. Now, when we look at the diagram, we said that if the vector π had length of three, then the vector π΅πΆ had length of five. And as a result, the vector πΈπΆ will have a length of two using these units.

And as πΈπΆ is two-thirds the length of π pointing in the same direction as π, it must be two-thirds π. And looking at the diagram, this seems very reasonable. So we can substitute this value into our expression for πΈπΉ. πΈπΆ is two-thirds π. And remember that we know that πΆπΉ is just π. So πΈπΉ is two-thirds π plus π. So we found the vector πΈπΉ in terms of the vectors π and π as required.

Just as a quick note before moving under part b, you might be worried that the ratio of π΅πΈ to π΅πΆ being three to five doesnβt mean that π΅πΈ has a length of three and π΅πΆ has a length of five. For example, maybe π΅πΈ has a length of six and π΅πΆ has a length of 10.

But in this question, weβre not told any unit. So weβre not given any length. So we can pick our units and we can pick them to make π΅πΈ be three units long and hence π΅πΆ be five units long. If you prefer, you can make π΅πΈ three π₯ units long for some π₯ and π΅πΆ five π₯ units long for some π₯ and then πΈπΆ will be two π₯ units long. We still find that πΈπΆ has two-thirds the length of π and hence is two-thirds of π.

Okay, letβs now move on to part b, where we have to find the vector π΄πΈ; thatβs this vector here in terms of the vectors lowercase π and lowercase π. Again, the approach is to write the vector π΄πΈ as a sum or difference of vectors. But here, there are many parts from π΄ to πΈ and itβs not clear which one is the best.

For example, we could travel via the point π΅. The vector π΄πΈ is the sum of the vectors π΄π΅ and π΅πΈ. And this is nice because we know that the vector π΅πΈ is the vector lowercase π. Unfortunately, itβs not particularly obvious what the vector π΄π΅ is. Alternatively, we could go via the point π·. The vector π΄πΈ is the sum of the vectors π΄π· and π·πΈ. But again, itβs not clear how easy itβs going to be to find these two vectors, π΄π· and π·πΈ.

Iβm going to use the solution where we go via πΉ, writing the vector π΄πΈ as the sum of the vectors π΄πΉ and πΉπΈ. Looking at the diagram, we can see that the vector π΄πΉ appears to be pointing in the opposite direction to the vector π and so is a multiple of that vector. And of course, the vector πΉπΈ is just the opposite of the vector πΈπΉ which we worked out in the first part, part a.

Iβm going to choose this path because I think itβ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 π΄πΈ is π΄πΉ plus πΉπΈ this vector πΉπΈ is the opposite of the vector πΈπΉ. And you worked out that the vector πΈπΉ was two-thirds of π plus π in part a. So really, we just have to find the vector π΄πΉ in terms of the vectors lowercase π and lowercase π. And as we said before, this vector π΄πΉ points in the opposite direction to the vector πΆπΉ, which is π. And so, itβs a multiple of the vector π, some number π times π. And as it points in the opposite direction to π, we know that this number π must be negative.

Now, weβd like to say something about the relative lengths of the vectors π΄πΉ and lowercase π so that we can find the value of π much like we found the value of π in the first part a of the question. This is where we use the fact that π΄π·πΈπΉ is a parallelogram. Notice that we havenβ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 π΄π΅ and πΉπΈ and the transversal π΄πΆ, we can see that these two angles are corresponding and hence equal. And with the transversal π΅πΆ, 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 π΄π΅πΆ β thatβs the big triangle β and triangle πΉπΈπΆ β thatβ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 π΅πΆ is five π₯ and the length of the corresponding side πΈπΆ is two π₯. So the ratio of corresponding sides of these two triangles is five π₯ to two π₯ or just five to two. And so, if you let the side π΄πΆ have length five π¦, then the length of the corresponding side πΉπΆ to have the ratio five to two must be two π¦. And this means that the vector π΄πΉ whose length we really want to find must be three π¦.

Now weβre ready to find the value of π in π΄πΉ equals π times π. The vector π΄πΉ has a length of three π¦, whereas the vector lowercase π has a length of two π¦. So π΄πΉ is three over two times as long as π. But π is negative as π΄πΉ points in the opposite direction to the vector lowercase π. And so, the vector π΄πΉ is not three over two π, but negative three over two π.

Now, we can substitute this for π΄πΉ in our expression for π΄πΈ. And of course, we do the same for the vector πΉπΈ. We get negative three over two π minus two-thirds π plus π. Letβs simplify. We distribute the minus sign over the terms in the brackets, making sure that we get minus π here. And we combine negative three over two π with the minus π to get minus five over two π. And so our final and fully simplified answer is that the vector π΄πΈ is negative two-thirds π minus five over two π.

To summarize then, in both parts π and π, 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.