### Video Transcript

π΄π΅πΆ is a triangle. π·, πΈ, and πΉ are points on the triangle such that π΄π·πΈπΉ is a parallelogram and
the ratio of the side lengths π΅πΈ to π΅πΆ is three to five. The vector π΅πΈ is π and the vector πΆπΉ is π. Part a) asks us βwork out the vector πΈπΉ in terms of π and π.β And part b) asks us βwork out the vector π΄πΈ in terms of π and π. Simplify your answer as much as possible.β

Letβs look at our diagram. Weβre told in the question that the vector from π΅ to πΈ is represented by the
lowercase letter π. And we can identify the vector from π΅ to πΈ on the diagram; itβs this vector
here. Weβre also told that the vector from πΆ to πΉ is represented by the lowercase letter
π. And again, we can identify this vector on our diagram; itβs 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, weβre going to have to use some other information that weβre given in
the question, for example, that π΄π·πΈπΉ 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·πΈ is parallel to the side π΄πΉ and the
side π΄π· is parallel to the side πΉπΈ. Weβll 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΅πΈ to π΅πΆ is three to
five. So if π΅πΈ is three units long, then π΅πΆ is five units long.

Our task in part a is to find the vector πΈπΉ in terms of π and π. We can locate this vector πΈπΉ on our diagram; itβs here pointing from πΈ to πΉ. And to find πΈπΉ 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 πΆπΉ is. This is π and so we write πΈπΉ as πΈπΆ plus πΆπΉ.

So we travel from πΈ to πΉ via the point πΆ going along the vector πΈπΆ and then
along the vector πΆπΉ which we know to be π. All we have left to work out then is what the vector πΈπΆ is in terms of π and
π. Well, we can see that the vector πΈπΆ points in the same direction as the vector
π. And so, it must be a multiple of this vector. So we can say the πΈπΆ 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΅πΆ had length of five. And as a result, the vector πΈπΆ will have a length of two using these units.

And as πΈπΆ 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 πΈπΉ. πΈπΆ is two-thirds π. And remember that we know that πΆπΉ is just π. So πΈπΉ is two-thirds π plus π. So we found the vector πΈπΉ 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΅πΈ to π΅πΆ being three to five doesnβt mean that π΅πΈ has a length of three and
π΅πΆ has a length of five. For example, maybe π΅πΈ has a length of six and π΅πΆ has a length of 10.

But in this question, weβre not told any unit. So weβre not given any length. So we can pick our units and we can pick them to make π΅πΈ be three units long and
hence π΅πΆ be five units long. If you prefer, you can make π΅πΈ three π₯ units long for some π₯ and π΅πΆ five π₯
units long for some π₯ and then πΈπΆ will be two π₯ units long. We still find that πΈπΆ has two-thirds the length of π and hence is two-thirds of
π.

Okay, letβs now move on to part b, where we have to find the vector π΄πΈ; thatβs this
vector here in terms of the vectors lowercase π and lowercase π. Again, the approach is to write the vector π΄πΈ as a sum or difference of
vectors. But here, there are many parts from π΄ to πΈ and itβs not clear which one is the
best.

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

Iβm going to use the solution where we go via πΉ, writing the vector π΄πΈ as the sum
of the vectors π΄πΉ and πΉπΈ. Looking at the diagram, we can see that the vector π΄πΉ 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 πΉπΈ is just the opposite of the vector πΈπΉ which we
worked out in the first part, part a.

Iβm going to choose this path because I think itβs 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΄πΈ is π΄πΉ plus πΉπΈ this vector πΉπΈ is the opposite of the vector πΈπΉ. And you worked out that the vector πΈπΉ was two-thirds of π plus π in part a. So really, we just have to find the vector π΄πΉ in terms of the vectors lowercase π
and lowercase π. And as we said before, this vector π΄πΉ points in the opposite direction to the
vector πΆπΉ, which is π. And so, itβs 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, weβd like to say something about the relative lengths of the vectors π΄πΉ 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·πΈπΉ is a parallelogram. Notice that we havenβt 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 πΉπΈ and the transversal π΄πΆ, we can
see that these two angles are corresponding and hence equal. And with the transversal π΅πΆ, 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΅πΆ β thatβs the big
triangle β and triangle πΉπΈπΆ β thatβs 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΅πΆ is five π₯ and the length of the
corresponding side πΈπΆ 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΄πΆ have length five π¦, then the length of the
corresponding side πΉπΆ to have the ratio five to two must be two π¦. And this means that the vector π΄πΉ whose length we really want to find must be three
π¦.

Now weβre ready to find the value of π in π΄πΉ equals π times π. The vector π΄πΉ has a length of three π¦, whereas the vector lowercase π has a
length of two π¦. So π΄πΉ is three over two times as long as π. But π is negative as π΄πΉ points in the opposite direction to the vector lowercase
π. And so, the vector π΄πΉ is not three over two π, but negative three over two π.

Now, we can substitute this for π΄πΉ in our expression for π΄πΈ. And of course, we do the same for the vector πΉπΈ. We get negative three over two π minus two-thirds π plus π. Letβs 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΄πΈ 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.