### Video Transcript

In the diagram, ππ΄ is equal to
πΆπ΅ and the line ππ΄ is parallel to the line πΆπ΅. The vector from point π to point
π΄ is equal to vector π, and the vector from point π to point π΅ is equal to
vector π. The point πΎ lies on the line π΄πΆ
such that the ratio π΄πΎ to πΎπΆ is equal to one to three. Find an expression for the vector
π΄πΎ in terms of vectors π and π.

Now the first thing to mention is
that, in typeset papers, vectors like π and π are written in bold casing. Well, we canβt really do that very
much in handwriting, so we tend to underline them instead. So Iβm just gonna underline the
vectors π and π like this in all of my working out.

Now the first thing to notice in
the question is that the distance ππ΄ is equal to the distance πΆπ΅ and that the
line ππ΄ is parallel to the line πΆπ΅. And this tells us that this line
from π to π΄, representing vector π, is the same length and direction as this line
from πΆ to π΅, representing the vector πΆ to π΅. In other words, the vector from πΆ
to π΅ can be written as vector π.

Now the question asks us to find
out about this vector here from π΄ to πΎ in terms of vectors π and π. Now at this stage, we donβt
actually know how to do that. But what you might spot if you look
carefully is if we start at π΄, we can find our way down to point πΆ using vectors
π and π. And we can do that in three
stages. We can go from π΄ to π and from π
to π΅ and finally from π΅ to πΆ. And I picked that route
specifically because it used vectors π, π, and then vector π again.

Now remember, vector π is going in
this direction. So the journey from point π΄ to
point π is going negatively along that vector, so thatβs the negative of vector
π. Then to get from point π to π΅, we
go positively along vector π, so thatβs adding vector π. And then to get from point π΅ to
point πΆ, weβre going negatively along vector π, so thatβs negative vector π.

Now I can gather like terms. Iβve got negative π and another
negative π, so thatβs negative two π, and Iβve got a positive π. So I can rewrite that as π minus
two π, or negative two π plus π. So weβve expressed the journey from
π΄ to πΆ in terms of vectors π and π.

But we only actually wanted to
express the journey from π΄ to πΎ in terms of π and π. And if you remember, the question
told us that the ratio of the distances of π΄πΎ to πΎπΆ is one to three. And that means the distance from πΎ
to πΆ is three times bigger than the distance from π΄ to πΎ. And in fact, it means that the
distance from π΄ to πΎ is only a quarter of the distance from π΄ to πΆ.

Look, this distance π΄ to πΎ here
is only one out of one plus three β thatβs four β so one out of four parts of the
distance from π΄ to πΆ. And we can write that like this:
the vector from π΄ to πΎ is only a quarter of the vector from π΄ to πΆ.

But we just worked out that vector
π΄πΆ can be represented by the expression π minus two π. So we can replace π΄πΆ in here with
π minus two π. And if we multiply through by a
quarter, that gives us that vector π΄πΎ is equal to a quarter of vector π minus a
quarter times two times vector π. Well, a quarter times two is the
same as a quarter times two over one. And we can divide top and bottom by
two, so cancelling that down. We end up with one times one on the
top over two times one on the bottom. Thatβs a half. So that gives us our answer. The vector from π΄ to πΎ in terms
of vectors π and π is a quarter π minus a half π.