# Video: Pack 1 β’ Paper 1 β’ Question 19

Pack 1 β’ Paper 1 β’ Question 19

04:16

### Video Transcript

In the diagram, ππ΄ is equal to πΆπ΅ and the line ππ΄ is parallel to the line πΆπ΅. 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 πΎ lies on the line π΄πΆ such that the ratio π΄πΎ to πΎπΆ is equal to one to three. Find an expression for the vector π΄πΎ 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 canβt really do that very much in handwriting, so we tend to underline them instead. So Iβm 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 ππ΄ is equal to the distance πΆπ΅ and that the line ππ΄ is parallel to the line πΆπ΅. And this tells us that this line from π to π΄, representing vector π, is the same length and direction as this line from πΆ to π΅, representing the vector πΆ to π΅. In other words, the vector from πΆ to π΅ can be written as vector π.

Now the question asks us to find out about this vector here from π΄ to πΎ in terms of vectors π and π. Now at this stage, we donβt 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 πΆ 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 πΆ. 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 thatβs the negative of vector π. Then to get from point π to π΅, we go positively along vector π, so thatβs adding vector π. And then to get from point π΅ to point πΆ, weβre going negatively along vector π, so thatβs negative vector π.

Now I can gather like terms. Iβve got negative π and another negative π, so thatβs negative two π, and Iβve got a positive π. So I can rewrite that as π minus two π, or negative two π plus π. So weβve expressed the journey from π΄ to πΆ in terms of vectors π and π.

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

Look, this distance π΄ to πΎ here is only one out of one plus three β thatβs four β so one out of four parts of the distance from π΄ to πΆ. And we can write that like this: the vector from π΄ to πΎ is only a quarter of the vector from π΄ to πΆ.

But we just worked out that vector π΄πΆ can be represented by the expression π minus two π. So we can replace π΄πΆ in here with π minus two π. And if we multiply through by a quarter, that gives us that vector π΄πΎ 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. Thatβs a half. So that gives us our answer. The vector from π΄ to πΎ in terms of vectors π and π is a quarter π minus a half π.