### Video Transcript

The diagram shows a quadrilateral π΄π΅πΆπ·. The vector π΄π΅ is equal to six π, the vector π΅πΆ is equal to three π, and the
vector πΆπ· is equal to two π minus seven π. The point π lies on the line π΄πΆ with π΄π to ππΆ is equal to two to one. The point π lies on the line π΄π· with π΄π to ππ· is equal to one to three. Find the vector ππ in terms of π and π, giving your answer in its simplest
form.

Letβs first consider where this point π is. It lies somewhere along the line π΄πΆ, dividing it up into the ratio two to one. This means that the point π is two-thirds of the way along the line π΄πΆ. Letβs consider also where the point π is. Weβre told it lies on the line π΄π· and divides the line π΄π· in the ratio one to
three. This means that the point π is one-quarter of the way along π΄π·.

Weβre asked to find the vector ππ in terms of π and π. We need to consider a route that we can take to get from π to π and then consider
what each part of the route is in terms of π and π. There are a number of different possibilities for the route that we could take. Iβm going to choose to get from π to π by going via the vertex π΄ of the
quadrilateral. So ππ is equal to ππ΄ plus π΄π.

I now need to find expressions for each of these two vectors in terms of π and
π. Letβs think about the vector ππ΄ first of all. Now, we need to be careful with the direction here. So ππ΄ is equal to two-thirds of the vector πΆπ΄ if I was travelling almost right to
left across the page. However, I donβt know what the vector πΆπ΄ is directly. So I need to go around the shape. Instead of going across the shape, I can get from πΆ to π΄ by going around the edge β
so going from πΆ to π΅ and then π΅ to π΄.

I do have expressions for each of these two vectors. But as Iβm travelling in the opposite direction to the vectors themselves, the signs
need to be changed. As the vector π΅πΆ is three π, the vector πΆπ΅ will be negative three π. And as the vector π΄π΅ is six π, the vector π΅π΄ will be negative six π. So we have that ππ΄ is equal to two-thirds of negative three π minus six π.

The numbers in this vector can be simplified. Two-thirds multiplied by negative three is negative two and two-thirds multiplied by
negative six is negative four. So ππ΄ simplifies to negative two π minus four π. So thatβs giving me the first vector in the route I want to take. And now, I need to think about how to find the vector π΄π.

Remember π divides the line π΄π· in the ratio one to three. And therefore, the vector π΄π is one-quarter of the vector π΄π·. I donβt have an expression for the vector π΄π· directly. But I can get from π΄ to π· by going around the other three sides of the
quadrilateral. I can go from π΄ to π΅ and then from π΅ to πΆ and then from πΆ to π·. And I do know the vectors for all of these. Iβm also travelling in the same direction as each of these vectors have been
specified. And therefore, the signs are the same. I have one-quarter of six π β thatβs for π΄π΅ β plus three π β thatβs for π΅πΆ β
plus two π minus seven π, which is for πΆπ·.

Simplifying this vector gives one-quarter of eight π minus four π. And then dividing both of the coefficients by four, this fully simplifies to two π
minus π. Iβve now found both of the two vectors I was looking for: ππ΄ and π΄π. So I can substitute the expressions into this sum in order to find the vector
ππ. ππ is, therefore, equal to negative two π minus four π plus two π minus π.

Finally, Iβve been asked to give my answer in its simplest form. Negative four π plus two π is negative two π and negative two π minus π is
negative three π. The vector ππ is, therefore, equal to negative two π minus three π.