### Video Transcript

Triangle πππ is shown below. π is π point on ππ such that ππ to ππ is equal to one to two. π is π point on ππ such that ππ to ππ is equal to one to three. The vector ππ is equal to six π, and the vector ππ is equal to two π minus two π.

Part a) Work out the vector ππ in terms of π and π. Give your answer in its simplest form.

A vector is a way of describing a journey. They can describe a direction and a distance. We arenβt always given a vector to describe a specific path. So weβll need to find alternative ways of doing that same journey.

The first vector weβre given is ππ. Thatβs the journey from π to π as shown. Remember, the direction matters, so we add an arrow to show that weβre travelling from π to π, not from π to π. And thatβs six π. The vector from π to π is this journey as shown. And thatβs two π minus two π. Weβre trying to get from π to π. Thatβs this path. We donβt have a vector that takes us directly on this journey, but we can use some of the information in the question to work it out.

Before we do though, letβs find a suitable path that we can use. To get from π to π, weβll first travel from π to π. Then weβll travel from π to π. We say then that the vector ππ is equal to the vector ππ plus the vector ππ. We know the vector that will take us from π to π. We need to go in the opposite direction. So letβs consider what the vector that will take us from π to π would be. To change direction, we change the sign. Thatβs basically multiplying by negative one. And the vector ππ is negative six π.

We are also told that the ratio π to π to π to π is one to two. This means we can split the line π to π into three equal parts, where ππ represents one of those parts. This means the vector ππ must be a third of the vector ππ. Thatβs a third of negative six π. A third of six is two, so a third of negative six π is negative two π.

Next, weβre going to need to find the vector ππ. But we donβt know what the vector ππ is yet. Instead, weβll have to travel around the triangle. Weβll go from π to π, and then weβll go from π to π. Thatβs six π minus two π minus two π. And the reason that weβre subtracting two π minus two π is because weβre travelling against the direction of the vector. Instead of travelling from π to π, weβre travelling from π to π. And we said that, in that case, we need to change the sign. We need to multiply by negative one.

When we expand this bracket, we get six π minus two π. But then minus minus two π is plus two π. And this simplifies to eight π minus two π. Weβre told that the ratio of π to π to π to π is equal to one to three. This time then, weβre going to split the line ππ into four equal parts. And ππ, the vector that we need, represents one of these parts. That means ππ is a quarter of the vector ππ. Itβs one-quarter of eight π minus two π.

We can simplify this expression by expanding just like in algebra. A quarter of eight is two, so a quarter of eight π is two π. And a quarter of two is one-half, so a quarter of negative two π is negative a half π. And we can now see our vector ππ is negative two π plus two π minus a half π. Negative two π plus two π though is zero. So the vector ππ is negative a half π.

Part b) Use your answer to a) to explain why ππ is not parallel to ππ.

Lines are parallel when they travel in the same direction. If vectors are parallel, that means theyβre scalar multiples of one another.

Letβs begin then by finding the vector ππ and the vector ππ. We just calculated the vector ππ is equal to negative a half π. And weβre told in the question that the vector ππ is two π minus two π. These are not multiples of one another. Thereβs no number I can multiply the ππ or ππ by to get the other vector. Since theyβre not scalar multiples of one another, the lines ππ and ππ cannot be parallel.