# Video: AQA GCSE Mathematics Higher Tier Pack 2 β’ Paper 3 β’ Question 24

Triangle πππ is shown below. π is π point on ππ such that ππ : ππ = 1 : 2. π is π point on ππ such that ππ : ππ = 1 : 3. The vector ππ = 6π. The vector ππ = 2π β2π. a) Work out the vector ππ in terms of π and π. Give your answer in its simplest form. b) Use your answer to (a) to explain why ππ is not parallel to ππ.

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### Video Transcript

Triangle πππ is shown below. π is π point on ππ such that ππ to ππ is equal to one to two. π is π point on ππ such that ππ to ππ is equal to one to three. The vector ππ is equal to six π, and the vector ππ is equal to two π minus two π.

Part a) Work out the vector ππ 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 arenβt always given a vector to describe a specific path. So weβll need to find alternative ways of doing that same journey.

The first vector weβre given is ππ. Thatβs the journey from π to π as shown. Remember, the direction matters, so we add an arrow to show that weβre travelling from π to π, not from π to π. And thatβs six π. The vector from π to π is this journey as shown. And thatβs two π minus two π. Weβre trying to get from π to π. Thatβs this path. We donβt 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, letβs find a suitable path that we can use. To get from π to π, weβll first travel from π to π. Then weβll travel from π to π. We say then that the vector ππ is equal to the vector ππ plus the vector ππ. We know the vector that will take us from π to π. We need to go in the opposite direction. So letβs consider what the vector that will take us from π to π would be. To change direction, we change the sign. Thatβs basically multiplying by negative one. And the vector ππ 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 ππ represents one of those parts. This means the vector ππ must be a third of the vector ππ. Thatβs a third of negative six π. A third of six is two, so a third of negative six π is negative two π.

Next, weβre going to need to find the vector ππ. But we donβt know what the vector ππ is yet. Instead, weβll have to travel around the triangle. Weβll go from π to π, and then weβll go from π to π. Thatβs six π minus two π minus two π. And the reason that weβre subtracting two π minus two π is because weβre travelling against the direction of the vector. Instead of travelling from π to π, weβre 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. Weβre told that the ratio of π to π to π to π is equal to one to three. This time then, weβre going to split the line ππ into four equal parts. And ππ, the vector that we need, represents one of these parts. That means ππ is a quarter of the vector ππ. Itβs 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 ππ is negative two π plus two π minus a half π. Negative two π plus two π though is zero. So the vector ππ is negative a half π.

Part b) Use your answer to a) to explain why ππ is not parallel to ππ.

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

Letβs begin then by finding the vector ππ and the vector ππ. We just calculated the vector ππ is equal to negative a half π. And weβre told in the question that the vector ππ is two π minus two π. These are not multiples of one another. Thereβs no number I can multiply the ππ or ππ by to get the other vector. Since theyβre not scalar multiples of one another, the lines ππ and ππ cannot be parallel.