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

Triangle π‘‹π‘Œπ‘ is shown below. 𝑃 is π‘Ž point on 𝑍𝑋 such that 𝑍𝑃 : 𝑃𝑋 = 1 : 2. 𝑄 is π‘Ž point on π‘π‘Œ such that 𝑍𝑄 : π‘„π‘Œ = 1 : 3. The vector 𝑍𝑋 = 6𝑏. The vector π‘Œπ‘‹ = 2π‘Ž βˆ’2𝑏. a) Work out the vector 𝑃𝑄 in terms of π‘Ž and 𝑏. Give your answer in its simplest form. b) Use your answer to (a) to explain why 𝑃𝑄 is not parallel to π‘Œπ‘‹.

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Video Transcript

Triangle π‘‹π‘Œπ‘ is shown below. 𝑃 is π‘Ž point on 𝑍𝑋 such that 𝑍𝑃 to 𝑃𝑋 is equal to one to two. 𝑄 is π‘Ž point on π‘π‘Œ such that 𝑍𝑄 to π‘„π‘Œ is equal to one to three. The vector 𝑍𝑋 is equal to six 𝑏, and the vector π‘Œπ‘‹ is equal to two π‘Ž minus two 𝑏.

Part a) Work out the vector 𝑃𝑄 in terms of π‘Ž and 𝑏. 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 aren’t always given a vector to describe a specific path. So we’ll need to find alternative ways of doing that same journey.

The first vector we’re given is 𝑍𝑋. That’s the journey from 𝑍 to 𝑋 as shown. Remember, the direction matters, so we add an arrow to show that we’re travelling from 𝑍 to 𝑋, not from 𝑋 to 𝑍. And that’s six 𝑏. The vector from π‘Œ to 𝑋 is this journey as shown. And that’s two π‘Ž minus two 𝑏. We’re trying to get from 𝑃 to 𝑄. That’s this path. We don’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, let’s find a suitable path that we can use. To get from 𝑃 to 𝑄, we’ll first travel from 𝑃 to 𝑍. Then we’ll travel from 𝑍 to 𝑄. We say then that the vector 𝑃𝑄 is equal to the vector 𝑃𝑍 plus the vector 𝑍𝑄. We know the vector that will take us from 𝑍 to 𝑋. We need to go in the opposite direction. So let’s consider what the vector that will take us from 𝑋 to 𝑍 would be. To change direction, we change the sign. That’s basically multiplying by negative one. And the vector 𝑋𝑍 is negative six 𝑏.

We are also told that the ratio 𝑍 to 𝑃 to 𝑃 to 𝑋 is one to two. This means we can split the line 𝑍 to 𝑋 into three equal parts, where 𝑍𝑃 represents one of those parts. This means the vector 𝑃𝑍 must be a third of the vector 𝑋𝑍. That’s a third of negative six 𝑏. A third of six is two, so a third of negative six 𝑏 is negative two 𝑏.

Next, we’re going to need to find the vector 𝑍𝑄. But we don’t know what the vector π‘π‘Œ is yet. Instead, we’ll have to travel around the triangle. We’ll go from 𝑍 to 𝑋, and then we’ll go from 𝑋 to π‘Œ. That’s six 𝑏 minus two π‘Ž minus two 𝑏. And the reason that we’re subtracting two π‘Ž minus two 𝑏 is because we’re travelling against the direction of the vector. Instead of travelling from π‘Œ to 𝑋, we’re travelling from 𝑋 to π‘Œ. 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 𝑏 minus two π‘Ž. But then minus minus two 𝑏 is plus two 𝑏. And this simplifies to eight 𝑏 minus two π‘Ž. We’re told that the ratio of 𝑍 to 𝑄 to 𝑄 to π‘Œ is equal to one to three. This time then, we’re going to split the line π‘π‘Œ into four equal parts. And 𝑍𝑄, the vector that we need, represents one of these parts. That means 𝑍𝑄 is a quarter of the vector π‘π‘Œ. It’s one-quarter of eight 𝑏 minus two π‘Ž.

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

Part b) Use your answer to a) to explain why 𝑃𝑄 is not parallel to π‘Œπ‘‹.

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

Let’s begin then by finding the vector 𝑃𝑄 and the vector π‘Œπ‘‹. We just calculated the vector 𝑃𝑄 is equal to negative a half π‘Ž. And we’re told in the question that the vector π‘Œπ‘‹ is two π‘Ž minus two 𝑏. These are not multiples of one another. There’s no number I can multiply the 𝑃𝑄 or π‘Œπ‘‹ by to get the other vector. Since they’re not scalar multiples of one another, the lines 𝑃𝑄 and π‘Œπ‘‹ cannot be parallel.

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