Video: GCSE Mathematics Foundation Tier Pack 1 β€’ Paper 1 β€’ Question 27

GCSE Mathematics Foundation Tier Pack 1 β€’ Paper 1 β€’ Question 27

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Video Transcript

Consider the regular hexagon 𝐴𝐡𝐢𝐷𝐸𝐹. 𝑂 is the centre of the hexagon. The vector 𝑂𝐴 is equal to π‘Ž, and the vector 𝑂𝐡 is equal to 𝑏. Find the vector 𝐸𝐡 in terms of 𝑏.

Vectors can seem really intimidating. But I like to think of them as being a bit like the tube system in London. Sometimes I want to take a tube from one place to another, but there won’t be a tube that does that exact journey. Instead, I’ll have to take two separate tubes. I’ll still end up getting to my destination, just not in a direct line.

In the case of this question, the journey I want to take is from 𝐸 to 𝐡. We can see that, to get from 𝐸 to 𝐡, we can perform two journeys. The first part of that journey takes us from 𝐸 to 𝑂, and the second part of that journey takes us from 𝑂 to 𝐡. In vector form, we say that the vector 𝐸𝐡 is equal to the vector 𝐸𝑂 plus the vector 𝑂𝐡. We’re already told that the vector 𝑂𝐡 is given by 𝑏. So we need to work out the vector that describes 𝐸𝑂.

Now a little knowledge of the geometry of a regular hexagon is useful here, though this can be inferred from the diagram. Not only are the opposite sides in a regular hexagon parallel and of equal length, but also its radii β€” a bit like a circle, the radius of a hexagon is the line from its centre to each of its vertices β€” are also parallel to two of its sides and of equal length.

Now if a vector is parallel and of equal length, then we can just use the same vector to describe it. Since the line 𝐸𝑂 is the same length as the line 𝑂𝐡 and it’s travelling in the same direction, we could say that it’s parallel. In fact, it’s called co-linear because it lies on the same line. But what that tells us is that the vector that describes 𝐸𝑂 must be the same as that which describes 𝑂𝐡. The vector 𝐸𝑂 is simply 𝑏. We can simplify this expression by collecting like terms to get that the vector 𝐸𝐡 is equal to two 𝑏.

Find the vector 𝐴𝐡 in terms of π‘Ž and 𝑏.

There are no direct labelled journeys that will take us from 𝐴 to 𝐡. Instead, we’ll need to find an alternative route. To get from 𝐴 to 𝐡, we could first go from 𝐴 to 𝑂 and then from 𝑂 to 𝐡. In vector form, we say that the vector 𝐴𝐡 then is equal to the vector 𝐴𝑂 plus the vector 𝑂𝐡.

We’ll need to be a little bit careful when we’re trying to find the vector that describes 𝐴𝑂. The line itself is labelled with the vector 𝐴, but we’re going against the direction marked. A change in direction leads to a change in sign. In this case, the vector that describes 𝐴𝑂 is negative π‘Ž. We were told that the vector 𝑂𝐡 is simply 𝑏, so the vector 𝐴𝐡 is given by negative π‘Ž plus 𝑏. It’s useful to know that that could also be written as 𝑏 minus π‘Ž.

Find the vector 𝐡𝐹 in terms of π‘Ž and 𝑏.

The vector 𝐡𝐹 might look a bit tricky to begin with. However, we said earlier that if a vector was parallel and of the same length, then it would have the same vector. To get from 𝐡 to 𝐹, we could go from 𝐡 to 𝐴 and then from 𝐴 to 𝐹. In vector form, we can say that the vector 𝐡𝐹 is equal to the vector 𝐡𝐴 plus the vector 𝐴𝐹.

It’s worked out that the vector that takes us from 𝐴 to 𝐡 is negative π‘Ž plus 𝑏. To get from 𝐡 to 𝐴, we’ll follow the same vector, but in the opposite direction. Remember, a change in direction leads to a change in sign. So the vector 𝐡𝐴 is π‘Ž minus 𝑏. We can see that the line 𝐴𝐹 is parallel to the line 𝐡𝑂. The vector 𝐴𝐹 is travelling in the same direction as the vector 𝐡𝑂. It’s against the direction marked. A change in direction means a change in sign. So the vector 𝐴𝐹 is given by negative 𝑏. Simplifying by collecting like terms, we get that the vector 𝐡𝐹 is equal to π‘Ž minus two 𝑏.

Now it’s important to notice that that isn’t necessarily the only route we could’ve taken to get from 𝐡 to 𝐹. Instead, we could’ve started our journey by travelling from 𝐡 to 𝑂, then from 𝑂 to 𝐴, and then from 𝐴 to 𝐹. That would have been negative 𝑏 plus π‘Ž minus 𝑏, which once again simplifies to π‘Ž minus two 𝑏.

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