# 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 π΄π΅πΆπ·πΈπΉ. π is the centre of the hexagon. The vector ππ΄ is equal to π, and the vector ππ΅ is equal to π. Find the vector πΈπ΅ in terms of π.

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 wonβt be a tube that does that exact journey. Instead, Iβll have to take two separate tubes. Iβ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 πΈ to π΅. We can see that, to get from πΈ to π΅, we can perform two journeys. The first part of that journey takes us from πΈ to π, and the second part of that journey takes us from π to π΅. In vector form, we say that the vector πΈπ΅ is equal to the vector πΈπ plus the vector ππ΅. Weβre already told that the vector ππ΅ is given by π. So we need to work out the vector that describes πΈπ.

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 πΈπ is the same length as the line ππ΅ and itβs travelling in the same direction, we could say that itβs parallel. In fact, itβs called co-linear because it lies on the same line. But what that tells us is that the vector that describes πΈπ must be the same as that which describes ππ΅. The vector πΈπ is simply π. We can simplify this expression by collecting like terms to get that the vector πΈπ΅ is equal to two π.

Find the vector π΄π΅ in terms of π and π.

There are no direct labelled journeys that will take us from π΄ to π΅. Instead, weβll need to find an alternative route. To get from π΄ to π΅, we could first go from π΄ to π and then from π to π΅. In vector form, we say that the vector π΄π΅ then is equal to the vector π΄π plus the vector ππ΅.

Weβll need to be a little bit careful when weβre trying to find the vector that describes π΄π. The line itself is labelled with the vector π΄, but weβre going against the direction marked. A change in direction leads to a change in sign. In this case, the vector that describes π΄π is negative π. We were told that the vector ππ΅ is simply π, so the vector π΄π΅ is given by negative π plus π. Itβs useful to know that that could also be written as π minus π.

Find the vector π΅πΉ in terms of π and π.

The vector π΅πΉ 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 π΅ to πΉ, we could go from π΅ to π΄ and then from π΄ to πΉ. In vector form, we can say that the vector π΅πΉ is equal to the vector π΅π΄ plus the vector π΄πΉ.

Itβs worked out that the vector that takes us from π΄ to π΅ is negative π plus π. To get from π΅ to π΄, weβll follow the same vector, but in the opposite direction. Remember, a change in direction leads to a change in sign. So the vector π΅π΄ is π minus π. We can see that the line π΄πΉ is parallel to the line π΅π. The vector π΄πΉ is travelling in the same direction as the vector π΅π. Itβs against the direction marked. A change in direction means a change in sign. So the vector π΄πΉ is given by negative π. Simplifying by collecting like terms, we get that the vector π΅πΉ is equal to π minus two π.

Now itβs important to notice that that isnβt necessarily the only route we couldβve taken to get from π΅ to πΉ. Instead, we couldβve started our journey by travelling from π΅ to π, then from π to π΄, and then from π΄ to πΉ. That would have been negative π plus π minus π, which once again simplifies to π minus two π.