# Video: Pack 3 β’ Paper 2 β’ Question 20

Pack 3 β’ Paper 2 β’ Question 20

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### Video Transcript

ππ΄π΅ is a triangle. π is the point on π΄π΅ such that π΅π to ππ΄ is equal to two to three. ππ΄ is equal to three π’. ππ΅ is equal to π£. ππ is equal to π multiplied by two π’ plus π£, where π is a scalar quantity. Find the value of π.

I like to think of vectors 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 single straight line.

In the case of this question, the journey I want to take is from π to π. The journey Iβll need to actually take is to go from π to π΅ and then π΅ to π which is part of the journey from π΅ to π΄. Weβll start then by considering the journey from π΅ to π΄.

There are no vectors to describe that journey. So weβll need to travel on alternative way. Weβll follow the lines which are described with vectors. The first line weβll follow takes us from π΅ to π. Then, the second line takes us from π to π΄. That gives us the entire journey from π΅ to π΄.

Letβs begin with the journey from π΅ to π. We can see that the arrow going from π to π΅ is marked with the vector π£. If we want to travel from π΅ to π, this is the opposite direction. So we change the sign of the vector. The journey from π΅ to π is, therefore, negative π£.

Next, we want to go from π to π΄. Thatβs given as three π’. Our total journey then is given by π΅ to π and then π to π΄. Itβs negative π£ plus three π’, which could be written as three π’ minus π£. Remember though we didnβt want to go all the way from π΅ to π΄. We actually only wanted to travel part of this journey to π. Weβre given that π΅π to ππ΄ is two to three.

That means the journey of π΅ to π΄ can be split into five total parts. π΅π is two of these five parts. Itβs two-fifths of the journey from π΅ to π΄. Since π΅ to π΄ is given by three π’ minus π£, π΅π must be given then by two-fifths of three π’ minus π£. Now that we have the journey from π΅ to π, we can write down the entire journey from π to π.

To get from π to π, weβll travel from π to π΅ and then π΅ to π. The vector ππ΅ is given as π£. And we calculated the vector π΅π to be two-fifths of three π’ minus π£. So the journey ππ is given by π£ plus two-fifths of three π’ minus π£.

Next, weβll expand this bracket. Two-fifths multiplied by three is the same as two-fifths multiplied by three over one. Two times three is six and five times one is five. So two-fifths multiplied by three is six-fifths. That means then that two-fifths multiplied by three π’ is six-fifths π’. Two-fifths multiplied by negative π£ is negative two-fifths π£.

Finally, we want to simplify this expression by collecting like terms. To do this, we need to make the denominators for this vector π£ the same. π£ is the same as five-fifths π£. Five-fifths π£ minus two-fifth π£ is three-fifths π£. Our expression for the vector ππ simplifies then to three-fifths π£ plus six-fifths π’.

The problem is this doesnβt look much like the expression given in the question. To make the vector three-fifths π£ plus six-fifth π’ look like what weβve got in the question, first, weβll change the order. Three-fifths π£ plus six-fifths π’ is exactly the same as six-fifths π’ plus three-fifths π£.

Next, to find the value of π, we need to factorize this expression. The highest common factor of six-fifths and three-fifths is three-fifths. So three-fifths goes on the outside of our bracket. Three-fifths multiplied by two π’ is six-fifths π’ and three-fifths multiplied by π£ is three-fifths π£. So the inside of our brackets becomes two π’ plus π£.

Notice now that our expression is the same form as the one given to us in the question. Since π is the number on the outside of the bracket, itβs the number multiplying two π’ plus π£, π must be equal to three-fifths.