Video: Pack 3 β€’ Paper 2 β€’ Question 20

Pack 3 β€’ Paper 2 β€’ Question 20


Video Transcript

𝑂𝐴𝐡 is a triangle. 𝑀 is the point on 𝐴𝐡 such that 𝐡𝑀 to 𝑀𝐴 is equal to two to three. 𝑂𝐴 is equal to three 𝑒. 𝑂𝐡 is equal to 𝑣. 𝑂𝑀 is equal to π‘˜ multiplied by two 𝑒 plus 𝑣, 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 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 single straight line.

In the case of this question, the journey I want to take is from 𝑂 to 𝑀. The journey I’ll need to actually take is to go from 𝑂 to 𝐡 and then 𝐡 to 𝑀 which is part of the journey from 𝐡 to 𝐴. We’ll start then by considering the journey from 𝐡 to 𝐴.

There are no vectors to describe that journey. So we’ll need to travel on alternative way. We’ll follow the lines which are described with vectors. The first line we’ll follow takes us from 𝐡 to 𝑂. Then, the second line takes us from 𝑂 to 𝐴. That gives us the entire journey from 𝐡 to 𝐴.

Let’s begin with the journey from 𝐡 to 𝑂. We can see that the arrow going from 𝑂 to 𝐡 is marked with the vector 𝑣. If we want to travel from 𝐡 to 𝑂, this is the opposite direction. So we change the sign of the vector. The journey from 𝐡 to 𝑂 is, therefore, negative 𝑣.

Next, we want to go from 𝑂 to 𝐴. That’s given as three 𝑒. Our total journey then is given by 𝐡 to 𝑂 and then 𝑂 to 𝐴. It’s negative 𝑣 plus three 𝑒, which could be written as three 𝑒 minus 𝑣. Remember though we didn’t want to go all the way from 𝐡 to 𝐴. We actually only wanted to travel part of this journey to 𝑀. We’re given that 𝐡𝑀 to 𝑀𝐴 is two to three.

That means the journey of 𝐡 to 𝐴 can be split into five total parts. 𝐡𝑀 is two of these five parts. It’s two-fifths of the journey from 𝐡 to 𝐴. Since 𝐡 to 𝐴 is given by three 𝑒 minus 𝑣, 𝐡𝑀 must be given then by two-fifths of three 𝑒 minus 𝑣. Now that we have the journey from 𝐡 to 𝑀, we can write down the entire journey from 𝑂 to 𝑀.

To get from 𝑂 to 𝑀, we’ll travel from 𝑂 to 𝐡 and then 𝐡 to 𝑀. The vector 𝑂𝐡 is given as 𝑣. And we calculated the vector 𝐡𝑀 to be two-fifths of three 𝑒 minus 𝑣. So the journey 𝑂𝑀 is given by 𝑣 plus two-fifths of three 𝑒 minus 𝑣.

Next, we’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 𝑒 is six-fifths 𝑒. Two-fifths multiplied by negative 𝑣 is negative two-fifths 𝑣.

Finally, we want to simplify this expression by collecting like terms. To do this, we need to make the denominators for this vector 𝑣 the same. 𝑣 is the same as five-fifths 𝑣. Five-fifths 𝑣 minus two-fifth 𝑣 is three-fifths 𝑣. Our expression for the vector 𝑂𝑀 simplifies then to three-fifths 𝑣 plus six-fifths 𝑒.

The problem is this doesn’t look much like the expression given in the question. To make the vector three-fifths 𝑣 plus six-fifth 𝑒 look like what we’ve got in the question, first, we’ll change the order. Three-fifths 𝑣 plus six-fifths 𝑒 is exactly the same as six-fifths 𝑒 plus three-fifths 𝑣.

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 𝑒 is six-fifths 𝑒 and three-fifths multiplied by 𝑣 is three-fifths 𝑣. So the inside of our brackets becomes two 𝑒 plus 𝑣.

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, it’s the number multiplying two 𝑒 plus 𝑣, π‘˜ must be equal to three-fifths.

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