Video: Pack 4 β’ Paper 2 β’ Question 17

Pack 4 β’ Paper 2 β’ Question 17

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Video Transcript

πππ΄, πππ΅, and π΄π΅π are straight lines. Four ππ equals ππ΄. π΄π equals three over two π΄π΅. Vector ππ΄ equals π. Vector ππ΅ equals π. πππ is a straight line. Prove that π is the midpoint of ππ΅.

So the first thing I want to actually draw your attention towards is actually some notation. So as you can see, Iβve actually written here π with a line underneath and π with a line underneath. And we can actually write these instead of bold π and bold π. The reason Iβve done this is one: itβs easy for me to write; two: itβs easy for you to see. And then also when youβre doing an exam, itβs actually easy for you to write and be clear for an examiner. Okay, it also means the exact same thing.

Now, letβs move on and actually solve the problem. So to start off with, what we can look at is that four ππ is equal to ππ΄. So therefore, this will tell us that the vector ππ is going to equal to a quarter of the vector ππ΄. So therefore, this is gonna be equal to a quarter π cause weβve already been told that the vector ππ΄ is equal to π.

So next, what weβre gonna do is to take a look at the vector π΄π΅. So thereβs going from π΄ to π΅. Well, to go from π΄ to π΅, using the vectors that we know, we go from π΄ to π and then π to π΅. So vector π΄π΅ is equal to π΄π plus ππ΅, which is gonna be equal to vector ππ΅ minus vector ππ΄. And thatβs because we actually went in the opposite direction of our way. So we actually went from π΄ to π, which can also be written as negative ππ΄. So therefore, this is gonna be equal to π minus π. And thatβs because weβre told that vector ππ΄ is equal to π and vector ππ΅ is equal to π.

So now, what we can do is actually use a bit of information that tells us that π΄π is equal three over two π΄π΅ because, therefore, vector π΄π is gonna be equal to three over two vector π΄π΅. And weβve actually just found vectors π΄π΅. So therefore, we can say that vector π΄π is gonna be equal to three over two π minus π. So great, we found that. And now, what we can do is actually move on to the next step.

Now, the next step is actually to take a look at our π because what we can say about the vector ππ is this is gonna be equal to π, which is just a constant, multiplied by the vector ππ΅. And we know that because weβre told at the beginning that πππ΅ is actually a straight line. So therefore, as we know that vector ππ΅ is equal to π, we can actually say that vector ππ is gonna be equal to ππ. And itβs this π that we want to find because if we find this π, itβs actually gonna help us to prove that π is the midpoint of ππ΅.

Well, now, letβs actually consider vector ππ. Well, vector ππ is gonna be equal to vector ππ΄ plus vector π΄π. Well, these are actually both vectors that we know. So therefore, we can say that vector ππ is gonna be equal to π plus three over two π minus π.

So now, what we can actually do is actually consider the vector ππ. And what Iβve done to highlight this is actually drawing it with a dashed pink line. Well, the vector ππ is actually gonna be equal to vector ππ minus vector ππ. And weβve actually chosen this rule because these are both the vectors that we know cause weβve found them out. So therefore, we can say that this is actually gonna be equal to π plus three over two multiplied by π minus π then minus a quarter π. And this is because a quarter π was actually ππ, weβve found that earlier on.

So therefore, if we actually expand the bracket, what weβre actually gonna get is π plus three over two π minus three over two π minus a quarter π. And this is gonna give us the expression three over two π minus three-quarter π. And we got the three-quarter π because we had π and then we subtracted one and a half π from it, which gives us negative a half π. And then, we subtracted another quarter π from it, which gave us negative three-quarter π. So great, weβve actually done that. Now, we found vector ππ.

Well, next, what we can actually do is move on and start to think about ππ because what that ππ can do is actually help us to think about the π that weβre looking for earlier. So if we actually travel from π to π, so we take the vector ππ, then weβre actually gonna go from π to π and then π to π. So therefore, itβs gonna be equal to vector ππ minus vector ππ cause like we said we went π to π, which is actually the negative of π to π. So therefore, this is gonna give us that the vector ππ is equal to ππ minus a quarter π. And thatβs cause our vector ππ was ππ and vector ππ is actually a quarter π.

So therefore, what we can actually say is that because πππ is a straight line, then ππ is gonna be a multiple of ππ. And we can actually use that to actually help us to find out what our π is, which therefore will enable us to prove that π is the midpoint of ππ΅. So therefore, as we already said, the vector ππ is gonna be equal to π, the vector ππ, where π is actually a multiple. So weβre saying itβs gonna be a multiple of the vector ππ.

So now, we need to consider what possibly that could be. Well, if we take a look at the coefficients of our πs, then, weβve got negative three-quarters and negative a quarter. We actually have to multiply negative a quarter by three to get to negative three-quarters. So you can see that actually the vector ππ is three times the vector ππ. So we can actually say that vector ππ equals three vector ππ.

But as weβre trying to find π, thereβs actually a more useful way that we can think of this. And thatβs that a third of vector ππ is equal to the vector ππ. And weβve already proved this with the π coefficient because actually a quarter or negative quarter is actually a third of negative three-quarters. So therefore, if itβs actually a third with the coefficients of π, it must be the same with the coefficients of π.

So what we can say that actually π must be a third of three over two. And we can rewrite it like this: a third multiplied by three over two is gonna be equal to π. So therefore, we can say that three over six is equal to π because you actually have one multiplied by three is three and three multiplied by two is six. So this actually gives us a value of π of a half.

Okay, great, but how thatβs gonna help us? Well, using this bit of information, so π being equal to a half, we can say that therefore vector ππ is equal to a half vector ππ΅. And thatβs because we found out earlier that vector ππ was equal to π vector ππ΅.

So therefore, what we can actually say is weβve actually proved that π is the midpoint of ππ΅.