GCSE MATHS Vectors FULL WALKTHROUGH GRADE 5-9

hey guys I hope your maths revision is going well today I was going to do a video which is every single thing that you need to get a Grade 9 in vectors uh for your GCSE math so this will go all the way from the foundation grade five questions all the way up to grade nine Vector proofs that are quite tricky so without further Ado let's just go straight through it I also have a link in the description which will be this PDF so you can follow along with me or even better answer the questions yourself and then use this as a mark scheme so here we're starting with the foundation content they can give you vectors and expect you to add them together so if you add two vectors all you're doing is adding the top numbers adding the bottom numbers so you've got on the top two plus one which is three on the bottom three plus five which is equal to eight and if you are multiplying vectors like this you multiply the top and the bottom so two a literally just becomes two times two three plus three one five so you multiply everything inside the brackets so this now becomes two times two which is four two times three which is six plus three times one which is three three times five which is fifteen and then again you're just adding the top and bottom numbers so you should get four plus three which is seven six plus fifteen which is twenty one cool next thing they can ask you to do is translate shapes by vectors so just so you were the top number tells you how far left and right you're going if it's positive you're going to the right so this is telling me to go to the right by three and the bottom number tells me how far up and down we're going positive numbers mean up negative numbers mean down so this negative means we're going to down by two okay so you just take each point and you go in this case three to the right so one two three and then you go two down one two so that'll be right here yeah and then you repeat for the other ones so here going three to the right you're going to go one two three one two you can probably see where the next point is going to go because you have to be three points above but it'd be up here okay and then again you're drawing joining it with uh straight lines as best you can anyway perfect now we're on to some kind of vector geometry problems which are quite interesting so the diagram shows a parallelogram like this says OA is 2A and B is 2b and find in terms of a the vector d a so if I have to describe essentially how vectors work when you do geometry it's just like a game okay here are the rules you can only go along lines where you know the vectors right so I can go from o to a and I can go from o to B but I can't go from o to D unless I know what o to D is now because it's a parallelogram when you draw lines from the corners the actual size are the same length so I actually know that this is 2A and this is 2b okay and that's important for the next question but I cannot go along these sides here so when it says go from D to a I just need to go along vectors that I already know and you can take any path you like and you'll get the exact same answer so if I knew what d2c was I could do D to C then C to a and now give me the same answers going from D to O and O to a so in this case because I only know OD and O A I'm going to go D to O is 2A plus another 2A because I'm going from o to a okay so that would be 4A in total then it says find in terms of A and B the vector a b and again I can't just go straight to the right because I don't know what the vector is so in this case if I label this with a red color so the first one's in blue I need to go from a to O and then from o to B now here's the second rule so the first rule is you can only go along vectors that you know the second rule is that if you go in the opposite direction to a vector all you do is make it negative so can you see from here the arrow on this Vector is pointing from o to a but I'm going from a to O So that's going to be a negative 2A instead of a positive and then I'm going along this Vector so it's a plus 2B and then I can't simplify it because just like in algebra you collect A's and B's but in this case I only have one a one lot of a and one lot of B and again for Vector AC if I do this in like an orange color we're going from a to O and then o to C okay because again we can only go along vectors that we know and I don't know what AC is now in this case since we're going in the opposite direction to 2A we're going to make that negative 2A and we're also going in the opposite direction to 2B so it'll be negative 2 B for that question okay so that's kind of the basics that they can ask you for but you do need to use your kind of knowledge of geometry as well so they may give you certain shapes for example a parallelogram where you needed to for this question it was essential to know that the O is actually right in the middle of the shape okay then we get onto some more types of vector geometry questions okay Vector proof is slightly different Vector proof is show that okay but this we have Vector geometry but they also give us the information that m is the midpoint of a b so we need to work out in terms of A and B the vector a b so again I cannot go straight from a down to B because I don't have a vector that goes from A to B okay so I need to go again I'll label this with a blue color I need to go from a to O and then from o to B now I'm going against or in the opposite direction to 5A so that is going to be a negative 5A and I'm going in the same direction as 3B so be a plus 3 B so negative 5A plus 3 B now what about the vector a m well this is where it gets a bit more interesting sometimes they give you that a letter is the midpoint of a b so what does that mean imagine if this was just a normal question with a length right so imagine if the length a b was let's say 10 centimeters and they're asking me what would the length of a to m b if it's the midpoint well you would half it right because in the middle is going to be halfway along so in other words it's going to be five centimeters okay with vectors it's actually the exact same logic vectors do tell you lengths okay so if we use red for this question m is going to be half of a to B because it is the midpoint of a b so in just the same way that if we had a normal length of let's say 10 centimeters we would half it if we have a vector we would half the vector because again vectors are basically lengths so this is going to be a half and then all of a b which is minus 5A plus 3B and then I would personally expand the bracket so give me minus five of uh to a plus three over two B okay and that is what I would do for that question in terms of can you leave it as decimals you're more than welcome to leave it as decimals it's just that we tend to leave it as fractions as like notation okay now here's an interesting question it says how about fine in terms of A and B the vector o to M so to go from o to M I actually have two options I can go from o to a and then a to M or I can go from o to B and then B to n both vectors will give me the same result okay now I'm actually going to add on here that we now know the vector a b and a m a m is just minus five over two a plus three over two B and likewise this overall Vector here is negative 5A plus 3B okay the reason why I'm drawing that on is because I want to show you that if you do it either way you're going to get the same answer if I go from o to a and then a to M all that's going to be is because I'm going along these two vectors all I do is add them so what we have is we have 5A Plus minus 5 over 2A plus 3 over 2B okay because again we're just going halfway between A and B now if I went from o to B I'd have 3B and then from B to M I would subtract minus 5 over 2 A plus 3 over 2B which would give me the exact same answer so let's work this out properly now we have 5 minus five over two so that's 5 minus 2.5 which is still 2.5 so that would give me still 5 over 2A if I do that in a bit more detail that is going to be if we do 5 minus 5 over 2 we have to put that over 1 times top and bottom by two we get 10 over 2 minus 5 over 2. which is five over two and then of course just add three over two B because there's no I mean there's nothing else is there so 5 over 2A plus three over two B and then hopefully we stick to the blue color from now on now we're getting on to some basic Vector proofs okay now the difference is is that they'll either say find the value of something or they'll say show that and both those come under the grade eight or nine topic of vector proof so we have a bunch of this information it's written most of it's written on the diagram my first step in any Vector question is to write on the diagram any information they haven't put on there it says p is the point on a b so set a p to p b is one to three so in other words it's going to be roughly here is point P okay and again that's one to three the ratio so I'm definitely gonna try and keep that in the back of my mind and this is op is equal to K so going from all the way here to here K 3A plus b so we need to find the value of K right so what we want to do here is find another expression for o to P so if I look where OTP is it's going to be o to P's like this and then I need to know what this is so the first thing I need to do is actually work out what A to B is but if I write this out fully we have op equals a plus some fraction of a b and what fraction would that be well this is where they introduce ratios right it's saying that the ratio of a to p compared to P to B is one to three so in other words this is one part and this rest of it is three parts so can you see that AP is just a quarter of the total length right because it'd be one over three plus one so it's a quarter of a b and now all I need to do is work out what a b is so a b if I write that let's say in a red color make it a bit clearer I need to go from a to O and from o to B so I need to go negative a Plus positive B so now if I sub that into here we get a plus a quarter minus a plus B let's scroll down a bit expand our brackets if we get a minus a quarter a plus b now a whole minus a quarter would give me three quarters a plus and I've got the quarter over here plus a quarter T okay now if we look at the answer it's saying this is equal to K times 3A plus b so in other words what they've done is they've factorized out something in order to get 3A plus b on the inside of the bracket so what I'm going to do and here's my little trick is I'm going to just write 3A plus b because I know that's what my answer is now on the outside is going to be a term where If I multiply three I'll get three quarters and If I multiply one I'll get one quarter and there's a few ways to do this uh one way is to just simply divide the two things because that's what we're trying to work out right I'm basically saying three times something is equal to three quarters what would that something be so one thing you can do is just move this three to The Other Side by dividing both sides by three and you might be or you might be able to just look at this and see but hopefully you can see that the answer is a quarter three quarters divided by three is a quarter if you're still not convinced the other way I can do it is if I look at the term B I know that one times something gives me a quarter divide both sides by one I get a quarter so in other words the number outside the bracket must be a quarter and it doesn't say that K has to be a whole number it just says find the value of K so in this case k is equal to one quarter so hopefully those methods make sense if they don't again just let me know in the comments and I'll go through it again but that's how you would do that kind of question next we have again another well in this case a proper Vector proof because it's a show that and it says show that c m and d are on the same straight line so how do I show that something's on a straight line well before we get into that let's first label up our diagrams any information that's missing so OA that's labeled o b that's labeled C is the point on is such that OC to ca is four to one so in other words it's on this line o to a right I know it's saying OC to C to a it's four to one so basically OC is four times longer than C A so I'm going to say it's roughly here okay which means OC is going to be 4A and C to a is just one a okay I'll go through that again if that doesn't make sense and M is the midpoint we've already got that D is the point such that o to B and to O uh to O to D is three quarters cool so we're trying to show that c to M to D is one straight line now what you can't do is draw a line with a ruler and say a it's on a straight line what does it mean when two things or two vectors are on the same straight line or in this case three points are kind of gave it away there what you're going to try and show is that cm and C D are parallel to each other because when vectors are parallel to each other they let they lie on the same straight line okay so you're going to try and show that cm and CD are parallel okay that's fantastic but how do I show that they're parallel well what you're going to do is show that cm is a scale is some fraction of C D okay so basically what we're going to do is say cm is equal to some Factor we can call it K if you'd like of CD that's what we're trying to show okay so how do we do that well what we're going to do is find an equation for CD an equation for cm and show that we can write them in this form and that's what you do anytime you have a straight line you can show that this fraction this segment here and if you imagine it right if we have a straight line long here cm is going to be a fraction of the total length CD right that makes sense so how do we do that well I'm going to work out CM first now cm is if I label this I'm just going to label it in red for now we're going to go from C to a and go from a to n okay so there's a few things we need to work out first firstly C to a if I look at this ratio here C A is one so it's four times smaller than OC so in other words this length here this whole length is 5A then C to a is a fifth of that because all I do is I change that to a fraction which is 1 over 4 plus 1 which is five so in other words this is just a okay actually I'm going to change that to a underline okay the other way to think of it is if OC is four times bigger than c a right then this side gets 4A and this side will get 1A and it still adds up to 5A in total right an AM is since it's the midpoint is half of a b so I need to work out a b so a b and I probably should write this somewhere slightly different slightly other side a b is equal to a half sorry it's going to be negative 5A because it's equal to a to o plus 3B and again the reason it's negative 5 is because we're going in the opposite direction to it so it's negative 5A plus 3B okay so c m now is just equal to a plus a half minus 5A plus 3B let's simplify this a bit so expand the brackets we have a Plus actually be a minus in this case 5 over 2 a plus three over two B now if I want to combine these two we basically have a 1 in front of this a so one over one so I need to double it so what I'm going to do is make that 2 over 2 a minus five over two a and the reason why I've done that is so I can actually subtract the two vectors from each other so 2 minus 5 is negative 3 over 2A plus three over two B okay so I've done that so far that's all well and good now I need to find out what CD is so I'm actually going to get rid of some of this stuff on the screen just to make it a bit cleaner again in your exam you can't really do this unfortunately but you can kind of join pencil so now I need to work out an expression for c d Okay so I'm going to write just CD equals equals cool and now we need to work out how to get from C to D so again we can go from C to a which we worked on the previous part or the previous little section we did is a then we go from A to B which again we have right here and then we just need to work out B to D now if I look at the ratio here it says from o to B is three parts and from o to D is four parts so in other words this B to D should have just one part right because again it takes three to get from o to B and the total thing OD is four which means the only part left is going to be B so from B to D is just B let's add that all up so we have a plus minus 5A plus 3B plus b now let's gather that up we have minus 4A plus 4 B now there's a few different ways we can kind of solve this next part okay the first way which again once you get a Little Bit Stronger at vectors you can just look and say okay if I want to write CM as a fraction of CD I can just do CM so this Pi in front of a divided this part in front of a and this part in front of B divided this part in front of B and see if they're the same and if they're the same cool now I think that that's a bit tricky for your first time so instead we're going to now substitute into this top equation bit longer tiny bit longer but it should make a bit more sense so we have minus three over two a plus 3 over 2B equals K times c d which worked as minus 4A plus 4 d you can probably already see that they're actually going to be the same but let's expand the brackets and there's a very useful and important technique I'm about to show you which you can and you have to use for certain questions and it's a very effective tool so it's good that we learn it on a slightly easier topic now if you look at this equation we can't collect A and B's we can't combine them which means everything in front of a on the left hand side should equal everything in front of a on the right hand side right because they're equal to each other so this is the important Technique we can use we can compare A and B so if I look at a on the left hand side we have minus 3 over 2 in front of a the right side we have minus four k do the same thing for B another goal here is to show that it's the exact same number so K is the same value now first of all we can cancel out these two parts and if we divide both sides by four we get three over eight equals K and on this side divide both sides by 4 we get three over eight equals okay so in other words cm is equal to 3 8 of c d right now because we can write them as a they call it a scale multiple of each other but essentially because we can write one as a fraction of another they must be on the same line okay if my value for K was different for both of them it means that there's no way I can write CM as a fraction of CD so that would be your proof here so they lie on same line because one can be written as a fraction of the other God I hate writing okay got a couple more questions and they're getting slightly more tricky so diagrams is a parallelogram cool we've done parallelograms we should be cool and let's think about the information so we have o a good we have OB good D is the point um such that it's two to one cool we that's fine is the midpoint cool show that a d and e on the same straight line same principle so we're going to show that a d can be written as a fraction so k a e right and the reason why I picked these letters is because the full line is a to e right that's the full line so we're trying to show that a d is a fraction of AE a question I get is also can I do d e compared to a e yes you can right that is completely fine so either way I'd just take the first part as a fraction of the whole line so now we need to work out what a d is and we need to work out what a e is and do the same process perfect so a d let's work out that what's a good way to do that uh we could go from a to c and then C to D or we could go from a to O and then o to D okay so let's try that so A to O because we're going against the 2A is going to be a negative 2A and that's going to be plus a fraction of O to C right now what fraction is that well it says OD to DC is two to one so this side OD is two parts this side is one part so it's in total it's two-thirds of O to C and you might notice that once you've done kind of one of these show it's a straight line questions the others are very very similar which means with vectors how to get good at it do at least one of each type of vector question which again is shown in this video and you'll have a very good grasp on it again sometimes they do throw in some kind of unknown questions which you can't really predict but you can kind of scrape together a couple of marks at least if you've seen a bunch of other questions plus two-thirds and what is O to C well o to C would be let's say we're going to go from o to a and then a to c now you might say but we don't know a to c remember it's a parallelogram which means opposite sides are equal in length so in other words this side here is also 2B so OC is just 2A plus 2 b Plus 2B let's open this up we have minus 2A plus four over three a plus four over three B now we need to work out what minus two plus four over three is so that'll be minus two over one so we need times top and bottom by three to make the denominators the same so give me negative 6 over 3 a plus four over three a plus four over three B and again vectors is kind of like a game it's just not a very fun one but it is a game nonetheless there you go that's my equation for a do you know so the other one we need to work out is a all the way down to e so again what I'm going to do is clear my diagram and we're going to work out what a is so a let's work that out okay so how do we go from a to e again there's a few different ways I think the quickest way is to go from a to c and then C to E I think that's the fastest way to do this so we're going to go from a to c and then C to e okay cool now a to c we already know is 2b from the previous part and C to e this is going to be an interesting one is a half right because e is the midpoint so actually let's write this down first so 2B plus a half c e sorry c b okay so now what is CB well again opposite sides of a parallelogram are equal so this way is 2A so this is half of minus 2A because I'm going in the opposite direction okay and that's a really key thing so remember I've actually drawn these arrows in the same direction because if they're parallel they they still point in the same direction otherwise there's something called anti-parallel if they go in the opposite directions so this is going to be a half of negative 2A so AE you just have three lines equals two B minus a okay cool this should be fun now all we need to do is show that we can write a D in terms of AE so we're going to do the same tactic so I'm going to put the A's first now so I'm going to write a d which is minus two thirds a plus 4 over 3 B equals minus a plus 2B okay and you can probably already see potentially the relationship between them okay you can potentially see it it's just two-thirds okay but again we're going to do the same thing we're going to compare A's first we have negative two over three on the left hand side and we have negative one on the right hand side I've just realized that I haven't put the K there so let me rewrite this so it's going to be K times AE right and a is this so it's going to be negative Ka plus 2 KB so just so you're aware what I did is I've times AE by K in order to get that so just times everything here by by K my bad so minus K is what it equals to so in other words K is equal to two thirds okay so that's fine so far but again we need to compare it with b as well we have four over three in the left hand side and we have 2K on the right hand side divide both sides by two so you you can literally divide the four by two if you're dividing by two so K again is equal to two thirds so in other words a d is just equal to two-thirds of AE which means it's on a straight line easy as okay again the question although it's a completely different shape and it looks different the steps were identical right so let's keep going this was a submitted question that we had which I thought was quite interesting it tells me that becd is a straight line and then it says find K because we have a missing letter here and this is for three marks it's an AQA question in case you're interested but again it's very similar to any Edexcel or OCR or educast questions that you get as well so how are we going to do this well they've told me already it's a straight line which means I know and again whenever I see the white straight line I take the first two bits so BC must be written as a fraction of BD now K is already used so I'm going to use a different letter what letter will I use well in a level maths we use Greek letters but I've been told that's a bit confusing but we can use any other letter so what should we use we can use a let's say y again the letter doesn't matter it doesn't mean anything but we know from all the previous parts that BC must be up to Green as a fraction of BD because they told me it's a straight line so what we're going to do once again we're going to work out BC workout b d and then we're going to have well I'm going to show you what we're going to have it's three marks it's not too shabby not too bad so B to C well we're going to go from again we can't just go in a straight line from B to C because we don't have a vector that goes that way so we're going to go from B to a and then from a to c so that's going to be Ka plus 3B and that's it I can't do anything else so now we're going to go from B to D again I'm just going to delete this stuff here let's go from B to D again I can't just go straight here and I can't go from C to D so I have to go from B to a and then a to d remember the rules you can only go along vectors you know and you if it's in the opposite direction you'd make it negative in this case it's in the same direction nice and easy so we have Ka plus 6A plus 7B and I'm going to do the same thing that I've done in all previous questions which is I'm going to compare A and B once I've subbed it into this equation okay so we have BC which is Ka plus 3B is equal to again y if you used mu or Lambda or any other letter in the alphabet so long as it's not a b or k because we've already got a B's and K's you get confused you're completely and utterly fine Open Bracket Ka plus 6A plus 7B right now what I'm going to do is expand the brackets on the right and remember the goal here is just to find what K is I don't really I don't actually call IT wires uh so let's call it k y a plus six y a plus seven y B and now we're going to do the same thing if we compare A's on the left hand side we have k and on the right hand side I have k y and 6y so that's KY I wrote the Y's in a different way let's do that plus six why I can't solve that because there's two unknowns there's K and Y so let's look at B so if I look at B I have 3 on the left hand side and I have seven y on the right hand side so now I know that Y is equal to three over seven because I just divide both sides by seven so sub that into here and I'm left with k is equal to three over seven k plus six times three over seven so I'm going to expand this bracket and I'm going to move the 3 over 7K to the left hand side now one this would be one minus three over seven which of course is going to be 4 over 7 k equals 18 over 7. I'm going to times both sides by 7 which will cancel the divide by sevens and that gives me 4K equals 18. I'm going to divide both sides by 4 which gives me 18 over 4. well in other words it gives me 9 over 2. or in other words four and a half which again because these are decimals I should probably leave this in decimal as well so let's just do 4.5 that's it okay perfect okay so two more questions we're definitely into grade nine territory now and once again we have o a n we have O and B and we have avv are all straight lines a n is equal to two oh a cool and M is the midpoint of O B so first of all again we're going to label the information that appears here so what we'll do is we'll get our pen and this length here is a vector a and this whole thing here is the length is the vector B and it says that a p is equal to k a b like has a scalar quantity given that npn is a straight line find the value of K okay so what we're going to do is we're going to work out AP a b now if you notice if I do for example AP then it's going to be a fraction of this so let's start with that we we already have the equation we need essentially right that it's a fraction of a b but we should work out what a B is first because that will give us a good foundation to start with so we have to go from a to O and then we have to go from o to B so that's minus a Plus B okay and for a to p um we can't really do anything right because we know it's going to be just this part here so what we can do is work out another equation for a to p so all we have to do is work out another expression for a p now one other thing you should know about kind of maths exam questions is they never give you information that is useless we are going to use all of this and if you notice I haven't used this an thing yet and I also haven't used the fact that npn is a straight line so how will I use that well I'm going to use the fact that if npn is a straight line then MP is some fraction so we'll call that y again selected in the last part of M n okay so let's get started so MP let's work out an equation for that that's going to be from n to O So because the whole thing is B this would be minus a half B because a it's the midpoint and for the second part is since it's the midpoint it's going in the opposite direction okay so we're going this way so that'd be minus a half B and then we're going to go o to a so plus a plus a p and if you remember AP is just some fraction of a b so that's K and what was our equation for a b it was minus a plus b now equals again we've used the letter Y but any letter will do and N is a bit more interesting we have to go from M to O So again we're going to have minus a half b and then we're going all the way from o to n now if you look at the information here a n is double O A since o a is a a n must be 2A says plus 3A and then we're going to expand the brackets and do exactly like we've done before so I'm going to write the A's first here so we're going to have a and then expand this bracket here we're going to have minus Ka so a minus Ka minus a half b plus KB right KB comes from this bracket minus K it comes from the bracket and then we have the A and the Heart minus of B I've just written it so that the A's are together and the B's are together which is of course completely fine then we have minus a half y b plus 3 y a now we're going to do what we've done a few times before now we're going to compare everything in front of the A's so on the left hand side we have 1 minus k equals 3y then if I compare the B's we have minus a half B so minus a half plus k equals minus a half y so if I times this through by negative 2 then minus a half times negative 2 would give me plus one right times it by two is minus one times by the negative plus one minus two k equals y and I'm going to sub this into this equation because remember I care about K I don't really care about y so we have 1 minus k equals 3 1 minus 2K so if 1 minus k equals 3 minus 6K I'm going to add 6K to both sides and I'm going to subtract 1 from both sides so this one's going to this side this minus 6 goes this side so it gives me 5K equals 2 so K is equal to fifths which is my final answer there not too bad five marks as well so again if you practice this you should note that we're just doing the same thing again and again again what we've done is comparing things the last three questions I believe it was so yeah should be kind of fun well not fun but interesting anyway so oabc so it's the final question so this is the kind of Pinnacle of grade nine I would say now I would like to point out that there are going to be questions out there that are different to these slightly but you're going to use the same techniques so if you want to send those in I will have my email in the description for you to email in the questions and I'll make either a video on it or I do it in those Tuesday live streams which will be running until the end of exams and we can go through it in that sense so this will give you all the skills needed to get a Grade 9 but there will be questions out there that have some surprising things because they have to give you unseen questions every now and then anyways that was a lot of uh blabbering so we have all of the vectors labeled already so that's all done there's a point on OB such that OD Toby is so two parts three parts and is the point on BCS or studies Okay cool so we need to work out the vector d e in its simplest form so what I mean the best thing to do is just start by writing something up for d e so for d e all right I can't really go straight there so what I'm going to try and do instead which might be interesting is there anything we could do that's interesting so it looks like we're going to have to do is we could go up here and then to here okay so that's kind of the short route so we're going to say that that's DB Plus B and now we have to write out what the hell DB is so DB is going to be some fraction of OB and if you look over here this is the point on OB such that OD to DB is two to three so this is two this is three so it's going to be three-fifths of OB because we're looking for the three part section and until there's five parts so it's going to be three-fifths of OB and then we can work out what OB is a bit later and b e is going to be well if you look there's one part for this there's four parts for here so in other words it's one-fifth of B C okay now we have to do is work out what OB is and BC so if I do that up here we have OB it's not really that tricky it's just a plus b right nice and easy and for BC I can't just go straight down here because I have no vectors that take me straight there so I have to go this way this way this way so we have minus B right minus a because again I'm going in the opposite direction and then plus 3B so if I simplify that quickly that would be minus a and then minus B plus 3B would be plus 2B and now we can substitute into d e and that's basically going to be our answer once we simplify there isn't actually anything too bad about this this was reported to be quite a tricky question for a lot of students and I'm guessing it's well actually I know why but it's because of these kind of lines in the middle and then it splits off over here so students got very confused and a bit overwhelmed so there were no kind of ratios you had to use there were no hidden letters or anything like don't have to introduce any weird letters or anything like that so it was relatively okay question assuming you kind of are comfortable with vectors which a lot of people aren't which is understandable and one fifth BC is minus a plus two fifths B plus two B two fifths is what happens when we expand it so I was getting ahead of myself so we have three fits a plus three fits B minus a fifth a plus two fifths B and now we have to actually simplify since they're all over five anyway it's quite nice so three fifths minus one fifths that's two-fifths a then we have three fifths B plus two fits B so that's just going to be plus b right because it's a whole and that brings us to the end of this video it's been quite long but this should cover pretty much every single technique and skill you need to get grade nine in GCC vectors now what you need to do is go and practice so hopefully you've kind of printed this off or like written on this PDF separately and now you can check your answers and now you know you're going wrong and you've got some good understanding if you didn't do that you just watched the video then I'd recommend downloading these questions and doing them in let's say a week's time okay so you would have forgotten this but then you can go through it and again watch it again to get it marked if there's any other topics you'd like me to do this for then do let me know but Circle theorems is probably going to be the next one and completing the square as well so I'll see you later and I hope you have a great day