[Music] in this video we're going to take a look at the topic of vectors now you've probably already seen colum vectors which look like this a column Vector describes a journey from a start point to an end point and we can draw these onto a grid so let's take a grid and draw this column Vector this Vector tells us to go six to the right and then three down so a diagram of this column Vector would look something like this in this video though we're not going to use column vectors at all and we're certainly not going to count how many squares we go left right up or down for this topic all of the vectors are going to be renamed using algebra for example this Vector here to get from the start point to the end point could simply be named with the letter A your exam paper will use bold letters to indicate that they're representing vectors so if the letter is not in bold it won't be a vector when we handwrite maths though it's very difficult to write in bold so many people will underline the letters that represent vectors like this now we're going to take a closer look at how we write down vectors so imagine we had three points Point O point x and point Y we could describe the journey to get from o to X using a vector and we're going to call this one a to write this down using Vector notation we would say o x with an arrow above it and we could read this as the vector from o to X and we name that one a so it's equal to a now let's add a second Vector to this diagram this time starting at o but going to Y and we'll call this one vector B since it's different to the vector a this time we started at o and we went to Y so this is the vector from o to Y which is equal to B if we decided to do these vectors bit in the opposite direction so starting at X then going to O that would be the vector from X to O if we were to do this though we need to reverse the direction of this arrow and if you reverse the direction of the arrow you need to change the sign of all the terms of the vector so where this Vector was actually positive a it will now be negative a so the vector from X to O would be negative a in a really similar way the vector going from y to O goes in the opposite direction of that Arrow so instead of B it will be negative B let's revert this diagram back to its original state and we're going to now consider the vector from X to Y the vector from X to Y would look like this now you might think we're just going to give this another letter for example C but it's actually possible to describe this Vector using A's and B's we need to consider a way of getting from X to Y now we're not going to go there direct along the arrow we just drew but we could go there using the other vectors that we already have so if we start at x with this Green Dot here and go along to O we've just gone along the vector from X to O So that's X O then if we go from here down to the point Y we've just done the vector from o to Y so o y we started at X and we've ended up at y so we have done the vector from X to Y we just went an indirect route from X to O and then o to Y and it doesn't actually matter how you get from X to Y as long as you go across vectors you already know so we could say that the vector from X to Y is equal to the sum of the two vectors from X to O and O to Y now we know what those two vectors are so to get from X to Y we're going to go from X to O first and X to O goes in the opposite direction of that Vector a so it's negative a and then we go from from o to Y which was positive B so we add B so the vector from X to Y is A + B now some people will write this the other way around it's considered slightly neater to write B minus a rather than negative a plus b let's try and apply this to some slightly more complicated situations so in an exam question you could be given a diagram that looks like this and then you'll be given all of the information that's represented on that diagram for instance o q is 5 a and you can see that on the diagram to get from o to Q is 5 a QR is 3 B minus a and once again that's on the diagram and so is from o to P being equal to B then sometimes they give you extra information that you need to use for example the vector from o to Q is the same as the vector from s to R we can see the vector from o to Q on the diagram that's 5 a so the vector from s to R which goes from s to R must also be equal to 5 a now let's use this diagram to answer some questions so sometimes a question might say we out the following in terms of A and B and we're going to start with the vector from Q to P so we need to start at Q so I'm going to place this Green Dot on Q and we need to go to P but we want to travel along vectors that we already know you can see we could go there direct but we don't know that Vector yet so instead we're going to go from Q all the way down to O which is the vector qo and then from o all the way across to P which is the vector op so we could say that QP is equal to the vector from Q to O plus the vector from o to p and we know both of those vectors the vector from Q to O is the opposite direction of that 5 a vector so -5 a and the vector from o to P is just B so it's plus b so this is the vector from Q to P - 5 A + B which if you wanted you could rewrite as B minus 5 a but I've left it like this which is also acceptable now since we found this one I'm going to mark it onto the diagram as well so to get from Q to P I add an arrow going from Q to p and I call it - 5 A + B now let's try and do another one this time the vector from P to R so you can see that we can't get from P to R directly but we could get there by going from P to Q first so the vector p q and then from Q to R so we add to this the vector QR now we do know both of those vectors from P to Q is the opposite direction of that Arrow -5 A + B so we need to do the opposite sign of those terms so instead of5 5 a it's positive 5 a and instead of positive B it's netive B then we went along the correct direction for the arrow from Q to R so that must just be 3 B minus a so we add to this 3 B minus a now just like when we have algebra we can simplify these vectors by collecting like terms we have 5 a - 1 a which is 4 a and we have B + 3B which is positive 2B so the vector from P to R is 4 a + 2 b let's put this onto our diagram as well now we're going to do one more Vector on this diagram we're going to do the vector from P to S so to get from P to S we'll take that Green Dot and we'll go from P to R first so that's the vector p r and we're going to add to this the vector from R to S so r s we can see PR we just found that one it was 4 a + 2 B but going from R to S is in the opposite direction of that Vector with 5 a on it so it must be 5 a then we collect like ter terms again here we've got 4 a - 5 a which is a and then we just have plus 2 B so from P to S is a plus 2 B now let's have a look at a second question so in this one we've got oabc which is a parallelogram and here it is and we've been told some information about two of its vectors o a is 2 a and o is 6B and you can see both of those on this diagram then we're going to answer some questions so this one says work out the following in terms of A and B and we're going to start with these two vectors vectors BC and AC now looking at the vector BC on the right hand side of this parallelogram that will actually be the same as the vector from o to a since these two lines are parallel and the same length the journey from o to a is exactly the same as the journey from B to C so BC is also 2 a in a very similar way the vector from o to B is the same as the vector from a to c so AC is 6B next we're going to add some more information to this question m is the midpoint point of AC so halfway along the line AC we're going to find the point M and we need to work out the vector from a to M so to get from a to c was 6 B so to get from a to M must be half of that Vector so we could write the a m is half of the vector from a to c and the vector from a to c was 6B so it's half of 6B which is 3B so the vector from a to m is 3B now what about if we ask for the vector from o to m this time we're going to bring that Green Dot back and we're going to consider a journey from o to M trying to go along vectors we already know so to get from o to M we could go along the vector o a first so we have the vector from o to a and then we go along the vector from a to M so we add to this the vector a m now o to a we can see on the diagram that's 2 A and A to M we just found out that was 3B so we add to this 3B so the vector o m is 2 a + 3 B now let's add some more information to this question the point n is the midpoint of OC now we don't actually have the line OC drawn on so I'm going to add it to the diagram as well and the point n must be halfway along this so somewhere like here next we're going to try and work out our Vector for o n so since n is in the middle of the points O and C the vector from o to n must be half of the vector from o to C so we could write that o n is half of OC the problem is we don't know the vector from o to C we're going to need to work that out first so we can write 1 half and then put a bracket and then inside this bracket we'll work out the vector o c so let's bring the Green Dot back and to get from o to C we could go from o to B which is 6B and then from B to C which earlier we knew was the same as O A so 2 a so we've now got half of 6B plus 2 a and for this you can just expand the bracket just like you would with algebra half of 6B is 3B and half of 2 a is 1 a so o n is 3 b + a now let's have a look at another Vector this time we're going to look at the vector from n to B so let's bring the Green Dot back we'll start at n and we need to find a journey to get to B so I'm going to say we go from n to O first so that's the vector from n to O and then we go along o to B so we'll add to this the vector from o to B now the vector from n to O we don't know but we do know the vector from o to n n we worked it out in the previous part so the vector from n to O must be the same as this one just with opposite signs so instead of 3 + a it's - 3B minus a then we need to add to this the vector from o to B and you can see that's on the diagram that's 6B then we just collect these like terms we've got -3b + 6 B which is 3B and then - A so from n to B is 3 b - A now that we've at midpoints we're going to extend this and look at questions with ratios so let's take this diagram here and the basic information is given to us o a is 3 a and OB is B you can see those on the diagram but we're going to add even more information we're going to say that OC is five lots of OB and there's a point x which is on the line AB there's a point Y that's on the line BC and then these two ratios are true ax to XB is 3: 1 and by y to YC is 2: 5 and then we're going to work out some vectors again in terms of A and B so we're going to start with the vector ax so we were told that OC is five lots of O So OC is the vector from o all the way to C so this really long line here that must be five times as long as the vector from o to B and you can see o to B in the diagram that's just B so the whole length from o to C must be five lots of b or 5 b it's then quite easy to see that B to C must be 4 B since the total length of the line is 5 B and OB is 1 B BC must be 4B so we put a 4B here next we're going to look at this part here where it says ax to XB is in the ratio 3: 1 so X must be some point on the line ab and since the ratio is 3 to 1 is going to be closer to B than it is to a so something like this where the distance from a to X is three parts of the ratio and from X to V is one part of the ratio then we can do a similar idea with the point Y so the ratio of b y to YC is 2: 5 so we can place a point Y on the line BC it's going to be closer to the point B since that has a smaller part of the ratio so somewhere like this where from B to Y is two parts and from y to C is five parts then we'll return to the question we're trying to find the vector from a to X let's take a closer look at the line AB so we wrote down that X was a point on this line so we end up with these ratios 3 to 1 now if this line has four parts of the ratio in total from a to X is three of those parts so it represents 3/4 of the line since from X to B is only one part of a total of four parts from X to B is one4 of the line this means we could write the vector going from a to X is 3/4 of the length of the whole line from A to B which we would write as 3/4 of a now we can just work out the vector a and then do 3/4s of it so so just like before in the previous question we write 3/4 then a bracket then inside this bracket we'll find the vector a to get from A to B I'm going to bring the Green Dot back and we'll go from a to O first which is the vector a o that goes against the direction of that Arrow so - 3 a and then from o all the way across to B which is just B so plus b we can now expand this bracket we need to do 3/4 of -3 a to do this you'll multiply the 3 by -3 which is 9 so it's 94 of a and then 3/4 * B is just 3/4 B so this is the vector from a to X you may be tempted to try and convert these fractions into decimals but in most cases writing your answers as fractions is going to be easier in vectors questions now let's have a look at another Vector this time the vector from X to Y I'm going to bring the Green Dot back and we need to find a journey from X to Y I'm going to suggest we go from X to B which is the vector XB and then we go from B to Y so we add to this the vector b y now we just need to work out these two vectors so let's do the vector from X to B first so we're looking at this part here in the same way that the vector from a to X was 3/4 of the vector from A to B the vector from X to B will be just one4 of the vector from A to B so it's 1 quter of the vector AB we'll use this idea with b y as well to get from B to why we're looking at this part here which is two parts out of a total of seven parts of the line from B to C so this is 27th of the vector BC now we just need to work out the vectors A and BC well we actually did AB a moment ago so it's 1/4 of what we did before which is -3 A + B + 27 of the vector from B to C now we actually wrote the vector from B to C down right near the start of this question it was 4 B so 27th of 4 B now we just need to work this out so we can expand the bracket in the same way we did before we do 1 * -3 so 3/4 a and then 1/4 * B so 1/4 B and for this last part here we'll multiply 4 and 2 which is 8 so it's 87th of B now we do need to simplify these by collecting the like terms there's only one term in a so 3/4 a but for the B ones we can add these together when we add fractions we need a common denominator so we're going to make these both out of 28 so 1/4 B would be the same as 7 over 28 B and 8 over 7 B will be the same as 32 over 28b so we have 3/4 a and now we can add these together we do 7 + 32 which is 39 so it's 39 over 28b next we're going to take a look at a property of parallel vectors so if we take a vector from A to B and let's call it for Sake 2 a + 5 b then we draw another Vector from C to D which is parallel to this so it goes in the same direction but it's longer this could be 4 a + 10 B and we'll do a third one from E to F that's longer still but also parallel so 6 a + 15 B so we could write down the vector from A to B is 2 a + 5 b and the vector from C to D is 4 a + 10 B but we could factorize this one here there's a common factor of two so if you factorize out a two we could write it as two lots of 2 a + 5 B but this part here 2 a + 5B is exactly the vector AB that's 2 a + 5B as well so we could replace that 2 a + 5B with the vector from A to B in a similar way the vector EF is three lots of the vector AB so what we've written here is that the vector CD and the vector EF are multiples of the vector ab and this is not by coincidence it's because they're parallel so we've observed that if two vectors are parallel then one vector will be a multiple of the other Vector the opposite is true if two vectors are not parallel one vector will not be a multiple of the other Vector these weren't the only ways we could have written these down if CD was twice as long as AB then we could write that AB was half as long as CD if EF was three times as long as AB then we could write AB is 1/3 of EF and you can even write down statements linking the vectors CD and EF CD D would have been 2/3 of EF and EF will be 3 over2 lots of CD so the really important thing to understand here is that if one vector can be expressed as a constant multiple of another one then they must be parallel let's use this in some questions so for this question we're going to have a diagram and then all of this information which is actually on the diagram and we're going to answer this question show that the vector OC is parallel to the vector a d so we just learned that if two vectors are parallel one of them must be a multiple of the other one so we're just going to work out these two vectors and hope that that's the case so let's start with the vector from o to C to get from o to C we could go across from o to B first so the vector o b and then from B all the way across to C so we add to this the vector BC now we know both of these we can see them on the diagram OB is 9B and BC is b - 5 a if we simplify this we've got 10 B minus 5 a so OC was quite simple but what about a d so to get from a to d we'll bring back the Green Dot and we definitely can't go there direct we might need to do a bit of a tour around this diagram so we'll start by going from a to O first so the vector a o and then we can actually go across o to C here since we worked that out previously so plus OC and then from C to D so we add to this the vector CD now we need to write all those vectors down so from a to O is the opposite direction of that Arrow so minus 4 4 a and then from o to C well that's the one we worked at in the previous part 10 b - 5 a and then from C to D that's on the diagram that's 20 b - 6 a then we just need to simplify so we've got -4 a subtract 5 a subtract 6 a which is going to be -5 a and then we've got 10 b + 20 b which is + 30b now we could reverse the order of this so rather than -5 a + 30b is 30b subtract 15 a so remember we were trying to show that these two vectors are parallel and we have both of the vectors OC is 10 b - 5 a and a is 30b - 15 a and if you look carefully you can probably see that if you multiply the oc1 by 3 you get the A1 3 lots of 10 B is 30b and 3 Lots of- 5 a is -5 a so we can write down that the vector from a to d is three times the vector from o to C therefore they are parallel next we're going to look at another important fact about vectors we're going to take two parallel vectors again so ones that look like this but not only are they going to be parallel they're going to form a perfect straight line so for instance in the bottom left we could call this point a those two vectors meet at a common point which we could call B and then there's a point at the top which could be C so the first of these two vectors is the vector from A to B and the second one is the vector from B to C so we know that these two vectors are parallel but but also they both contain the point B we can use this information to write down a statement about how to check that three points form a straight line we would say that if a b and c form a straight line then the vector a and BC must be parallel notice again here we have the point B in both of those vectors it doesn't have to be the point B that's in common though we could also look at the vectors A and AC and they must be parallel notice this time both of those vectors have the point a and it could also be the point C so the vectors AC and BC must also be parallel so if you want to prove that three points form a straight line you need to find two vectors going along that line that involve one point in common and those two vectors must be parallel if they're not parallel it's not a straight line let's have a look at how we can use this in a question so if we take this diagram here and all of this information about the vectors and the question we're going to answer is show that pqr is a straight line so we know from what we just did that to show pqr as a straight line we need to find two vectors one of those letters must appear in both of those vectors and we need to show that those resulting vectors are parallel so the options we have are we could find the vector PQ and QR or we could find PQ and p r or we could find p r and QR when I look at this diagram I can see that PQ is going to be easy to work out so I'm going to do that one first to get from P to Q I'm going to go from P to O first and then once I've done this I'm going to go from o to Q we can see that both of these vectors are actually given in the question so to get from P to O we go against that Arrow so - 6 a and from o to Q we go along the arrow so plus 9 B so now we need to decide if we're going to do PR r or QR to go with this and looking at the diagram I think QR would be easier so we're going to work out the vector from Q to R and at first it looks like we don't have all the information we need but we are told in the question that op is equal to two lots of Qs and we do have op so if op is double Qs then Qs must be half of op if op is 6a then Qs must be half of this which is 3 a so we can put from Q to S onto our diagram at 3A we can now get from Q to R by going from Q to S and then s to T and then T down to R and we have all of these vectors on the diagram now so from Q to S is 3 a and from s to T we go against that Arrow so we need to change the signs of those terms so it'll be negative a and plus 4 B and then from T to R we go with that Arrow so it's - 4 A minus B then we need to collect like terms here we've got 3 a subtract a subtract 4 a which is -2 a and then we have 4 B subtract B which is plus 3 B so we've now found the two vectors we need and we just need to check that they're parallel so if we look at - 6 a + 9 B and -2 a + 3 B we can see that these are parallel because one of them is a multiple of the other if we take three lots of QR we'll get PQ so we can write that PQ is equal to three lots of QR and then we write a statement something like therefore pqr is a straight line it's worth remembering in this question that we weren't just showing they were parallel we specifically chose vectors that had a common point in this case Q and now we're going to have a look at one final question so in this question we have this diagram and we're given this information o a is 9 a o is 4 B OC is 5 B minus Ka where K is some constant and OB to BD is in the ratio 2 to 3 and the question for this one says given that a is parallel to CD find the value of K because we've been told that a is parallel to CD we're going to try and work out vectors for ab and CD first let's start with a to get from A to B we can go from a to O and then o to B and we know both of those vectors they're given in the question to get from from a to O is - 9 a since we go in the opposite direction to the arrow and to go from o to B is just POS 4 B so that's the vector a done what about the vector CD to get from C to D I'm going to go from C all the way back to O and then o all the way to D now we know the vector from o to C that's 5 B minus k a so the vector in the other direction will be the same one but we reverse those signs so it'll be k a - 5B then we want to find the the vector from o to D but we don't have this Vector we do have part of that line from o to B but we need the part from B to D but conveniently in this question we've been given those in a ratio it's 2: 3 so if we write out OB to BD in the ratio 2: 3 and underneath the OB we put the vector we know which is 4 B then we can see that this here has been multiplied by 2 b 2 * 2B is 4 B so we do 3 * 2 B which is 6B so the vector from B to D must be 6B so now we want the vector from o all the way to D we can add those two together which is a total of 10B so it's plus 10 B there's some simplifying we can do here since we have two vectors with B's - 5B + 10 B is POS 5B so we've now written down the vectors for a and CD remember in the question it said that these two vectors were parallel we know that if two vectors are parallel one of them is a multiple of the other now since we don't know what the multiple is in this one we're just going to call it n so we could say that AB is n lots of CD where n is some multiple that we don't know at the moment so let's write in the vectors AB is 9 a + 4 B and this equals n lots of CD which is Ka a + 5B if we keep the left hand side the same but expand the right hand side we do n * Ka which is nka and n * 5B which is 5 NB now at the moment this may look overly comp complicated but remember that n and K are just some numbers that we need to find what we're going to do now is use a technique called equating coefficients because these two vectors are equal to each other we must have the same amount of A's on the left and on the right and we must also have the same amount of B's on the left as on the right so if we look on the left hand side we have -9 a and on the right side we have nka this means that -9 must be the same as NK so we write that down as an equation -9 = NK we can do the same thing with the B's on the left hand side we have four B's and on the right hand side five NBS so four must be the same as 5n so we write that as an equation two now the equation on the right hand side can solve really easily if you divide both sides by five you end up with 4 5 = n now that we know n we can substitute that into the other equation so it's 9s 4 FS of K we can then solve this equation if we multiply both sides by 5 we get -45 = 4K and then divide both sides by 4 we'll get -45 over 4 equal K so in this question we were asked to find the value of K and the answer is -45 over 4 thank you for watching this video I hope you found it useful check out the one I think you should watch next subscribe so you don't miss out on future videos and if you want to try some exam questions on this topic I've put some in this video's description