A LEVEL MATHS: Vectors Exam Practice LESSON 14

hey guys and welcome back to another one of the a-level maths uh tuition sessions I'm doing on YouTube uh so today we're going to be covering the vectors so specifically two D vectors uh today so the 2D vectors there's a few different kinds of questions that can get be given to so I've got a bunch of exam questions here and some questions from the most recent exam papers as well so first of all we have given that the point a has position Vector three I plus 4J and point B has position Vector minus four I plus 7j find the vector a b so one thing to note is that in a level we move on from using the column notation and we start using the IJ notation as well you're going to use this in mechanics especially as well so just you need to get very familiar with that uh what I would like to do here though is to make it a little bit easier is we can actually turn those vectors into column notation just for the purposes of calculating in this case I believe it's easier but in some other cases it might be a okay but we'll leave it as this for now so to get a to B what we do is we take Vector B and we subtract Vector a from it so we're doing this minus minus four minus 3 will give me minus seven minus 4 gives me three so if you wanted to write that out in IJ notations just minus 7 I plus 3 J nice and easy now Part B is it asks you to find this but it has the modular signs around it now the modulus signs means we're basically taking the magnitude of these vectors basically it's the size of the vectors okay so the way we do that is we use essentially it's Pythagoras right we do square root and we just Square the two components and add them together and then take the square root so I have the calculator here okay it's very yeah minus seven squared Plus three squared gives you square root 58 and generally speaking you can just leave magnitudes in third form and this is true for mechanics as well that's it so four marks really not too shabby I think question two is where it gets a bit more interesting so we have three I plus uh KJ now if we take the modulus of this we get 3 root 17s we need to find the value of K so remember when we take the modulus of something we Square the components and add them together so they're saying that this and once we square root it should equal 3 root 17. so to solve this symbol Square both sides so we get 3 squared it's K squared and if we were to square this we're going to get I mean to honest you can use a calculator you don't need to do in your head mental matters for losers whoops 153 nice and easy so then minus 9 squared from both sides so basically minusing 9 from both sides so that's 144. so K is equal to plus or minus 12. okay don't forget especially in a level we really need to remember the plus minuses whenever we square root a value so there are two values that it could have been here okay and as this is also going to be very important for you to remember remember there's always going to be two values especially because we took the modulus okay so when you take the modulus of a function for example it makes everything positive right so we need to keep in mind that there are always two values move to the next one again we have something that's very similar so let's just smash through this as quickly as humanly possible so we can get on to the more interesting exam questions that are coming up once again for the vector a b all you're doing is you're taking the vector B and subtracting the vector a so we have minus a 2 minus minus five and seven and again I prefer using the column notation instead of the IJ notation uh just for calculations because it helps me keep all the numbers in line so minus eight minus minus five would give me minus three two minus seven gives me minus five and so if we were to write down IJ notation it's minus 3i minus 5. J and then finding the modulus shouldn't be too tricky all you need to do is square the two components and add them together so again if any of you have any questions please let me know but I think so far it's not too bad again we're just going through the Basics at the moment but we have some exam questions coming up in a second wow I really so let's put the number in a here we go that makes more sense Route 34 and again in the exam in mechanics as well so mechanics and pure this comes up you can leave your answer in terms of certs you don't really have to round these okay here we are getting on something slightly new we have two vectors given that the resultant force of vector A and B is minus 2i minus three J find the values of X and Y so resultant vectors are when you add vectors together so what they're basically telling me here is that minus five a I sorry plus 7j plus x i plus y j should give me minus 2i minus 3j because again resultant vectors are when you add vectors together so what I'm going to do here is I'm going to gather the eyes together so we have minus 5 and we have X so I could write that as minus 5 plus x i and that's just to keep the eyes together okay and then I can keep the J's together by doing Plus 7 plus y J equals minus 2i minus 3j so hopefully that makes sense so far but again do let me know if it doesn't so you if you haven't come across this trick before the I and J components are completely independent of one another right so the number in front of I here should equal the number in front of I here and the number in front of J here should equal the number in front of J here oh so this is called comparing coefficients so if we compare I we have minus five plus x equals minus two add five to both sides we get X is equal to three nice and easy this comes up a lot when we do vectors in mechanics and in pure so it can come up in two sets of your exams we have seven plus y equals minus three so then Y is equal to minus ten so there we are we're done any questions about that if not that's fine okay I think I'd be fine so far so we have got some uh more exam questions coming up okay so we have a triangle a b c so this is where things get a bit more interesting and it can be good to actually draw out a diagram as well so for example we don't know what this triangle looks like exactly that's totally fine draw generic looking triangle call that a call that b call that c it tells me that the vector A to B is 6i plus 2J and Vector a to c is eight I minus five J and we need to find the vector b c while b c is just going to be AC minus a b okay just because of how the way uh vectors work so uh doing this we can use the column notation once again so AC was a and minus five minus a b is six and two two in other words 2i minus seven J okay hooked and then it just says find the length of the line a b so the length of the line um is what I mentioned before it's simply the modulus of the vector a b so this is going to be the square root so again they like to kind of ask you the same thing in a different way because it allows us to that's just root 40 I don't need a calculator because what that allows them to do is to determine whether or not you understand what you're doing right so instead of just asking you to work out the modulus they say find the length of the line because what they want to know is do you know what the modulus gives you it gives you the length of the vector right or the size of it yep so three forces act on an object so again this is a very stereotypical mechanics question find the resultant Force where a result force is just the sum of all of the forces okay so all we're going to do is literally add all of these together so again I'm going to do in column notation because it makes me feel better but you can just add up all of the I's and J's that's also completely valid of course is so minus five plus four is minus one plus three is two seven plus six gives me 13 minus five would give me eight or we can write that as 2i Plus 8j not to showy so far again another very stereotypical mechanics question but this was actually from a pure paper but again the comparison between the mechanics and the pure stuff it's basically the same okay very very very very very very similar so velocity of 7i minus five J find the speed of the car so here's where you need to recognize the difference between vectors and scalars if I take the magnitude of the Velocity I get the speed so it's simply just equal to the square root of 7 squared and that's it okay the same thing if they so they may give you like an acceleration as a vector but you can also work out the acceleration as an actual number where you just take the magnitude of it okay it's just that with velocity with no direction is speed route 74 nice it is just a second and then the bearing this is where things get a bit more interesting okay now we need to actually draw out what the uh Vector looks like so if we draw a little coordinate axes like this this goes seven across and five down so our Vector looks something like this where this part here is five and this part here is seven so we need to work out the bearing now bearing is always from North and North is straight up this is a 90 degree angle so we just need to work out what this angle is here now we have the opposite and the adjacent side so Theta is just equal to inverse tan opposite over adjacent so 5 over 7. which turn 35.5 degrees so the actual bearing is 90 plus 35.5 which is equal to 126 to the nearest degree right because remember bearings have to be to the nearest degree they're three digit numbers no more no less okay so that is it for the first section of the exam um questions so those are from a bit ago you may even see something like that in your textbooks but I'm going to get on to essentially the 2020 questions uh there's two of them for vectors let's go through that okay so relative to a fixed origin you have three different position vectors A B and C and we have to work out what a b is so again with a B nice and easy you're taking the position Vector of B right so OB and you're taking away okay so again we can use this in a column notation or the ijk notation either way it's completely and utterly fine so 3 minus three minus four minus 2 5 minus six three minus two gives me one minus three minus five gives me minus eight minus four minus minus six would give me two or again you can write that as I minus eight J plus two k now Part B says shows that the quadrilateral is a trapezium giving reasons for your answer so a trapezium is going to have two parallel sides right it's going to have so I mean assuming it looks like this it might something similar so one of the sides has to be or two of the sides have to be parallel not one of them okay so if I have a look at my Vector a b are any of these parallel to them well I look at C and I can see that to go from to go between these it's a factor of two like if you want to go this way I Times by two and if I want to go the other way I divide by two so in other words OC is equal to twice a b therefore they are parallel hence o a b c now if that's not convincing enough for you Let's uh go a bit deeper right so are there any other quadrilaterals that have two parallel sides well you may have like a parallelogram for example but if I have a look at the other kind of vectors none of them are parallel so the only quadrilateral that has only two parallel sides is a trapezium right a rectangle and a square have more than that they have two pairs of parallel sides parallelogram have two pairs of parallel sides Etc um kites rhombus Etc they only have they have no pairs of parallel sides so the only shape four-sided shape that has one pair of parallel sides is a trapezium and as we can see the only two sides that are parallel are OC and a b therefore [Music] it must be a trapezium I hope that that's convincing enough for you now here's where things get a bit more freaky um so the other 29 sorry 2020 question was honestly pretty hefty again it's only six marks though so pointy such that OC is equal to twice O A so it might be good to just add on this thing here where that is point C okay and then what we can also do is add um and it also says Point m is the midpoint of a b so let's draw um straight line uh cuts through c m C and M and cut o be at the point n so I did that perfectly so what we can do here is just drag these along slightly okay given that o to a is a so again I like to label up absolutely everything and Odes B is B find C to m in terms of a b well what we know is that o to C is twice o a so both half of this are the vector a right so to go from C to M what I'm looking at is I'm going to go from C to a and then go from a to n now I know what c to a is it's just going to be negative a right a to M I don't know but I know it's going to be half A to B Because m is the midpoint now how do I go from A to B well if I look at it a b is just going to be from a to O plus o to B so in other words it's going to be minus a plus B right because I'm going against the arrow here and then with it along there so in total it's going to be minus a it's a half minus a plus b so that gives me so minus a minus a further a is going to be minus 3 over 2A and this is going to be plus a half b and that's it for two marks now here's where things are going to get a bit more tricksy show that o n is equal to this thing where Lambda is a scalar constant so where Lambda comes from it comes from the ratio of this line here so o to n to n to B is going to be something like one to Lambda or Lambda to one okay so if you want to work out what o to n is that is the same is going from o to C and then C to n right because we need a few different ways to to get there to make this work so in other words we can make that o to C plus Lambda C to M so let's just work that out right so because they all are on the straight same straight line it's a scalar multiple so C to n is going to be just C to M multiplied by some value that I don't know yet okay so the reason why I chose C to M as well instead of kind of going along this thing is because we just worked out C to n okay so it's quite useful to use the previous parts so what we can do here is we can now write out what this is so o to C is obviously just going to be 2 a and then Lambda times all of this nonsense that we have out here okay and now all we need to do is expand this and show that it's the same thing so in terms of my A's I'm going to have 2 minus 3 over 2 Lambda a and for B's it's just going to be a half Lambda B which is exactly what it shows or not shows it asks for over here then lastly for part C it says prove that uh Owen to NB is two to one so because we don't know what o n is we can just call it some other letter mu B and we know it's going to be some something to do with B because it's along the line OB right so it's a scalar multiple of the vector B and what we can do is we can actually equate it to this okay so if we equate that out then what we can do is just think of the best way to put this if we compare the uh a terms we have two minus three over two Lambda equals zero if we compare the other terms we have mu equals a half Lambda okay so if we start with I think we start with this one because this one has two unknowns we have two equals three over two Lambda whoops so then Lambda is just going to be two times two four over three okay now if Lambda is four over three using this equation over here we will get mu is equal to half of that so that's two two thirds perfect so then from there if you have the ratio of o n so yeah just comparing that it's going to be two to one if you look the um mu is the actual uh it's the fraction form of the ratio right so o to n is the 2 the numerator n to B is 2 plus something gives me three so it's going to be one so in other words we have our final answer and it looks like everyone's okay with that in the chat so we can call it that for today so thank you very much for coming out