 Fine. So last class we talked about the introduction of vectors, which was primarily revolving around the type of vectors addition of vectors and I also talked about Resolving vectors into components right which also included the concept of direction cosines when we talked about a resolution of vectors in 3d space Okay, so today I'm going to begin with the scalar multiplication the scalar multiplication in vectors as the word itself Says you are multiplying a scalar and a vector over here So let's say lambda is a scalar quantity and a is a vector. So if you're multiplying it, what are you doing to the vector? you are either Scaling the vector up or scaling the vector down or You are reversing the direction and or you are reversing the direction of the vector. So if lambda is Lambda is Negative, okay, you are reversing the direction of the vector. Okay, so if lambda is positive There's no change in the direction No change in the direction. Okay, if Mod lambda is greater than one. Basically, you are scaling up the vector. That means you are elongating it Okay, and if model of this is less than one, of course has to be greater than zero That means you are scaling down the magnitude of the vector in short. What I want you to know is When you multiply a vector by a scalar quantity, you end up getting a vector parallel to itself Okay, in words of vectors we call a and a lambda a to be collinear to each other To be collinear to each other That's why in many problems solving if somebody says write a vector in the direction of a particular vector We normally choose that vector as lambda times that vector in direction of whose we want a vector Okay Now this reminds me this concept is very simple to look at but this reminds me of a very important concept Which I'm going to take up next that concept is finding the vector along the bisectors of Two given vectors so finding a vector along the bisector of two given vectors Okay, so let's say we have a Vector a like this and Is another vector be like this let us Prove one important thing that is going to be used as a result also later on for us if you're bisecting the angle between a and b that means let's say this angle and this angle are same Okay, you are bisecting the angle between a and b then this yellow vector This yellow vector. Let me call it as C vector can be represented as a lambda times a by mod a plus b by mod b Can we take up a minute or so to prove this because this result is an important result and This will be used to solve a lot of questions later on when we are dealing with vector equations of lines, etc Let me just check how many people has joined in Megha has joined in Nikita has joined in good afternoon Nikita your exams are over DPS Yes, good. And how about Aditya Arvind exams are over or they end on October 1st. Okay Do you have any paper tomorrow? No, okay. So today is the last class before vacations Let me remind you 30th of September. You have an online test J main 10 to 1 Don't forget to take that test Yes, anyone any success in this a lot of information can be gathered from the result itself This is nothing but a cap This is nothing but a cap. This is nothing but B cap if you're done You can unmute yourself and explain me the result. Okay, so let me try this out for you all So if you take a unit vector along a let's say I make a unit vector along So I'm making the same diagram now with a unit vector. So this is your a cap. Okay You got it with issue Okay So let's take a unit vector along B cap unit vector means vector of length unity So if I take a angle bisector of this Okay, can I say angle bisector would actually be a line connecting the co-initial points of a and b Let me call this point as o to the midpoint of To the midpoint of this Eximities of a and b isn't it? Do you all agree with me or not? Let's say I call this point as point P Okay, do you realize that the angle bisector? Right would actually be a vector which will pass through P also P is nothing But it's the midpoint of let me call this as M and N Okay, P is the midpoint of M and N. Do you agree with me or not? Of course because I made a Isosceles triangle over here in isosceles triangle the bisector of the angle of the two equal sides will actually touch the opposite side at its midpoint right now if I have to find out what is my position vector of P what will you say very simple since om is the vector Which is actually signifying the position vector of point M. So we can say B cap is the position vector of M Similarly a cap would be the position vector of N So P point will be a cap plus B cap by 2 according to the section formula. Okay. See this is one of the Good advantages of knowing position vectors so OP vector would be a vector which is which is going to be Let me not write OP cap because it's not a unit vector So OP vector will be nothing but a cap plus B cap by 2. Okay. Now C vector is Collinear to Collinear to OP vector. That means this C vector will actually lie parallel to the OP vector Right, if it is flying parallel to the OP vector, I can write C as some alpha times this by 2 Where alpha is some scalar quantity where alpha is some scalar quantity If that is the case, I can write C as alpha by 2 a cap plus B cap Alpha by 2 I can start calling as a different scalar quantity. Let me call it as lambda So it'll be lambda times a cap plus B cap Okay, and a cap is nothing but a by mod a B cap is nothing but B by mod B Okay, and therefore any angle bisector Vector could be represented as this. Okay, very important result. Please remember this as a very important Concept because there are problems which are directly asking you to find out the bisector of the angle of any two vectors Of course some information would be provided to you to get lambda and all Okay, by the way, I have a small question for all of you. What if I asked you? The external angle bisector vector for this You understand the meaning of external angle bisector. So now let us say my a vector is like this B vector is like this Okay, and this is the angle External angle. Okay, this is the external angle. So what is the external angle bisector vector for this? So is it a minus B by 2 So, yeah, yeah, what did you say Vidoshi? a minus B by 2 a Minus B by absolutely correct. So basically what is going to be changing here is that you're just trying to find out the Internal angle bisector of a and minus B now So all you need to do is your B will be replaced with minus B. So you can say external angle bisector is Basically lambda times or any scalar times a by a modulus a minus B by modulus B That exactly comes from a minus B a cap minus B cap by 2 formula. Is that fine? So this result also is important keep this in mind Any question so far? Aria is there. Shaughnak is there guys Please be on time because normally when I start with a new concept and you have not joined in You lose out on very important things Aria exam is over. Aria Shaughnak Are your exams over today only over or it was over like Okay, Shaughnak exam is over mid-October Okay, are you exam got over today only last week? Okay, enjoy. So let's have a question on this Let's take a question on this Hope you can read this question the question says find a unit vector C if Minus i plus j minus k bisects the angle between C and 3i plus 4j Question is clear So you have to find you have been given the bisector actually. This is your bisector This is your bisector of these two vectors vectors C and vector 3i plus 4j you have to find the C vector in this case you have to find this vector in this case Either you can type the answer as a response on the chat box or you can just speak out the answer I'll give you three minutes to do this time starts now Anyone any success? Hi Gargi. Good afternoon Okay, so let us assume that our C vector is X i plus y j plus z k Okay Now it's known that C is a unit vector that implies mod C should be equal to 1 which implies x square plus y square plus z square should be equal to 1 correct Now this given vector that is minus i plus j minus k should be lambda times A by a cap in this case a is your C okay, and C by C cap is nothing but C itself because C itself is a unit vector So you don't have to divide by under root of x square plus y square plus z square because that is already given to you as 1 It's a unit vector plus 3i plus 4j divided by 5 So essentially what have I used I've used the formula that the bisector of A and B let's say this is the angle bisector Okay, so C vector is given by let me not call it C because he's already been used here Let me call it as P vector P vector is a cap plus B cap times lambda Okay, that's what I have used over here. Now we can compare the coefficients since i, j, k are On both the sides you can compare the coefficients of these vectors So minus 1 is equal to lambda x plus 3 by 5 Okay, that's one equation Other equation will be 1 is equal to lambda y plus 4 by 5 and the third equation is minus 1 is equal to lambda z Correct. Let us try to get x in terms of lambda. So from here I got a response That's absolutely correct That's correct with Ushi. That's correct So we do she has already given the answer. Okay, so from here I can write Sorry, this is also lambda. So from here I can write Lambda times 5x plus 3 is equal to minus 5 right, so lambda is or x you can say x is Let me make x the subject of the formula x in terms of lambda Lambda x is minus 1 minus 3 lambda by 5 that is Minus 5 plus 3 lambda by 5 lambda is your x. Okay Now let me find y in terms of lambda. So lambda y is equal to 1 minus 4 lambda by 5 That means y is equal to 5 minus 4 lambda by 5 y 5 lambda, okay, and already z is equal to minus 1 by lambda now these these values of x Y and z we can use it in the equation over here, which was x square plus y square plus z square equal to 1. So let's use this Let me name it as 1 2 3 and 4 Okay, so I'm using 1 2 3 4 in 1 so using 2 2 3 4 in equation number 1 I would end up getting 5 plus 3 lambda by 5 lambda whole square 5 minus 4 lambda by 5 lambda whole square 1 by lambda square is equal to 1 Right, let us multiply throughout with 25 lambda square. So 5 plus 3 lambda square Plus 5 minus 4 lambda square plus 25 is equal to 25 lambda square Okay, now I'm sure when you expand this there will be a 25 lambda square which will get cancelled with this You will get other than that from here you will get 60 lambda and from here you will get minus 40 lambda And other than that you will end up getting 25 plus 25, which is 50 is equal to 0 Is that fine? Is there anything which I'm missing out? Oh, sorry 75 it'll be not 25 75 25 25 25 will be 75. Yeah So here we get 10 lambda is 75, which means lambda is equal to 15 upon 2 Okay, bus done our job is done. I just have to put this value of lambda And get my value of x y and z respectively correct So if I use it in the second equation lambda is 15 by 2 I'll get minus 5 plus 15 by 2 will be 45 by 2 divided by 5 into 15 by 2 Right that will give me 10 55 minus 55 by 75, which is nothing but minus 11 by 15 Okay, so x value is minus 11 by 15 Let's put in y also So y will be Let me put in the original equation. I think this will be much easier to find So y will be 1 minus 4 lambda by 2 that is 60 by 2 divided by 5 into 15 by 2 That's going to be 10 minus 50 by 75 That's 10 by 15 5 minus Yeah Minus 10 by 15 Hope I've not made any sign error. Just confirm 1 plus 1 is equal to 15 By 2 y plus 4 by 5 into 15 by 2 3 2 yeah y is correct. Yeah Next is z z would be minus 1 by lambda, which is minus 2 by 15 correct So having found out x y z my c vector would be x i Which is minus 11 by 15 i minus 10 by 15 or you can say minus 2 by 3 j minus 2 by 15 k Clear is that fine any doubt regarding this Okay, clear everyone shanak great So as you can see This concept proved very very handy for solving this problem Right and it has very wide application. So if you start solving your previous year questions also you realize that this concept has been widely asked at so many places Okay Now next concept, which I'm going to take up Is very vital because It will be a founding concept for your understanding of co-linearity and co-planarity of vectors and points That is the concept of linear combination. Okay, try to understand this linear combination term Because from it, I'm going to talk about linear dependency and linear dependency of vectors Okay, this concept is missing from your ncrt syllabus Okay So first of all, let me make you understand. What is the meaning of Saying that a vector is a linear combination of You know a system of vectors. So we say that a vector r Is a linear combination This word is very important because when I'm using this word, you should know what do I mean by using You know by the use of linear combination phrase. Okay. So this phrase is very important So we say a vector r is a linear combination of Let's say a system of vectors a1 a2 a3 Etc till an if if r could be expressed as Scalar times These vectors that means there exists There exists scalars m1 m2 m3 etc till mn. So let me write here where m1 m2 m3 etc till mn are scalar quantities Okay, if r could be expressed as scalar times a1 plus some other scalar times a2 plus some other scalar times a3 like this We actually say r is a linear combination of this system of vectors Okay Now some terms associated with it are linear Linearly independent And linearly dependent. So let me first take up the concept of when do we say A set of vectors or a system of vectors are linearly independent Again, see what I want you here to focus on Is the terminology which I am going to use it The terminology is going to be useful because when you read this in your textbooks or when somebody is explaining you a particular concept Probably he will make the use of these words like linear combination or linear independent So even for problem solving, I'll make use of such words Okay, there you should understand. What do I imply by use of these words? So when I say A system of vectors a1 a2 a3 etc till an are linearly independent Are linearly independent. What do I mean by this? What I mean is If you express a linear combination of these vectors, see again, I'm using the word linear combination So you should know what I'm talking about. So if you express a linear combination of these vectors as a null vector Okay Then All your scalar quantities involved over here must be zero That is all your m1 m2 etc till mn must be zero So Then only we say these vectors are linearly independent of each other Right to give you an example Vector i j and k cap are actually linearly independent They're actually linearly independent Okay, why? Because if you try to Find out three scalar quantities m1 m2 m3 Such that if you express this as a null vector Right, then it can only be possible if each of these components m1 m2 m3 are individually zero Getting my point. This is another way of saying K cannot be produced by K cannot be generated by some scalar times i plus some scalar times j Okay, where x and y here are scalar quantities So Yeah, so uh, so um, are linearly independent Uh, vectors are always perpendicular. No, no, no, not necessarily not necessarily Linearly independent means One vector cannot be expressed using the other two vectors. Let's see if I talk about linear independence of three vectors Remember linear independence can also be applied to two vectors. Okay, for example i i and j Okay, I'm just giving an example of i and j because they're simple to connect to I and j are linearly independent That means no matter what you do you can never produce i from some lambda time j vector No, no lambda would exist like this lambda Cannot exist here There's no lambda You can say this cannot happen for any lambda Belonging to real number Okay In other words if you find if you try to do this operation That means if you express the linear combination of i and j as null vector You will only conclude that m1 and m2 can be zero Right That's why we normally try to express a vector in terms of i and j and k because they are linearly independent One doesn't share any component with the other Right the purpose of using i j k to express any vector or to resolve any vector was the A basic principle that it is they are linearly independent of each other There's no component of one vector in line of the other That means there's no Dependency of these vectors on one another Getting the point So in the previous example, you would have seen that When I compared the coefficient, let me go back to the previous page When I was solving the problem of angle by sector Right so easily I compared my i with i j with j k with k Correct because I knew that if I bring everything one side I will get something like something i something j something k Correct and this is null vector and i j k are linearly independent. So all these something must be individually zero That actually led to these three equations that you have you actually solved Getting my point. This could only be possible Here actually I did not tell you what was the reason behind comparing i with j See you cannot do that if your vectors were linearly dependent on each other Are you getting my point? So Let me take that concept first before I talk about anything else. What is the meaning of linearly dependent? By the way, all of you please make a note of this When I say a system of vectors are linearly independent What do I actually mean by it? This is the mathematical definition But in a plain and simple word just try to understand that You cannot express any one of these vectors in terms of the other vector That is the meaning of linearly independent of each other Right hope you have noted this down Okay Let me now talk about vectors being linearly dependent Again a system of vectors a1 a2 etc That's it till a n are said to be are said to be Linearly dependent if A linear combination of these vectors when expressed as a null vector Okay implies all m1 m2 Implies all m1 m2 till mn are not zero All are not zero Just like in the previous case every Scalar quantity m1 m2 etc till mn was zero in this case at least few of them Right should not be zero for example i plus j And i and j These three vectors I've just taken a simple example. Let me take a 2 i and 3 j also. Okay. These three vectors are linearly dependent Okay, because if you try to Express a linear combination of these vectors As a null vector you realize that there will be some m1 m2 non-zero values also for which this is possible Okay, let me show you how in this case M1 i m2 will be nothing but 2 m2 3 m3 j Okay, this means m1 could be zero But 2 m2 plus 3 m3 equal to zero can lead to so many non-zero values of m2 and m3 for example here m2 could be uh, let's say three m3 could be minus two Okay, so as you can see not all of them are zero In other words, you could express one vector In terms of the other two that is the meaning here when you say they are linearly dependent For example i plus j could be expressed as half of 2 i And one third of 3 j Getting my point So here these three vectors are linearly dependent on each other Are you getting my point is the difference between linearly independent and linearly dependent clear in your mind Mathematically speaking i'm just going to state these two important Characteristic if a1 a2 a3 till an are linearly independent they would satisfy this property And if these vectors are linearly dependent Sorry linearly independent then they would satisfy this property Is that clear? What is important per se is the two theorems which are related to linear independence and linear dependence The two theorems are number one A pair of collinear vectors A pair of collinear vectors Collinear means what in your in your mind when you read this word collinear what should come in your mind parallel okay a pair of collinear vectors are always Linearly dependent This is the one which is of use to us directly in problem solving okay A pair of collinear vectors are always linearly dependent and second theorem that it we are required to know is that a triad of coplanar vectors Guys, let me tell you you'll get a direct question on coplanarity and collinearity in your school exams also. Okay A triad of coplanar vectors are always Linearly dependent Okay, and the opposite is also true. That means let's say I call it as 1a So 1b means a pair of non collinear vectors A pair of non collinear vectors are always Linearly independent Okay, and similarly to be A triad of Non-coplanar vectors You understand the meaning of triad right three A triad of non coplanar vectors are always Linearly independent a triad of Non-coplanar vectors are always linearly independent getting a point The conclusion here from the second 2a and 2b is that so from 2a and 2b you can say more than three vectors More than three vectors Are always linearly dependent. Okay, because the fourth vector Right Would actually become dependent on some of the vectors, right? Okay, so we say I don't know whether you have learned about cross product or not in school So any vector in 3d space, okay any vector in 3d space could actually be expressed as xa plus yb plus z a cross b Okay, so this vector becomes linearly dependent on these three vectors right But a b and a cross b Are linearly Independent let me tell you because they will never be coplanar Right, they are they are triads of vectors which are never be coplanar. I'm sure you would have learned cross product. This is always perpendicular to the plane containing a and b How many of you have not done vectors in school at all? Are you the physics concept? I think you should be knowing it, right? We all know that a cross b is perpendicular to the plane containing containing a and b So a b and a cross b forms a triad of linearly independent vector So by using these two are just like ijk see ijk also follows the same nature, right? Let's say a is i cap b is j cap then a cross b has to be k cap So any vector you can actually express as a linear combination of ijk. That is what we call as resolving vectors, isn't it? So the same way i've just given you a broader This is a broader understanding of your resolving of vectors. This is a broader interpretation of Resolution of vectors Okay, so resolution of vectors is nothing but it's a scope which comes under the concept of linear independence and linear dependence getting my point This is slightly deep it it requires a bit of you know deeper understanding. It doesn't come very naturally And that's why since this concept is slightly challenging. It is missing from your ncrt and you know cvc curriculum Okay, so these theorems Please keep them in mind Okay, i'll repeat them once again A pair of collinear vectors are always linearly Good evening, Anirudh We joined too early class started at four So Yeah, a pair of collinear vectors are always linearly dependent a triad of coplanar vectors are always linearly dependent A pair of non collinear vectors are always linearly independent And a triad of non coplanar vectors are always linearly independent Okay, and remember more than three vectors will always become linearly dependent And right four five six vectors. They'll always become linearly dependent. Okay Okay, no problem. Anirudh Fine, so where are we going to use this linear independent linearly dependent? So so far we have been talking about these concepts without any application to it. So now time to see where do we apply it So we'll be directly applying it to these theorems and why are these theorems will be studying the concept of Collinearity of two vectors Okay, by the way, this concept should not be new to you because when I was talking about scalar multiplication You already learned how to make a vector collinear to other Okay, so here the concept is if your vector a and b Are collinear Collinear means what I mean In your in your mind this image should come in your mind a is like this b is like this Okay, that means they are parallel to each other. By the way direction can be Ulta also no problem. Okay So if vector a and b are collinear Then a and b become linearly dependent Are linearly dependent, right? that means If you express x a plus or let's say m1 a since I have been using m1 m2 So let me continue with that trend So if you express m1 m2 Like this why am I writing a2 and a1? b and a I was talking about right, so let me use that If this is a null vector then m1 m2 Both are not both Should be at least both should not be zero okay In other words, you could express a as minus m2 b by m1 Which means you can express a as some lambda times b. This is the Simple and straightforward meaning that vector a and b are collinear That means one could be expressed as some lambda times b Are you getting my point? Which means that if a and b are not collinear, what will happen? So if a and b are not collinear, let's look at the other face of this concept If a and b are non collinear then a and b are linearly independent that means If you try to express If you try to express m1 a plus m2 b as a null vector Then the only conclusion is both a m1 and m2 are zero Getting my point. So this was the concept of collinearity of two vectors I'll give you a simple question Just a very very basic question to make you This to make this concept clear in your mind Are these two vectors? Let's say minus i plus 2j plus k and 2i minus 4j minus 2k Are these two vectors collinear? I think most of you can answer it within one second also See I only started getting the answer. Yes, they are collinear. Why? If you use this logic things will be quite fast. Okay, so use this. I recommend this to you always. So if you want to Ensure whether two vectors are collinear You should be able to figure out whether one could be expressed as a linear Or sorry one could be expressed as a scalar times the other Okay, so here you can say minus half times 2i minus 4j minus 2a. So there exists a lambda Right there exists a lambda Belonging to real number for which this is possible Right if a and b were non-collinear you will never be able to find a lambda like this Right lambda will never exist such that you could express a as lambda b or b as lambda a Are you getting my point? Okay, if you want to use the other mechanism that means if you want to use this part that means m1 m2 both not zero Then you have to use the fact like m1 let m1 minus i plus 2j plus k Plus m2 2i minus 4j minus 2k let it be a zero vector Okay, if it is zero vector you will get minus m1 plus 2 m2 equal to zero i'm comparing the i coefficient This is your j coefficient and this is your k coefficient which will all mean the same thing to you actually Okay, they all mean the same thing. They all mean the same. They're all same Okay, so if you solve any one of them There can be definitely non zero values of m1 m2 existing Okay, for example m1 can be one m2 could be half like that Getting my point Anything that you would like to know more about this in fact, we'll be doing problems later on But concept-wise Any clarification that you need these types clr if things are clear I'm going to extend this concept of collinearity of two vectors to collinearity of three position vectors So i'm going to extend this to concept of collinearity of three position vectors three position vectors Also you can say three points So let's say we have points a b and c points a b and c Whose position vector is vector a vector b vector c Okay now if these three vectors are Collinear if these three vectors are collinear then Any one of these situations is going to be true That is either your ab vector would be some lambda times ac Or your ac vector would be some lambda times bc Or something like let's say ab vector ac vector is lambda times Or you can say so any of these three conditions would hold true That means if you make any two vectors by using these three points They themselves would be collinear Isn't this very obvious Because if they all line a straight line if you can draw a straight line through them So let's say if I choose ab vector like this And if I choose bc vector like this, let me choose a different color Aren't ab and bc themselves collinear? Correct. That means i'm using this concept Or aren't ab and ac vector themselves collinear. That means i'm using this concept So if any one of these conditions Please note any one you don't have to show all the three. Okay, any one is fine So if any one of these conditions Hold good Then you can safely conclude that your points abc are in a single line. That means they are collinear Right This is the easier part of the concept By the way, we are going to talk about the concept of collinear That is the concept of collinear This is the easier part of the concept By the way, what if they are non collinear? Right, what will happen if abc are not collinear are not collinear then what will happen? You'll never be able to find a lambda Such that you could do this or you could do this Or you could do this Okay, so no such lambda would exist No such lambda would exist such that one could be expressed as A scalar times the other Getting my point Now this is one way of looking at it Second way of looking at it is That is a slightly more challenging way. This is the easier way. So normally I call it as the easy way Easier method. Okay There's a slightly challenging method That is the second method If your vectors abc Are collinear If these three vectors are collinear Then if you express these vectors Linear combination as a null vector Okay Then it implies m1 plus m2 plus m3 should be zero but m1 m2 m3 Are not all zero Very important. Please note this down This is a concept which confuses a lot of people. Many people don't remember this also Many people just take the first method because this is an easy method. This method is slightly challenging But nevertheless, we should know it actually Okay, the concept is again. I'm repeating If your vectors abc are Collinear If your vectors abc are collinear And you express a linear combination of these vectors as a null vector Then m1 plus m2 plus m3 should be zero But m1 m2 m3 are not all zero Right one way m1 m2 plus m3 could be zero is when all of them are zero right zero zero zero But here they should not be all zero That is the idea here getting my point Okay, how is this property actually coming into picture? Let's let's try to understand this This property comes from the fact of section formula. Let's say ab And c are collinear That means We should divide the join of a and c in some ratio. So let's say x and y there are some value Remember x and y cannot be both zero. Okay So I can express b as xc Plus y a By x plus y If you cross multiply it will be x plus y B Is equal to xc plus y a Correct In other words, if you group your abc at one place, you will get y a Minus x plus y b Plus xc is equal to null vector Right now this is playing the role of m1 This is playing the role of m2 This is playing the role of m3 In other words, you are basically trying to do m1 a M2 b m3 c is equal to null vector So here you can see m1 m2 plus m3 Will actually become a zero value Because it is actually this term That is what I said over here But x y etc They are not zero X is not zero y is not zero Okay, that is the meaning of this term m1 m2 m3 are not all zeros Does it make sense Everyone If this makes sense to you I have a question for all of us Clear Let's have a question Very simple one To begin with So the question is quite clear The vectors this, this and this Have their initial point at 1, 1 Find the value of lambda So that these vectors terminate on one straight line Okay One thing let's have it as a norm Whenever you are saying some answer Please message me privately Okay So that others are not interested Okay, so that others are not influenced By that answer Okay, fine So anybody who is answering me Please message it privately to me Okay, it's fine as for now So those who did not understand the question It's basically like this There are three vectors All of them originate from 1, 1 Okay, so there is a vector like this Okay, there is a vector like this There is a vector like this So let's say this vector is 2i plus 3j This vector is 5i Plus 6j And this vector is 8i plus Lambdaj And all these vectors terminate On a line You have to find the value of lambda Okay Now plain and simple If all these vectors Eliminate on a line That means these vectors a, b, c Position vectors Okay, they must be Collinear How do I find the position vector of a Yeah, all of you are correct Those who have answered you are all correct Okay So how do I find the position vector of a Remember 1 comma 1 Is as good as saying i plus j Okay So this vector is 2i plus 3j Let me call this as Point p Okay So pa vector is 2i plus 3j So what is your point a then It's very obvious that pa means Position vector of a minus Position vector of p This is 2i plus 3j So oa will be nothing But op plus 2i plus 3j That is nothing but i plus j plus 2i plus 3j That's 3i plus 4j So this point is 3i plus 4j Similarly this point will be 6i plus 6i plus 7j And this vector will be 9i plus lambda plus 1j Now these 3 Vectors are Are On the same line Right, so can I use the fact that a, b is Some scalar times Yeah sorry So ab vector could be expressed as Some scalar times you can call it as Let's say beta Beta times any other vector You can say Ac or bc whatever you want Okay so beta times let's say ac Okay Where beta is some scalar quantity So what is ab vector Ab vector is b minus a Which is actually 3i plus 3j Is equal to beta times Ac vector is what 6i Plus Lambda minus 3j Hope you know how to find Vectors when their endpoints are given Okay just compare the coefficients 3 is equal to 6 beta That means beta is equal to half And 3 is equal to Beta which is Half times lambda minus 3 So lambda minus 3 is 6 So lambda is equal to 9 Pain and simple Please type clear if this is clear to you Alright let's move on to the next question If a and b are 2 Non-colonial vectors show that The points given by L1a plus M1b L2a plus M2b And L3a plus M3b Are collinear if this determinant Is 0 Just type done if you are done with it Okay Chirag Done That's another way to do it Aniruddha Exactly actually I was Going to do it that way also But first try to deal things In vectors Later on you would have a tendency To convert things into Cartesian But actually if you do that It sends a very wrong Signal to the examiner That you are not good in vectors That's why you are relying too much upon Cartesian As I already told you Vectors and coordinate geometry Are very closely linked To each other But once the question is given in vector Try to work it out In terms of vector only Of course you can make use of other Concepts like Determinant and all may be used Done anyone? Okay let me take this concept up Oh Gaurav is done great So let me take this concept up I'll solve this in two ways Okay If you are saying the vector A, B Are non-colinear Okay just like your I and J vector That means they are linearly A and B are linearly independent Right now If you are saying these three points Are collinear that means Let me call this point as point A B and C Okay So AB could be expressed As some lambda times Lambda times Any other Any two vector you can take Let me take it as AC vector So AB vector Is nothing but Vector B minus vector A So if you do that You get sorry B minus A So L1 minus L2 Should be some lambda times Lambda times AC vector That is C minus A That is L3 minus L1A What does it imply? It implies L2 minus L1 Is Lambda times L3 minus L1 Right Why because A and B are linearly Independent so whatever are the coefficients Of A on the left hand side of the equation The same must be the coefficient Of A on the right hand side Comparing the coefficient So I am comparing this with lambda into this term Similarly M2 minus M1 Is some lambda times M3 minus M1 Okay Now let us look at this determinant Given to you in the question Let us subtract Let us subtract Column number one from the other two columns So L1 M1 1 L2 minus L1 M2 minus M1 0 L3 minus L1 M3 minus M1 0 Okay what actually have I done I have taken the determinant And I have done this operation C2 is C2 minus C1 C3 is C3 minus C1 Okay we will do that method also Don't worry Now I can see that L2 minus L1 Is lambda times L3 minus L1 So you can basically write L1 M1 1 This you can write it as Lambda L3 minus L1 Lambda M3 minus M1 0 L3 minus L1 M3 minus M1 0 Correct Now take lambda common From C2 Take lambda common from C2 If you do that your determinant becomes L1 M1 1 L3 minus L1 M3 minus M1 0 L3 minus L1 M3 minus M1 0 That means these two columns have become Identical They have become identical That means this particular determinant Collapses And becomes a zero that's what we wanted to show This is one way And I have taken the easier method to do it Let us try to Do the same by using the other Method also which I said is slightly More challenging and normally people Don't like using that method So we'll use that method also to prove That this determinant would be zero By the way any doubt regarding This method This is clear so I will use The other method as well To prove it So if you're saying That these Three points are Collinear Can I say that Second method Can I say that M1 L1a M1b Plus M2 L2a M2b Plus M3 L3a M3b If expression is null Would result into M2 plus M1 M2 M3 Let me use X, Y, Z To be more I've already made use of M1 M2 M3 here so we can't use it again So let me use X, Y, Z So this should imply X plus Y plus Z should be zero But Not all X, Y, Z must be zero Yes or no So the meaning of Saying the three points This, this and this are collinear This is what we had discussed Okay Now if you look at this equation And you try to find out the coefficient Of A It is X, L1 Y, L2 Z, L3 Coefficient of B would be X, M1 Y, M2 Z, M3 Right Equal to null vector Now since A and B As A and B Are Non collinear Implies A and B are Linearly independent If they are linearly independent Remember these scalar quantities Must individually be zero Correct So this implies X, L1 Y, L2 Z, L3 must be zero And X, M1 Y, M2 Z, M3 must be zero So indirectly speaking You got three equations now One is this equation Other is this equation And other is this equation Basically you got a system of Homogeneous Linear equations And this system of Homogeneous linear equations Must have a non-trivial solution This must have a Non-trivial solution Why a non-trivial solution all of a sudden? Because remember Not all X, Y, Z must be zero Okay Now go back to your determinant chapter And try to recall The condition when a system Of homogeneous equation Has a non-trivial solution Non-trivial solution means Non-zero values of X, Y, Z should exist Okay When can that happen? That can only happen when the determinant Made by the coefficients of these equations Right So right now what are the equations that we have One, this is the coefficient 111 Here I have L1, L2, L3 Here I have M2, M1, M2, M3 So the determinant made by the coefficient Of these equations L2, L3, M1, M2, M3 111 Must be equal to zero And this is exactly what I Required to prove This is exactly what I required to prove This is what I wanted to prove Done. Are you getting this point? See both the methods I use both the methods to prove it The thing is the The process must be clear in your mind Is it clear? Anything that you Would like to ask me from this Especially the second method I am sure the first method is quite clear to everyone Anything that you would like to ask You can unmute yourself and ask me Clear? If this is clear, we will not talk about Co-planarity Of three vectors Okay By the way, some people ask me Sir, you talked about co-linearity Of three points. Why didn't you Talk about co-linearity of two points Right Let me tell you Two points are always co-linear Okay In a similar way, many people ask me Why don't you talk about co-planarity of two vectors Let me tell you Two vectors are always Co-planar Are you getting my point? Two vectors are always co-planar Many of us would think like Okay, let's say a vector is like this And other is like this. Let's say there is a cube Then you cannot form a plane Like a road is going like this And a flyover is being constructed over it Like this Of course it is not co-planar because There is a cube. But remember we are dealing With three vectors When you are dealing with three vectors You can bring one vector In plane of the other always By moving it parallel to itself Okay So three vectors We always talk about co-planarity Of three or more vectors Two vectors are always co-planar Getting my point Fine So please try to recall the concept Try to recall the theorem that If three vectors are co-planar They are linearly dependent If three vectors are co-planar They are linearly dependent So if vector A B Are co-planar Are co-planar Then A B C Are Linearly dependent In other words one vector could be expressed As Some scalar times the Other two vectors Getting my point Here X and Y pair Is a unique pair Why I am saying unique because There is only one way you can express C As X A plus Y B That means X and Y only one such pair Will exist. They cannot be multiple Such pairs Okay I am assuming here that A B C Are pair wise Co-linear Are pair wise Co-linear That means A is non-colinear to B B is non-colinear to C and C is non-colinear to A So please make a note of this Now people ask me What if they are non-co-planar What will happen So let me write that condition also If A B C Are non-co-planar Then expressing this Would lead to Inconsistent Values of X and Y Would lead to inconsistent Values of X and Y Let me try to explain this Through a simple example Let me try to explain this Through a simple example Let's say we have By the way if you want to copy this down You can copy it quickly Before I give you a problem Let me show you that X and Y pair is unique You cannot have More than one X Y pair For expressing C In terms of A and B So let me explain first Prove for uniqueness of X and Y Let's say C could be expressed Not only in terms of A and B like this But also C Can also be expressed as PA plus QB Correct In both of them represent the same vector That means XA plus YB Should be equal to PA plus QB Correct In other words X minus PA Y minus QB must be equal to a null vector Now since A and B are Non-co-linear It implies A and B are linearly dependent Sorry Linearly independent Linearly independent And if they are linearly independent What have I told you That individually both these terms must be 0 0 Just like I and J Remember Take the example of I and J in your mind Whenever you are confused with this So This two expression Must be individually B 0 That means your P becomes X And Q becomes Y That means this pair X comma Y Is unique It implies X comma Y pair is unique Okay Now let me cite an example What do I mean when I say When the vectors are non-co-planar There would be inconsistent The values of X and Y Will come up if you try to do this C as a linear combination Of A and B Let me give you a question Are these vectors Are these vectors Co-planar Co-planar means do they lie On the same plane Can you pass a plane Through these three vectors Okay Now let us try to Apply Let me go back to the previous page Let us try to apply this concept Let us try to express One of the vectors As a linear combination of the other two And try to see Whether do I get a unique value Of X and Y Or do I get an inconsistent value Of X and Y Let us try that out Let me express any one of them You have to express the first one In terms of the other two Any one you pick up Let us say I pick up the last one 4 I minus 2 J Minus 3 K And I try to express it as Let's say the first vector Okay Plus Y times The second vector Let me try to compare The I, J and K coefficients So 4 is equal to Minus 2 X minus 2 Y That means X plus Y is minus 2 Number one equation Second equation is Minus 2 X Plus 4 Y That means X minus 2 Y is 1 Okay That's the second equation Third equation is going to be Minus 3 is equal to 4 X minus 2 Y Okay First let me try to solve 1 and 2 And try to see Whether the third one is also satisfied By the same result Correct So if I solve this, let me subtract it So 3 Y is equal to Minus 3 Minus 1 So X plus 2 is equal to 1 So X is also equal to minus 1 Okay Let me check whether this satisfies it Minus 3 is equal to Minus 4 And this is going to be plus 2 No this is not correct That means here X Y value is Inconsistent X Y value is inconsistent It doesn't satisfy my third equation If it doesn't satisfy my third equation What conclusion do I draw? What conclusion should I draw? No They are Non-coplanar Is that clear everyone? Please type clear If it is clear Okay Now there is a simpler way to look at it There is a simpler way to look at it Right now Focus on these three equations All of you please Focus on these three equations If you look at these three equations Basically These equations are these three equations Now these are basically Three lines Okay So when I say if they are coplanar They should have a common solution That is what I mean to say Consistent Unique pair of X and Y That means these three lines Must have A common solution In other words they must be concurrent If they are concurrent Please try to recall your Straight lines chapter In your class 11 You have learned that If A1X Plus B1Y plus C1 Equal to 0 A2X B2Y plus C2 equal to 0 And A3X Plus B3Y plus C3 equal to 0 If these three lines are concurrent If these three lines are concurrent Everybody knows the meaning of concurrent Concurrent means All of them Will pass through a common point This is the meaning of concurrency Then remember we had Learned this that this determinant Must be 0 The coefficients of X and Y And the constant term Must be 0 You can actually use the same thing And make your life easy So let's check whether the determinant Formed by these coefficients So what are these coefficients 2, 2, 4 2 minus 4 minus 2 4 minus 2, 3 Is it 0? Let's check So let me expand it So 2 times Minus 4 Minus 2 times 6 plus 8 And 4 times Minus 4 plus 16 This is going to be minus 32 This is going to be minus 28 And this is going to be 48 I'm sure this is not 0 If they are not 0 Means they are not concurrent Means they will not have a Common value of X and Y satisfying it And if they don't have a common value Of X and Y satisfying it That means there is an inconsistency In the value of X and Y And hence these 3 vectors Are non-coplanar By the way One more important finding over here The finding is If you see these Numbers here 2, 2, 4 2 minus 4 minus 2 4 minus 2 and 3 You would realize that You could actually also write it as Let me write it here You could also write it as 2, 2, 4 2 minus 4 minus 2 And 4 minus 2, 3 Correct Basically I have tried to Transpose it but there are Actually a symmetric matrix So nothing happened So if this is 0 Let's say this is 0 You could actually You could actually make these numbers As the Let's say I put a minus 2, minus 2 And this I put as minus 2 Plus 4, minus Plus 2, sorry Minus 2, 4, minus 2 4, minus 2, minus 3 This will actually form The coefficients Of these 3 Vectors which were provided to us Correct So this brings me to one more important Theory that you will anyways Learn later on That's an easy method to Find the co-planarity of 3 vectors Easy method to check the Everybody please make a note of this Easy method to Check the co-planarity of 3 Vectors Let's say the vectors are A1 I, A2J A3K B1I, B2J, B3K And C1I C2J, C3K If you want to check Whether these 3 vectors are co-planar or not Then you just put these Coefficient of IJK In form of a determinant like this And if this determinant happens to be 0 Then yes they are co-planar Okay And if this determinant is not equal to 0 That means this is not equal to 0 That means it implies They are non-co-planar Okay This is a much easier and effective way to Solve Or prove the co-planarity of vectors Is that fine Clear everyone can we take a Problem on this Let's take a question on this Oh sorry this is a question on Co-linearity The question is in front of you The non-co-planar vectors prove that These 3 vectors are also Non-co-planar That's the first part of the question Second part of the question is If these vectors are non-co-planar Prove that A is 4 Yes there is a slight mistake in this question I wanted to check whether you are getting it right It's actually A equal to minus 4 You are correct Aniruddha Okay most of you are done Good So first part if ABCR Prove that these vectors are also Co-planar So if you want to prove that these Vectors are co-planar Now remember this is Very similar to saying that Your A is like your I cap B is your like J cap And C is your like K cap Which themselves are non-co-planar Correct So prove that 3i minus 7j Minus 4k 3i minus 2j plus k Co-planar Indirectly you are trying to prove that Okay I'm sure many of you Would have directly Made that determinant Correct But let me tell you You should also mention the fact that One could be expressed As Linear combination of the other two Which ends up giving you 3x plus y is equal To 3 Minus 2x plus y is equal To minus 7 And x plus 2y is equal To minus 4 Okay That means 3x plus y minus 3 equal to 0 Minus 2x plus y plus 7 equal to 0 And x plus 2y plus 4 equal to 0 So if this should have Consistent solution That means they must be concurrent So let us check The determinant formed by The coefficients over here So 3 1 minus 3 Minus 2 1 7 Okay, let's find this value So 3 4 minus 14 is Minus 10 minus 1 Minus 8 minus 7 And minus 3 Minus 4 minus 1 That will give you minus 30 Plus 15 Plus 15 And yes it is coming out to be 0 That means these 3 vectors are coplanar Okay That's the first part of the question Second part of the question If these 3 vectors are coplanar Prove that a is minus 4 So here you can directly use your determinant 2 minus 1 1 1 2 minus 3 3 a 5 This must be equal to 0 Let us expand this So you will get 2 times 10 plus 3 a Plus 1 You will get 5 plus 9 Which is 14 And plus 1 a minus 6 Which should be equal to 0 Which is nothing but 20 plus 14 Minus 6 and this is going to be 4 a Okay, I think it gives you Why did I write 4 a over here That's 7 a right Sorry Let me just connect this 7 a So this gives you 7 a is equal to minus 28 So a is equal to minus 4 Clear everyone, how does it work Any doubt regarding solution of this question Please punch clear Great Next concept that we are Going to talk about is the Concept of coplanarity Of 4 points Or 4 position vectors Coplanarity of 4 Points or 4 position vectors Now again remember I am not Going to talk about coplanarity Of 3 points because 3 points Will always be coplanar Getting my point You can always pass a plane To any 3 points So let's say A, B, C are 4 points A, B, C and D Are 4 points with the following Position vectors Okay, if they are Coplanar please note The vectors formed By choosing Any 2 of these points So let's say I make 3 vectors A, B, A, C and A, D These 3 vectors Are also coplanar So what are you doing Using your previous concept To handle this So if you know how to prove the Coplanarity of 3 vectors You can also prove the coplanarity Of 4 points Simple you just have to make 3 vectors out of those 4 Any 3 vectors you can make Not necessarily A, B, A, C, A, D You can make A, B, C Any Let me write it as Any 3 vectors Formed From These 4 points Will be Coplanar Are you getting this point Now I'm not going to talk about it Because you already know this condition This is already known to you So you can manage it through the use of The test of coplanarity of 3 vectors Correct? This is an alternate method also If these 3 points These 4 points are Coplanar That means There's one more way To actually prove their coplanarity That means if you choose Let's say X, Y, Z, W Scalar quantities such that This is equal to null vector Then this implies X plus Y plus C plus W should be 0 But all not 0 This is another way to prove it But this is less used This is more widely used So this is slightly complicated The second method is slightly complicated First method is more widely used Is that fine? Let's take a question on this So please read this question Question says Position vectors of the point A, B, C, D Are given to you Like this, this is your A This is your B This is your C And this is your D By the way, this is 4 plus 4i 4i plus 5j plus Lambda So if these 4 points A, B, C, D Lie on a plane They lie on a plane Find the value of Lambda Please type in your response once you are done Okay, what is that method I already discussed 2 methods Can you just do it And send me a pic on my whatsapp See, one method implies the other It's not like they are exclusive of each other There has to be a connect between both the methods So when I say method one Method two, it's not like Both are mutually exclusive of each other One is basically Expressing the same thing in its own way Oh okay, never mind After the class, you can share the method with me I'll be Interested in seeing what you have done Oh my god, that's a horrible answer So right now I don't know the answer But Let's solve it, let's see the response from other people So Aakarsh has already told me Some answer I don't know whether it's right or wrong Right now Aakarsh Let's wait for a few more people to answer Okay, RMR has a different answer All together Anybody else Now I realize that figures are quite ugly For you all Yeah Vidushi has some different answer Shaughnak is asking for a minute Okay Not getting confused No worry Chirag We'll deal with it Just let's give others one minute Okay, Shaughnak is getting a different answer Fine So let us make quickly Three position vectors Three vectors out of it A, B, A, C and A, D probably Okay So A, B will be what A, B will be B minus A vector So B minus A vector will be Minus I Plus 5J Minus 3K Okay, remember for A, B vector You just have to do O, B minus O, A That is position vector of B Minus position vector of A So I subtracted 2I minus 3I Which is minus I 3J plus 2J which is 5J And minus 4K Plus K which is minus 3K Okay A, C vector will be C minus A So C minus A will be Minus 4I Plus 3J Plus 3K Minus 3I Plus 3J And plus 3K A, C vector would be I Sorry, A, D vector will be I Plus 7J And you will have lambda plus 1K Okay Now these three vectors must be Coplanar For these three points to be Coplanar That means the determinant Formed by Minus 1, 5 Minus 3 Minus 4, 3, 3 1, 7, lambda plus 1 Should be 0 Let us expand with respect to the third Row So 1 into Hide this, hide this 15 plus 9 Which is going to be 15 plus 9 Is 24 Minus 7 Hide this, hide this Minus 3, minus 12 is 15 And finally, lambda plus 1 Times Minus 3 plus 20 Which is 17 Okay So let us simplify this 24 And this is going to be 105 Plus 17 Plus 17, lambda equal to 0 So this gives you the answer Lambda as Minus 146 by 17 Okay, all of you Started realizing where you went wrong Clear? Okay, now how would I solve this Problem if I had to use the previous method Sorry, if you have to use the second method Okay In second method, if you want These three vectors to be Coplanar, remember we had To express this As a null vector And this should imply X plus Y plus Z plus W should be 0 Right, let us do that as well So X A Let's write A vector here A is our 3i minus 2j minus k Y B will be 2i plus 3j minus 4k Plus Z Minus i plus j plus 2k Plus W Let me check the other vector It is 4, 5, lambda 4, 5, lambda 4, 5, lambda This must be equal to null Correct So if you compare the coefficients Of X, it will be Sorry, if you compare The coefficients of i, it will be 3x Plus 2y Minus Z plus 4w Be equal to 0, that's one equation Compare the coefficients of j Minus 2x Plus 3y Plus Z Plus 5w must be 0 Compare the coefficients of k Which is minus X Minus 4y Plus 2z Plus lambda w equal to 0 And the fourth equation is X plus Y plus Z Plus W equal to 0 Remember Here X, Y, Z W Not all 0 They are not all 0 That means this homogeneous equation Must have a non-trivial solution That's the meaning of not all 0 0, 0, 0 is of course a solution But we don't want that to happen It must have a non-trivial solution That means What is happening My good friends over here The 4 by 4 determinant Our here must be equal to 0 Getting my point I think for the first time You are dealing with a 4 by 4 Never mind So here also All the properties are valid 4 by 4 determinant You can do one thing You can take your column number 1 And subtract it from other columns So I can do Take your first column Subtract it from the other columns So what will happen This will remain same 3 minus 2 minus 1, 1 This will become a minus 1 5 minus 3, 0 Minus 4 3 3, 0 This will become a 1 7 Lambda plus 1, 0 Now let us expand with respect to the first Sorry, respect to the last row Expand With respect to R4 Ok So it will become 1 Only this part will be seen Which is actually the determinant Formed by Minus 1, minus 4, 1 5, 3, 7 Minus 3, 3, Lambda plus 1 This should be equal to 0 Let us expand this Let me expand with respect to the R4 Minus 3 Minus 28, minus 3 Minus 31 Minus 3 Minus 7, minus 5 Minus 12 Lambda plus 1 Minus 3, plus 20 Minus 17 93, plus 36 17, plus 17 Lambda equal to 0 Right So let us choose 17, Lambda is minus 146 Lambda is minus 146 by 17 Yet again Actually, I am happy to see this That is why I put this smiley So that is how you get the solution In an alternate way But by using a 4 by 4 Now do you realize why people hate the other method Why people don't write this Because it leads to a 4 by 4 Determinants to solve But nevertheless this determinant Thankfully we get all 111's In one of the rows Because this will always be there As an equation for you Is it clear or not Please punch clear If it is clear to you Questions Questions Let's take questions Let's take A slightly different question altogether Points A, B, C, D are related as Points A, B, C, D are related as And this where X, Y, Z, W are Scalers Some of any two X, Y, Z, W Is not 0 Prove that if A, B, C, D are Concyclic Then this equation must be true Anybody who is done can send Their Working to me personally on whatsapp Coincyclic means they are All these 4 points lie on a circle So if 4 points are Coincyclic they will lie on a circle So let's say these are your 4 points A, B, C, D So these are your position vectors A That means you can always make a Quadrilateral with them But anyways you don't need that They are just Coincyclic Let's say this is point B Anyone Anyone is trying at least If not I can start solving Okay Any idea? Let's talk about it The first thing that you would all Know If there are 2 chords Which are intersecting at a single point Okay Then P A into P B Is equal to P C into P D P A into P B is equal to P C into P D How many of you are not aware of this How many of you want me to prove This first Just let me know if you are not aware of this Just write not aware That means I have not got any response Means everybody is aware By the way the proof of it is very simple If you connect a triangle like this You know that this angle Is this angle right Let me call this as angle X This angle is angle Y This angle is angle Y And of course these 2 are ZZ each That means triangle A D is Similar to triangle Oh well done Midoishi Can you send me the solution Triangle P A D is Similar to triangle P C B Okay So I can say P A by P D is P C by P B So P A by P D Is equal to P C by P B Okay cross multiply So this implies P A Into P B is equal To P D into P C Correct Now let's go back to the same question Once again Now it has been given that X A plus Y B plus Z C Plus W D by the way let me remove these X Y because they already been used up In the question so that you are not confused With this X and Y on the X and Y given in the question So let me just It is this Is that fine So what is given to us is that X A plus Y B plus Z C Plus W D equal to a null vector Implies X Y X plus Y plus Z plus W is equal to 0 Okay So now if you look at this term Carefully What does this imply This implies that A B C D are Coplanar What else does it imply What else does it imply What are these X Y Z and W Does anybody have any idea What does this X Y Z and W Actually represent What does X Y Z and W Actually represent Exactly They actually represent this ratio This is your X Y Z and W terms How do you know that That's also very simple If you see this This point P This point P is nothing but X A plus Y B By X plus Y Okay The same point P is also Z C Plus W D By Z plus W Getting my point Now from here how do I conclude That X A plus Y B Plus W D is equal to a Null vector And X Y Z W is equal to 0 How do I conclude that Medushi So could you repeat one second My question is From here from these two expressions How do you conclude that This is 0 And this is equal to a null vector From here from these two conditions How is it evident that These two conditions that is mentioned in the question Are satisfied This and this are satisfied Anyone I mean this is an open question To all of you All I can conclude from here is this X A plus Y B By X plus Y Is equal to Z C plus W D By Z plus W Yes I know So from here how do I conclude From here how do I conclude That X A plus Y B Plus Z C plus W D Is going to be a null vector And X plus Y plus Z plus W is equal to 0 That is my question How do I get to these two conditions If these two conditions are proved That means the reverse is also True that means this ratio is Justified correct So once we have Justified this we have to come back To this conclusion also Because both imply each other I'll give you some time to prove this Think about it Now meanwhile let us say We take this ratio to be correct Then what do we conclude Then how do we use this We can use this equation now P A into P B is equal to P C into P D Okay So P A what is P A length Can I say P A length Would be I'll refer to this figure only here Can I say P A length will be X plus Y times Okay At this length This length itself is mod of A minus B Okay now you can see How these terms are making their appearance Okay Similarly P B Would be what By the way let me take a mod over here In case your X and Y is negative Okay P B would be what P B would be similarly Y times Mod A minus B Okay What about P C P C length would be again P C length would be W by W plus Z Times C minus D W by W plus Z Mod times C minus D P D length will be Z by W plus Z times C minus D If I use this formula now P A into P B Is equal to P C into P D You will see that I end up getting Mod X Y By mod X plus Y Square A minus B Square and Sorry this is going to be Z I am sorry I wrote a W Mod W Z by mod This C minus D Square Correct Now these two are same Y because X plus Y plus Z W is equal to 0 That means X plus Y is equal to Z Negative Z plus W Correct So modulus of these two quantities Would be the same So square will also be the same So you can cancel them off I think this is what we wanted to prove In this Mod X Y Square is mod Y W Z Mod C minus D square Are you getting my point? Now does this ring a bell In your mind that how this will lead To this? So if you take your Z plus W Let's say you consider this to be Correct then it will be negative X plus Y Isn't it? So if you take that Then what will happen This will become negative X C Plus W D X plus Y That means It will lead to X A Plus Y B plus Z C Plus W D equal to a null vector That means if you assume this to be true You can actually prove this And vice versa So ultimately by the use of These two you can prove this To be true And from this you can actually conclude That this is how the ratio Is basically divided So this is a very good problem Which actually tells you That the second method that we were using What was that X Y Z and W in that Getting my point Can we take up next question now? Let's take this question It's a geometry based question but we have to do it By using vectors Let P be an interior point of a triangle A B C And A P B P C P Meet the sites B C C A A B In D E F respectively Show that A P by P D is A F by F B Plus A E by E C Yes anyone please type done if you are done Let me sketch a figure meanwhile So A So P is some interior point Says that A P B P And C P meets the side Okay so A P B P And C P It meets the side B C in D E And F We have to show A P is to P D Is A F is to F B A F is to F B Plus A E is to A C Okay so here we can say clearly That A B C And P are co-planar So I have given a position vector To all these points So can I say X A plus C Plus W P Is equal to another vector Okay Then X plus Y plus Z plus W should be equal to 0 Yes I know Is this thing clear everyone How do we use this concept How do we use this concept to solve this Given problem Now let's look at this expression here And these two expressions Can I write from there that X A plus Y B Okay Is equal to negative Z C plus W P Okay Let me divide both sides with X plus Y Now we know from this equation That X plus Y Is Negative of Z plus W Correct So if I replace this The term by negative Z plus W Negative negative will go off Correct Which means X A plus Y B by X plus Y Is equal to Z C Plus W P By Z plus W Isn't this resembling section formula Correct If this is resembling section formula Tell me can I conclude anything about X A plus Y B by X plus Y Means this ratio They are taking it as X is to Y Okay And They are referring to as F point The same as Z C is to W P Z C C and P point Okay so C and P point They are referring to So this point and this point They are referring to This ratio here F is dividing the joint of C and P Externally In what ratio This is W is to Z I think Z is This is your Z and this is your W Okay But never mind we don't need that We just need this fact that A F is to F B which is required in our problem As you can see A F is to F B is required That is going to be Y is to X Okay Now I can write down The very same expression over here In a slightly different way I can write it as X A Plus Y W Sorry X A plus W P Equal to negative Y B plus Z C Okay Let me divide by X plus W on both the Sites That X plus W is Negative Y plus Z So if I replace this with Y plus Z Negative This and this will get cancelled off Okay What does this tell you Let's focus on A and P point A and P point And this is B and C point This is B and C point So this tells you that Point D Is nothing but Y is to Z Point P is Y is to Z And what is this ratio What is A P is to P D here A P is to P D Is such that this relation Must be satisfied by D This is the position vector of D remember This is the position vector of D X A plus W P By X plus W So what is this ratio By P D Can anybody tell me P by P D Let's say this is K Is to 1 This is K is to 1 Okay So this is the case of external division So according to this formula Your position vector of D should have Been K into P Minus 1 into A By minus 1 plus K Correct Compare these two If you compare these two You will come to know that W by X Is actually minus 1 W by X is actually Minus K Yes or no So K is minus W by X Correct And A P is nothing but K minus 1 is to 1 So I can say P D A P is to P D Or A P by P D Is K minus 1 is to 1 Minus W by X minus 1 Yes or no This is also a very important Result for us So one result is here Y by X Other result A P by P D Is minus W by X minus 1 Remember both of these results are required A P by P D is required here And A F by F B Is required over here Now I have to find A E is to A C How will I do A E is to A C Now all of you please focus here A E is to A C connects A and C So keep A and C together So write it like this X A plus Z C Is minus Y B Plus W P Divide by X plus Z Both the sides Divide by X plus Z Both the sides X plus Z X plus Z Is negative W plus Y So I can replace this with Negative Y plus W Negative negative gone So what does this say This says that Point E Point E coordinates This is nothing but X Sorry, sorry, sorry, sorry Opposite, X is to Z Okay So I can say A E by E C Is X by Z A E by E C Is equal to X by Z So this is your number 3 achievement A E by E C is Sorry, Z by X Not X by Z, Z by X Now let us try to see whether These 3 bracketed Expressions Satisfy this condition or not So AP by PD Let me write AP by PD Which is nothing but Minus W by X minus 1 Okay And let us do A F by F B Which is Y by X Plus A E by E C A E by E C Which is nothing but Z by X So it becomes Y plus Z by X Now they are claiming that These 2 are equal Are they? Let us check So if you Say minus W by X minus 1 Is equal to Y minus Z by X Basically you are trying to say Minus W minus X By X is equal to Y plus Z by X X X cancel That means you are trying to say X plus Y plus Z plus W is equal to 0 Which is yes it is true It is true from this expression Here itself it is mentioned That X plus Y plus Z is equal to 0 So yes Whatever is Asked here in this particular proof That is correct Okay So basically here we are using A very deep understanding of We are treating these X, Y, Z and W As the ratios In which these points are arranged In space Or cutting each other in space Or cutting each other in the plane Getting my point It is a very beautiful problem Which opens a lot of concept So I will just take one more problem here Sir can you explain How you got the ratios again Yeah sure see First of all I took these two terms together This and this And I send these two to the other side So I got this term You understand this expression right From here Now basically here They are pointing towards Position vector Of two points Here they are pointing towards Position vector of two points So there is a point Which divides A and B In some ratio And also C and P in some ratio That means that point Is collinear with A B As well as C and P So any point which is collinear With A and B and also with C and P has to be your F point Correct So if you just look at The left hand side of the expression It indicates to the fact that F is dividing A and B in the ratio Of Y is to X Now look at the reverse Just try to recall Your section formula in reverse If there is a point Whose position vector is this Don't you think that point divides A and B in the ratio of Y is to X That's how we wrote over here Y is to X That's how AF is to FB became Y by X Is it clear Aditya Aditya Can you similarly relate it to the other facts For example now when I write it as XA plus WP Is equal to YB Plus ZC YX plus W and Y plus Z respectively There is a point which divides A and P That means it is collinear with A and P So collinear with A and P is D Which is also collinear with C and B Are you getting it So from this understanding I came to know that D is dividing The join of A and P In some ratio here this ratio is external ratio That's why we got our answer Of AP by AD as Negative W by X minus 1 In similar way when I wrote XA plus ZC by X plus C Is equal to YB Plus WP by Y plus W It means There is a point which is collinear To A and C There is a point which is collinear To A and C and also to B and P Also to B and P Which has to be E point There is no other point There is no other point which is simultaneously collinear To A and C as well as B and P And this gives you a very simple Understanding that A and C are divided in the ratio Of Z is to X by E point So AE by EC Is equal to Z by X ratio Once I got that I just try to fit Into this formula And obtain something which is a fact According to the question So when I obtain these ratios I just put it in this And try to equate it to see whether It gives you a truth or not And it actually gave me a truth condition That means this is correct That means this is correct This is equal to this is correct And shown Okay It's a very good question actually if you see this Yes, Sridhi rightly said It's an impossible question if you haven't done This type of question before Okay, guys one thing we'll do is We are going to start with Scalar product next class So I'm going to stop it over here because Any other concept which I'm going to take up We'll just take the Flow of the class in the next session Okay, so I'll start afresh With the concept of Next class We'll start afresh with the concept of Dot product And vector product Or cross product what we call it We'll try to wrap this up In the next class Okay, I'll also try to Cover up scalar Triple product in this Okay, and in the next to next class We are going to talk about Next to next class We are going to talk about Vector triple product And some miscellaneous concept We'll include vector equation And vector equation System of reciprocal vectors All those stuff Langrange's identity I wanted to have one class With you on differential equation Online Okay Early October 5th, 6th or 7th Okay Because we don't have much time At our hands, so we have to complete 3D also and A differential equation And area under the curves So 3D I wanted to do with you In a physical class Differential equation needs Lot of problem solving, so I wanted to have An online session, okay So there would be one of these days In fact two of these days When we'll conduct a class 8th time reserving for the Deshera Voltage