 Good to go. So we are going to begin with a new chapter today that is vector algebra. Not a new concept to you. So in today's class you'll find that most of the things you're already aware of and in those spaces I'll be a little faster in order to save your and my time. So I'll just begin with a broader level understanding about vectors. Most of the physical quantities that we see around us that we see around us can be broadly classified as scalars and vectors. Scalars and vectors. Let me tell you there is a third quantity as well which is mostly referred to as tensor quantities. Now scalar and vectors themselves are tensor quantities. Normally we say a scalar is a tensor quantity of rank 0, vector is a scalar quantity of rank 1. There is something called dyad which is a tensor of rank 2, there is a triad which is tensor of rank 3 and so on. But I'm not going to talk about those because that anyways you're going to study later on in your undergrad courses. So meanwhile how do you define a scalar quantity, any physical quantity which can be represented only by stating its magnitude. So any magnitude is sufficient enough to state it. Example, mass, distance, speed, temperature, money etc. Here as those quantities which require magnitude, direction, also called as sense, right? Can I see these two parameters are sufficient enough for a quantity to be called a vector quantity? Yes, no? Yes, sir. Okay, what about current? Current has magnitude as well as direction, is it a vector quantity? Definitely not, right? So these two are not sufficient enough, they are necessary but they are not sufficient enough. You need a third parameter as well which we call as the vectors must follow the parallel gram law of addition which currents don't follow. For example, you must have all heard about KCL, correct? So let's say there are two currents coming up like this of 5 ampere and 3 ampere and let's say the resultant of these two is going in this wire, so you know that 8 ampere is going to flow in this wire. So it's just a algebraic sum that you are doing, okay? So it doesn't depend upon the direction of these wires. So for example, if I have 5 like this and let's say if I have 3 like this, okay? And there is an outlet over here, this still will carry 8 ampere. So it doesn't depend upon what is the direction of these wires, okay? These currents will still algebraically add up to give you 8, okay? So they don't follow the parallel gram law of addition, hence current cannot be called as vector quantities, okay? So remember a vector should have magnitude, direction and these two are not sufficient enough it must also follow the parallel gram law of addition or triangular law of addition whatever you call or polygon law of addition, okay? Examples are displacement, okay? Most of the vectors you would have already come across in your physics, displacement, force, velocity, okay? These are vector quantities. How do we represent a vector? How do we represent a vector? We represent the vector by using a small alphabet with an arrow on top, okay? So A with an arrow on top or if you don't put an arrow, mostly in books they don't put an arrow then they would write it in bold, okay? You would also state a vector by stating its initial and the terminal position. So for example, if a vector starts from A, ends at B and the sense is from A to B then we call it as AB vector. If the sense is from B to A, let's say if your vector is like this, then you would call it as BA vector. So here the initial point is your B and terminal point is your A. So this is called the initial point and this is called the terminal point, okay? Now the three important things a vector should have, what we call as a characteristic of a vector, okay? So vector has the following three characteristics. One, it has got a magnitude, okay? Which we represent as the vector written within two parallel lines. It actually gives the length of that line which you are using. It is synonymous with the length of the line that you are using to represent the vector. So longer the length of the line you have to make, the longer is the more magnitude that vector has. For example, if I make these two vectors where this length is double of this length, so let's say this length is 1 centimeter and let's say this is 2 centimeter. Then this vector that is vector B would be, let's say vector B is parallel to vector A. Then vector B will be double of vector A, okay? So it is double the magnitude of vector A. Direction is also the same in this case, okay? So magnitude is nothing but it gives you an idea about the strength of that vector quantity, okay? Second thing that it has is sense or direction which I have already stated, okay? Sense or direction. And the third thing is your support. Now what is support? Support is basically the infinite line from which the vector has been carved out, okay? So it is a line from which the vector has been carved out. The vector is taken. For example, if you say AB vector like this, this AB vector belongs to an infinite line like this, sorry, it belongs to an infinite line like this, correct? So from this infinite line, you have taken out a vector like this which you are calling as your AB vector. Please note that this infinite line will be called as the support. Not very important concept but you should be knowing about the names involved while you are talking about vector quantities. So this is called a support, okay? Next we are going to talk about types of vectors. So I will be slightly faster because you are only aware of all these concepts. Types of vectors. I will first start with zero vector also called as the null vector. What is the zero vector? Something whose magnitude is zero, okay? But direction is undefined. So a zero vector is a vector whose magnitude is zero. Direction is not defined. It is just like a zero polynomial, right? What is the degree of a zero polynomial? Undefined. What is the argument of a zero complex number? Undefined. In the same way, zero is also a vector quantity but its direction is not defined, okay? Why do we need it? For example, let's say you walked 5 kilometers towards north, okay? Then you walked from the same point back 5 kilometers south, okay? What is your total displacement? You will say zero, right? So you started from the same place, you are back to the same place, correct? So zero vector is your displacement here and the direction of this would be magnitude is zero and direction for this would be undefined, okay? Because if you tell somebody I had zero displacement, he has no way to figure out what was your direction of motion. You could have gone in any direction and come back to the same position. Next is unit vector. What is unit vector? Unit vector is basically a vector which is first of all represented by a cap, okay? It is always a vector which is in a direction of a particular vector, let's say vector a and its magnitude is unity, okay? So a unit vector in the direction of a vector a is nothing but that vector which has the same direction as the direction of a but has a magnitude of one, okay? So this vector will have the same sense or same direction as that of a, as that of a, okay? How do we find the unit vector? Unit vector is found by dividing the vector by its magnitude, okay? Some of the well-known unit vectors that you would have already come across would be your i cap, j cap, and k cap, okay? i cap is a unit vector in the direction of, in the direction of positive x axis. Positive x axis. If you want a unit vector in the direction of negative x axis, it will be minus i cap. j cap is the unit vector in the direction of positive y axis and k cap is the unit vector in the direction of positive z axis. Third type of vectors that we are going to talk about, co-initial vectors. What are co-initial vectors? Vectors which have the same initial point. For example, these two are co-initial vectors because they have the same initial point over here. Okay? Normally, in order to add vectors, we need to make some vectors co-initial. Okay, we will talk about this when we are talking about the parallelogram law of addition of vectors. Similarly, vectors can be co-terminus also. What are co-terminus vectors? Co-terminus vectors means they have the same terminal point. Let's say A, B. Okay? They have the same terminal point. Let's say P. Okay? Normally, you are talking about co-terminus vectors when you are dealing with the forces acting on a body. Okay? So when you draw two forces acting on a mass M, let's say one force is acting like this, other force is acting like this, they become co-terminus. Next. Equal vectors. So two vectors A and B are said to be equal. Are said to be equal. When number one, their magnitudes are same. Somebody is typing something. Oh, is it so? But I can hear my audio properly because I am in the middle of a problem. Huh? Even I am facing that problem. Oh, is it because of the bad weather? Because it's raining over here. There's some background sound from your side. Oh, okay. I understood what is the background sound. One second. Is it clear now? Yes, sir. Yes, sir. Okay. Yeah, let me see who are the submitters also there. Okay. Yeah, when are two vectors said to be equal when their modulus are on, when their moduli are same and secondly, their sense must also be the same. Direction must also be the same. Direction is also the same. Right? For example, A and B vector of the same length, same direction, then A and B vectors will be said to be equal. So vector A will be equal to vector B. Okay. Next. Like and unlike vectors. What's a like and unlike vectors? Two parallel vectors which have the same direction are called like vectors. So we say A and B are like vectors, are like vectors, when they have the same direction, when they have the same direction. Okay. Their magnitudes need not be equal. For example, A can be like this, B could be like this. Okay. So where we say these two are like vectors. Okay. Similarly, A and B would be called as unlike vectors. Okay. When these... No, remember that when they have same direction, I mean to say they are parallel and that sense is same. Unlike vectors, A and B would be parallel, but their direction would be reverse. We have reverse direction. Getting the point. So let's say A is like this. B is like this. Then in this case, A and B would be unlike vectors. Okay. Next. What was the number which I was at? Sixth. Okay. Seventh. Next is, this is very important. The concept of free vectors and localized vectors. Free vectors and localized vectors. What is a free vector? Free vector is something which has no fixed initial point and terminal point. For example, a vector A like this, if it is free, then if you draw it like this also, it will be the same. If you draw it like this also, it will be the same. They will all be A, A, A vector. Okay. So this is free to move. Okay. Now, for example, displacement. Okay. If I say I'm going five kilometers towards North. Correct. It doesn't matter whether I'm starting my journey from Bihar or I'm starting my journey from Punjab or I'm starting my journey from Bangalore. All these vectors, even if I go five kilometers North would be considered to be the same. Correct. Does it actually matter from where I'm starting my journey? It has no fixed. So free vectors, they don't have fixed initial point. So no specific initial point. And since there's no specific initial point, there's no specific terminal point also. Okay. But remember the magnitude and the sense must remain the same throughout. Okay. That means you can translate, you can take them parallel to themselves anywhere in space, but you can't change their direction and you can't change their magnitude. Look, by the way, let me tell you in maths, you always deal with three vectors. Okay. In maths, your vectors are free. Okay. So use in mathematics. In maths, we normally take a vector parallel to itself anywhere we want. Whereas localized vectors, they have fixed or specific initial points. An example of it would be a simple example of weight of a body. Okay. So let's say there's a body like this. Let's say it's weight force. It will always start from the center of mass of the body. Correct. It will always start from the center of mass. It cannot go anywhere. For example, I cannot draw a body over here and say this is the weight force of the body. Correct. It will be long because this is a localized force. It will always start from the body. If you attach a string to the body also. Okay. Its direction will always be opposite starting from the object and going the other way around. Isn't it? So tension is a localized force. Okay. So localized four vectors are mostly used in physics. Use in physics. Okay. So unless Intel stated, in fact, in all the cases that we are going to deal with, we are going to talk about free vectors, not localized vectors. Any questions so far? No, sir. Okay. Now I'm going to quickly jump into the actual operations on vectors. Operations on vectors. How many of you have taken bio? Me, Arthra, Samyukta, Likhet, Vihamsini, Rashmi. How many chapters are left? How many? Don't say sir, we haven't even started. Consigns, guys. I think Arthya has consigns, right? Yes. How much load do you have? Not, nothing left actually. Nothing left, all done? Yes, sir. Good, good, good. Okay, at any stage if you feel that you are worried about the preparation, let me know, this session is recorded, so don't worry about it. Or you can, I'm sure you would must be sitting with your bio book open in front of you. I can see that. How many of you have been sitting with the bio book in front of you? Nobody will admit right now, but I'm sure most of you would be. See, I don't mind because exam is important. I actually wanted not to have this session, but then I thought that you already have a lot of days left for, you already got a lot of preparation time, so I think you would have already prepared for it. Anyways, let's continue with the class. But in any stage you feel that you want to focus on your tomorrow's exam, you can just, you know, you can just be mute and just, you know, keep the session open in front of you. Fine. So operation on vectors, we'll first talk about addition. Addition of vectors. And I think you have done this a couple of times to find resultant. That's why we call it as finding the resultant of vectors. Also, okay. Addition is as good as finding the resultant of vectors. So when you add two vectors A and B, and you get a vector C from it, we call C as a resultant of A and B. Now, how do we add two vectors? There are multiple ways to do it. The first way that we are going to talk about is the parallelogram law of addition. So first method is a parallelogram law of addition. Suitable for those cases where your vector is localized. Okay. But it can also be applied to free vectors. Okay. So parallelogram law of addition is used when your vectors are localized. For example, if you want to, if there's a, if there are multiple forces acting on a body, that's it is a body of mass M and force F1 is acting in this way. Force F2 is acting in this way. And you want to find out the resultant of these two forces. Okay. Then we can use the parallelogram law of addition. This law says that if you have two vectors, let's say vector A is like this and vector B is like this. Let me draw it slightly. Yeah. If you want to add them or if you want to find the resultant of these two vectors, first make them co-initial. Okay. How do you make them co-initial? By connecting their initial points. So this is the initial point of A. This is the initial point of B. Let me make it co-initial. So if you make it co-initial, you have to either take A parallel to itself so that it comes over here or take B parallel to itself so that it comes over here. Okay. Any of the two you can do. So let's say I take B vector parallel to itself. So A is like this. B, I brought it like this. Okay. Remember, I have not changed their magnitude and direction. So this law says that complete a parallelogram, complete a parallelogram such that these two vectors form the adjacent sides of the parallelogram. Once you have completed the parallelogram, a vector originating from the common co-initial point, going till this opposite end, that vector would be called your resultant of A and B. Okay. Couple of things here to be noted that this A plus B will always be having a magnitude less than equal to the sum of the individual magnitudes of A and B. Okay. This is because of the triangle being formed over here. So as you can see, this is your vector A. Okay. So if you see this triangle, magnitude of A plus B is the length of the third side, which will always be less than the sum of the other two sides. Okay. This is also called the triangle inequality. Next method that we normally adopt to add vectors is called the triangle law or triangular law of addition. In this, what do we do is we connect the terminal point of one with the initial point of the other. So let me take the same two cases. This is your vector A and this is your vector B. Okay. So what do we do? We connect the terminal point of A to the initial point of B. You can do the other way around also. It doesn't make a difference. You can connect the terminal point of B with the initial point of A. So let me do that. So let's say A goes like this, B is like this. Okay. B is like this. Then a vector starting from the initial point of A and going to the terminal point of B, that vector would be your A plus B vector. That means you're completing a triangle by doing this. Okay. That's why it is called the triangular law of addition of vectors. Okay. And same inequality holds true in this case as well. Now, triangular law of addition is actually a very specific case of what we call as the polygon law of addition. Polygon law of addition is suitable for those cases where you have more than two vectors involved. Okay. More than two vectors involved. Okay. For example, if I have a vector like this A, any other vector B, and other vector like this C. Okay. If I want to add these vectors, what I do is, again, I start connecting the terminal of one with the initial of the other, terminal of one with the initial of the other. And this you can do it in any order. It doesn't have to start from A. You can start with B also, you can start with C also, but just keep connecting the terminal of one with the initial of the other. For that, you need to move these vectors around also. So let's say I connect A, then I connect B, then I connect C. Okay. So, Polygon law says that the resultant would be a vector connecting the initial point of the first vector to the terminal point of the last vector. And normally in your books and all, they would draw this with dotted lines. Okay. This line represents that. This is the resultant of all these. Okay. So, this vector will be your A plus B plus C. Okay. Okay. Couple of properties that we need to know about vector addition, properties of vector addition. First property is vector addition is commutative. It doesn't matter whether you're doing A plus B or B plus A. They are commutative. Secondly, they are associative. So it doesn't matter whether you add A plus B and then add the resultant to C or whether you add A to the resultant of B and C. Okay. So this is associative. Okay. Third, zero vector is called the identity vector. Okay. So zero is called here. Zero is the additive identity. These terms have actually evolved because vectors form a very important vertical under linear algebra. Okay. So you need to know this however there's no direct uses of it. Then there's something called additive inverse. So my negative of a vector is called its additive inverse. So we say my negative A is the additive inverse, additive inverse of vector A. Okay. And I've already told you the triangle inequality that is A plus B mod will always be less than equal to mod A plus mod B. Equality will hold true when your A and B are in the same direction. Okay. Equality will hold when your A and B will be in the same direction. Okay. Now, when we talk about subtraction of vectors, we don't have to deal separately with subtraction of vectors because subtraction is actually covered under addition of A with negative of B. Now, what is negative of B? Negative of any vector is nothing but, it's a vector with the same magnitude. Let's say negative of a vector, let's say I call it a vector as vector P. Okay. Negative of a vector P is a vector which is having the same magnitude as P but direction is reversed. But the direction of, but the direction is reverse of P. Okay. I'm going to talk more about it in the scalar multiplication. The direction is reverse of P. Okay. For example, if I say vector A is going 10 kilometers towards north, then negative A is nothing but going 10 kilometers towards south. Okay. Simple as that. So when you're subtracting two vectors, you're adding A with the negative of B. Okay. So let's say, these are my vectors A. Okay. And this is my vector B. I want to find A minus B. I want to find A minus B. So what do I do is, first of all, I reverse the direction of this vector. So that means, let me take the reverse of vector B which is like this. Then I can follow any of the two laws that we have seen, whether triangle law of addition or parallelogram law of addition doesn't matter. So let's say I follow the parallelogram law of addition. So first I have to make them co-initial. Okay. So this is the initial point of A. This is the initial point of B minus B. So I have to make them co-initial. So let me do it like this. Okay. This is my A and this is my minus B. Then I complete a parallelogram like this. Okay. This vector here would be your A minus B vector. Okay. Now let me show you something very interesting. Had you completed a parallelogram with A and B itself, let's say I complete a parallelogram with A and B itself. Okay. Of course, this diagonal will represent A plus B. Okay. This is your A plus B vector. Okay. And you would be happy to know that in the same parallelogram, had you connected the other diagonal, as you can see, this length and this length are exactly the same. Now the direction has to be from the B point towards the A vector here. This vector will now represent a A minus B vector. So in the same parallelogram, one of the diagonal represents A plus B. Other diagonal represents A minus B. Okay. Now, I'm going to begin with something which is very important for problem solving, position vectors. Okay. What are position vectors? First of all, before I talk about them, let me tell you, these are the concepts which are going to connect vectors which are going to relate vectors to coordinate geometry. That's why it becomes very important that we understand this concept of position vector very, very carefully because many of the geometrical theorems or proofs can be done through vectors if we're able to understand what is position vector. Okay. Now what is position vector? Position vector is a vector which connects the position of a point with a reference position. Okay. With a reference position which we normally call as the zero vector also or the origin. So what is origin in case of coordinate geometry is the reference point or reference vector in case of vectors. So a vector which connects O to P like this, this vector would be called as the position vector. This vector would be called as the position vector of P. Position vector of P. Okay. And normally we write it like this. Let's say I call it as vector P. So this is your P vector. Okay. Now let me tell you, this is a very, very, I would say a rare way of representing a point's position vector. Normally they just say P having a position vector of this. They will never show you the origin. Okay. Origin is up to you to choose. See, it's just like mentioning a point. Let's say there's a point one comma two. Do I tell you that where is the origin? No, right? You choose your origin and accordingly you make a point one comma two. So origin position is never specified. It is up to you to choose your origin position, correct? Wherever you choose the origin position, relative to that you choose one comma two, correct? In the same way, if I say the position vector of a point is, let's say P vector, I would never say where is my origin. It is up to you to choose, you know, what is the position of the origin. Okay. I will just specify something like this. P has a position vector of P. Okay. So I will say something like this. P has a position vector of P. Q has a position vector of Q. Okay. Now let me ask this as a question. What will be P Q vector then? Can somebody tell me what is P Q vector in terms of small P and small Q? Sir, are you asking the distance between P and Q? No, no, I'm talking about the vector P and Q. Come on. I think you've done this chapter in school, right? Okay. I've got some response. Q minus P. Shreya says Q minus P. Why not P minus Q? How do you know it's Q minus P? Sir, they have the same initial point origin. So we need to take minus P in order to do it. Okay. See, let's say I take the origin over here. Okay. Now it's up to me where I take this origin. Okay. So when I say position vector of P is P, what does it mean? It means that a vector connecting, a vector connecting O to P is this P vector, correct? A vector connecting O to Q is your Q vector, correct? Yeah. Now, if you want to think as if you want to travel from P to Q, okay? And the only roads available are this road and this road. See the motion of my cursor. You can only take the known roads, this road and this road, correct? So if you want to go from, or you can use your vector addition. See, vector addition is nothing but trying to figure out alternative way to accomplish that vector path. Okay. So what I normally tell people who have not been introduced to vector addition is that, let's say you're trying to go from P to Q. So I'll take the path from P to O first. Okay. Now when you're going from P to O, you are traveling minus P vector, correct? Because you're going in the reverse direction of P. Now you're going from O to Q. O to Q means you're going in the direction of Q. So basically this will become Q minus P as you rightly said. Okay. Other way of looking at it is, let's say I consider that O P plus P Q will give you Q. Look at the way the arrows have been framed. O P plus P Q will end up giving you O Q. Okay. So P plus, let's say P Q is unknown vector. This will be equal to Q. So from here also we end up getting the same answer. Now in order to facilitate remembering of this, normally I say if you want to go from A to B, it's nothing but position vector of B minus position vector of A. Just remember this formula. Okay. We call this, I call this as destination minus source. So wherever you're going, that is your destination. From where you are coming, that is your source. So A, B is nothing but destination minus the source position vector. Okay. Now if this concept is clear, we'll take few problems on this, but before that I would like you to solve a simple question. Let's say this is a regular hexagon, A, B, C, D, E, F. Okay. Let's say AB vector is A, BC vector is B. Okay. Find the following vectors in terms of A and B. One is CD vector. So tell me what is CD vector? D vector, EF vector, and FA vector. In terms of A and B, express these following vectors. Remember it's a regular hexagon. I'm sure at least D vector and EF vector you can say very easily, correct? Yes. What is that? D would be minus A. Correct. D, it is minus A. EF will be? Minus B. CD, if you can find out, the negative of CD will be FA, right? But how do you find CD? Sir, can we say that the sum of this whole thing will be like zero? So it doesn't help you because CD and minus FA will cancel. AB and D will be cancelling. BC and EF will be cancelling. So how does it end up giving you the value of CD and EF? A, F. Okay, yes. Okay, let me help you out. Would you all agree with me that this vector here would be 2B vector because this is in the same direction of B and double of the vector B because of geometry? Okay, should I show you how it is double the length? I'm sure you can figure it out. This is 60 degree, correct? So this will be, let's say this length, whatever is this length, X. This will be X by two. Similarly, here also it will be X by two, okay? And this is X itself. So X, X by two, X by two will make it two X length, okay? So since the direction is the same and the magnitude is double of that of B, I can call this vector as 2B, correct? Now let us use the polygon law of addition. So can I say AB plus BC plus CD will be equal to AD. AB plus BC plus CD is equal to AD. AB is A, BC is B, CD, not known to me, but AD is 2B. So can I say CD vector will be nothing but take it on the other side, it will become B minus A. So CD is B minus A. So FA would automatically be A minus B. Clear? My question's gone. What is this? Just give me a moment, Friday. Okay, let's start with this question. Can you all read this? Yes, sir. Yeah. Two forces AB and AD are acting at the vertices of a quadrilateral AB CD and two forces CB and CD at C prove that the resultant is four EF, where E and F are the midpoints of AC and BD respectively. Let's draw it first. So let me call this as A, B, C and D. So AB and AD act on A. Similarly, CB and CD act on C. Okay. Now they're saying the resultant of AB, AD, CB, CD. That means AB plus AD plus CB plus CD. We have to prove that this is equal to four EF. E is the midpoint of AC, AC midpoint is here. Okay. F is here. Let's say F is here. So it's four times EF. Can you prove this? Any idea? Okay, the best thing here would be take one of the points as origin. Okay, let me call this as O, okay? So let me change the name over here also. From AB, let me make it as OB. And from AD, let me make it as OD. Okay, now this choice is completely yours, right? Choosing of origin is completely upon your whims and fancies. Fine. Now, why have I done that so that I can call this as position vector B and I can call this as the position vector D and I can call this as your position vector C. Okay, remember when you say position vector of B is B, that means you are indirectly referring to this vector. This is your B vector, this is your D vector and vector O to C, that would be a C vector. Okay, but that is not desired actually. So I'm not bringing into a picture. Now, let's say if I talk about the left hand side. Left hand side of this expression is OB, which is B vector. OD, which is D vector. And CB, what is CB? CB, as I told you, destination minus source. Destination minus source will be B minus D, correct? CD will be again, D minus C. That means your left hand side of the expression actually ends up giving you, I think some Ds will get canceled off. Why do I write B minus D? It is B minus C, right? This is C, yeah. So your left hand side will be 2B plus 2D, or you can say minus 2C plus 2D, okay? That is nothing but twice of B minus C plus D. This is your left hand side of the expression. Now let us check out what is EF. EF means position vector of E minus position vector of F, correct? Now how do I figure out the position vector of E? Now here comes your better understanding of position vectors. Position vectors, as I told you, is something which relates vectors to coordinate geometry, okay? So before I move on to this question, this question was actually a medium for me to tell you how we can use position vectors in the same way as we use coordinates. So let's say if I have a point A and B, whose position vectors are vector A and vector B respectively, and let's say there's a point C somewhere over here whose position vector is C. So that C divides the join of A and B in the ratio of M is to N, right? You would be surprised to know that C vector could actually be written as MB plus NA by M plus N, very much similar to the, does this remind you of section formula? Does this remind you of section formula? There's a halo on top of this smiley, okay? Section formula, correct? How does it come out first? It's a very simple explanation. So let's say this is my reference vector O. If you connect this, connect this, or let me just pull out this N from, now see here, if you're saying position vector of A is A, that means this vector is A, this vector is B, okay? Now OC vector is what is your vector C? We have to figure that out. How do I get this C vector? Very simple. We all know that this entire vector will be, let's say I give this as the direction of this vector, so the direction is this direction. AB vector completely is nothing but B minus A, okay? So what is AC vector? AB vector is B minus A. So AC vector is nothing but a vector in the same direction as AB, but its length is M by M plus N times B minus A, right? Remember, the entire length, if you take it as M plus N, AC is just the M part of it, it's M by M plus N times B minus A. So A to C is known to you now, which is M by M plus N times B minus A, correct? Now, can we now apply the addition of vectors? Can I say C vector is the sum of OA and AC? So C vector is OA, which is your vector A. AC vector is the same vector which we had figured out a little while ago, that is M by M plus N times B minus A. Take the LCM. So it is M plus N times A plus M times B minus A, which makes it as MA plus NA plus AB. Plus MB minus MA, okay? So these two gets canceled by M plus N. So the answer is MB plus NA by M plus N, right? Very similar to the section formula that we used to have in our coordinate geometry. Getting this point, okay? So in a similar way, if I want to get a midpoint of two vectors, all you need to do is take the end position vectors, add them and divide by two. So position vector of E would be zero plus C by two, zero plus C by two, which is nothing but C by two. So this is a position vector of E. What's the position vector of F? Position vector of F will be B plus D by two, okay? Is that clear? So position vector of E will be C by two, while position vector of F will be B plus D by two, okay? So EF vector would be destination minus source, that is B plus D by two minus C by two. So take a two common, so it is B minus C plus D. Now, what I'm claiming is that four EF should actually give me the same as my left hand side. By the way, four EF will be two times B minus C plus D. And we can clearly see that these two expressions are same. So left hand side and right hand side are same, and hence proved. Is that fine? So guys, let me tell you here, position vector was, in fact, position vector along with the addition of vectors was a very important tool in our hands to solve these kind of questions where normally a coordinate geometry problem is being addressed. Is that fine, everyone? Clear with this? Yes, sir. Yes, sir. Okay. Time to take a slightly complicated problem. In this series was it, this is seven days. Okay, anyways, yeah, 14, nine. Read the question properly. If O is the ortho-center and O dash is the, sorry, O is the circum-center, O dash is the ortho-center of triangle ABC. Prove that. O A plus O B plus O C is O O dash. Any idea? Let's make a diagram first. This is your O dash. This is your O. So is it option A, like first option? It is not option. We have to solve these three questions. Aditya. I think I was trying to figure out. These are not options. Then prove that, say prove that question. I can understand exam time now. Okay, first let me just draw a perpendicular like this. O D. Okay, I'll help you with the first part. Acha, how many of you remember from your last year that in any triangle, the distance of the ortho-center, so let me just draw a separate case over here. Let's say this is your ortho-center O dash, okay? How many of you know that O dash A is two R cos A where R is the circum-circle radius, where R is the circum-circle radius? How many of you know this? If you're not aware of this, not an issue, I can just explain you how this figure actually comes. So let's say this is your 90 degree over here, right? And this is also 90 degree, okay? So we all know that this angle, okay, since this is angle B, this angle is going to be 90 minus B, correct? Okay, and since this angle is A, this angle will be 90 minus A. Everybody agrees to this? Let's say O A dash, let O dash A be X. So let's say this is X, okay? Now, since this is 90 minus A, can I say this length would be X sine B? The opposite side to B, we call it as small B. Now, since this is 90 minus A and this hypotenuse is small B, can I say the same length O A dash, sorry, O A dash, I'm saying. A, P, let me call it as, okay? AP have already shown that it is X sine B. Can I say the same AP length can also be written as B, B cos A. Now, from here I can say that X sine B is B cos A, correct? From our sine rule, we know that B by sine B, by the way, all of you should remember sine rule as well. A by sine A is B by sine B is C by sine C and this ratio is actually two R. Very few people know that this is two R, okay? So from these two, I can say that small B is two R sine B. So when I place that small B with two R sine B, sine B, sine B gets canceled, so X is equal to two R cos A. This result should be remembered because it is heavily used in properties of triangles, okay? Now, why am I referring to this result all of a sudden in this question, okay? Do you realize that OD and O, let's say I call this as, okay, let me call it as AO dash. Can I say OD vector will be parallel to AO dash vector because both are perpendicular to the same line BC, correct? Now, we know that this is the circum center, so this is your rate, circum circle radius, okay? Since this is your angle A, can I say this will also be your angle A? Now, that means OD magnitude is nothing but R cos A, correct? Now, OD magnitude is R cos A, AO dash magnitude is two R cos A, and both are parallel also. So can I conclude from here that, from these all I can, can I conclude, can I pass a judgment that AO vector itself will be twice of OD vector? Yes or no? What do you say about this? Again, if you want to look at the figure, you can have a look at it. This vector and this vector are parallel and this vector is double the length of this vector and hence I can make this conclusion that AO vector is double of OD vector, okay? Is that fine? No problem with respect to this? Fine, if this is okay, now let us move on to the requirement. The requirement is OA plus OB plus OC, I have to prove it as OO dash, okay? Now see it, let me start with LHS, LHS is OA, okay? OA plus OB plus OC. Now, OB plus OC, can I say it is two OD? Can I say this is two OD? All of you agree with me on this or not? Now you must be wondering how it is two OD, very simple. Again, you can see that D is the midpoint of B and C, isn't it? D is the midpoint of B and C. So if I consider O to be your origin, this is position vector B and let's say this is position vector C, then D would have a position vector of B plus C by two. In other words, OD vector would be OB vector plus OC vector by two. So OB plus OC would have been two OD vector. Okay, so from here I can say OC plus OB will be two OD. Yes or no? Okay? Yes sir. And two OD is what? Two OD is AO dash, okay? Let me write it in a slightly different way. I'll write it AO dash plus AO dash is nothing but O, O dash. It's very obvious because if you're going from O to A and then from A to O, it is as good as going from O to O dash. Done, first part of the proof is done. Clear? Such a huge, such a huge problem but the concepts involved were your concepts that you had already learned in your past. Try the next one, prove that O dash A plus O dash B plus O dash C is equal to two O dash O. Again, shift your origin to O dash. Let me use black pen over here, okay? Now O dash B plus O dash C vector. O dash B plus O dash C vector. Do you all agree that it will be two O dash D vector? Yes or no? Yes sir. Now, O dash A. O dash A, you already proved that. So these two are taken care of. O dash A, you already proved a little while ago that O dash A is two D O dash, sorry, two D O. O dash A is two D O, correct? That means if you add them up, that is O dash A plus O dash B plus O dash C is equal to two O dash D plus, in fact, two you can take common, D O. Okay, so O dash D, O dash D plus D O, will you what? O dash O, so it'll be two O dash O. This also done. Last one at least you try. A O dash, O dash B, O dash C. You can use the result of the second one in the third one. Any idea? See, very simple here also. Now, this term over here, let me write it separately, A O dash, you want to prove A O dash plus O dash B plus O dash C is equal to A O dash, you can write it as two A O dash minus A O dash, correct? O dash B, O dash C, okay? So I'm starting from my LHS. Now, if you reverse the order of A O dash, you can write this as plus O dash A. This result is already known to us. What is that result? That result is two O dash O. So this will become two A O dash plus two O dash O. Take two common, A O dash O dash O is nothing but A O. So two A O, done. This is your RHS. Now, this question is saying that is equal to AP, where AP is the diameter through A of the circum-circle, which is obvious, because A O is going to be acting like the radius, okay? So twice of the radius will be nothing but the diameter. Okay, that's what this question is trying to prove. Okay, so third part also done. Let me take you down to the part where I approved it. Any question here, please ask me. Any doubt, any question, any concern? No, sir. Okay, let's move on to the next concept. Yeah, simple question based on section formula, done? Sir, one minute, just five seconds. Shreya is done. Sir, done. So let me discuss it. We divide the point P and Q. P has position vector i plus 2j minus k and Q has position vector minus i plus j plus k, okay? If the division ratio is two is two one, internally, okay? Then my formula is, okay, I'll just write down the formula for internal division. Let's say this is P vector, this is Q vector. So it'll be two Q plus P by three. For external division, it will be two Q minus P by two minus one, which is one, okay? So just plug in the values two Q plus P by three and this will be, so this will give you a minus i plus 4j plus k by three, okay? And external division will give you a minus 3i, zero j plus 3k. Is that fine? Is this what we are getting? All of you? Yes, sir. Fine, so now we are ready to take up the concept of resolution of vectors. First, we'll talk about resolution of vectors in 2D. So let's say we have a vector a, okay? I can resolve these vectors along two perpendicular directions and normally we choose those two perpendicular directions as your direction of the x-axis and the y-axis, okay? So this vector can be resolved along x-direction and y-axis direction, okay? Now, how do we resolve it? Very simple. See, let's say the magnitude of this vector is mod a, okay? And let's say this angle is theta, correct? So if I drop a perpendicular from here, can I say this length would be mod a cos theta, okay? So let me name it as o, a, p. So o, p length will be mod a cos theta and ap length will be mod a sin theta, correct? At the same time, we know that if the coordinates of a is x, y, x actually could be written as mod a cos theta, y could be written as mod a sin theta, okay? Now, think as if you want to reach from o to a. So you can take a path from o to p and then take a path from p to a, correct? O to p path would be nothing but mod a cos theta into a unit vector along the x-axis, which we know as i cap, okay? Pa is nothing but the length of pa, that is mod a sin theta, and a unit vector along ap. Now, this unit vector is the same as the unit vector along the y-axis, which is j cap, okay? So your oa vector, which is nothing but your vector a, could be written as op plus pa and op plus pa is nothing but mod a cos theta i, mod a sin theta j, okay? And since your x and y are your mod a cos theta and mod a sin theta respectively, you can write this like this, okay? So one thing is for clear over here is that the position vector of a, the position vector of a, that means the vector connecting the origin to the point a, which is nothing but this vector, is actually the coordinates of that point along with i and j, unit vectors attached to them. Are you getting my point? So if I say the position vector of a point is three i plus four j, what does it mean? That means the vector connecting this point with the origin, this is three i plus four j and this coordinate of this point is actually three comma four, okay? So this is how we can resolve a vector along the direction of x and the y-axis respectively, which we call as the resolving of vectors. Is that fine? This can be very well known to you, okay? Now the problem comes when we are trying to resolve a vector in 3D. So let's talk about resolution of vectors in 3D, okay? So let's say this is my right-handed coordinate system. This is your x, this is your y and this is your z, okay? And let us say we have a vector like this, a vector. This point is a. Now, before we understand the concept of resolution of vectors in 3D, we have to understand something called direction cosines. Have you heard of this term in your school? Yes. All of you have heard this? Yes, sir. How do you define direction cosines? It's the cos of the angle made by the vector with x, y, z. Absolutely correct. So basically let's say this is your vector a and this is your positive x-axis, positive y-axis, positive z-axis. If you draw a plane which is containing oa and ox, I hope you can imagine a plane which passes through ox and oa, then this angle which we call as angle alpha, okay? If you draw a plane through oa and oi, let me call this plane as containing this angle beta. And if you draw a plane through oz and oa and that plane contains this angle gamma, then cos of these angles basically are called the direction cosines. These are your direction cosines. Normally we address them by simple alphabets to make our life easy, L, M, N, okay? Later on, this will be a very useful concept for understanding the equation of plane and line in three-dimension, okay? So in 2D, we have the concept of slope. So in 3D, the same concept of slope is now converted to direction cosines concept. So there's nothing called slope of a line in 3D. The equivalent concept of slope in 3D is direction cosines. Are you getting my point? Okay. Now, in order to resolve this vector along 3D, okay? Let me again use polygon law of addition, okay? So let me drop a perpendicular from a on the xy plane. Okay? So let's say, yeah. Let me drop a perpendicular, let's say a dash, okay? And let me connect a line parallel to a dash with the x-axis, let me call it as p, okay? Now, let's say the coordinate of this point is xyz, okay? So can I say this length is x? Can I say this length is y and this length is z? Okay? You will all realize that your length x is nothing but the length of this vector into cos of the angle of, cos of the angle between the vector and the x-axis. y will be nothing but mod A into cos of the angle made with the y-axis and z length will be nothing but mod A into cos of the angle made with the z-axis, correct? So in order to reach from O to A, I have to take a path. You can see the figure over here. From O to A, if I want to reach, I can take a path O to P, P to A dash, A dash to A, okay? So it's O to P, P to A dash, A dash, A dash to A, okay? O to P, O to P is nothing but A dash cos alpha into I, okay? A dash cos alpha times I vector, I is the unit vector along x-axis. P A dash, this length over here, you can see the motion of my pen over here. P A dash, let me slightly put it to the left. Yeah. P A dash will be mod A cos beta in times j cap and A dash A vector would be mod A cos gamma into k cap, okay? So if you take mod A common, you'll get something cos alpha I cos beta j cos gamma k. And this is nothing but your vector A. This is nothing but your vector A, okay? So basically you have resolved the vector A along the i, j and k components. So you can write this like this also, okay? One thing that's very clear from this derivation is that if you look at this term, A by mod A actually represents cos alpha I cos beta j cos gamma k, which is nothing but saying that A cap, remember A by mod A is A cap is li plus mj plus nk. Now this is a very useful relation for us. This relation says that the direction cosines are nothing but the i, j and k components of the unit vector along that given vector A. So if I have a vector A like this, then a unit vector along this A, this is nothing but let's say this vector has direction cosines l, m, n. Then this vector is actually li plus mj plus nk. Very useful information, we'll be using this later on very frequently. Is this clear guys? Direction cosines represent nothing but the i, j and k components of a unit vector, of the unit vector along the direction of A. Is this clear? Yes, sir. Now if this is clear, let me ask you some questions. Hope you can read this question. Let me know once you're done. Okay, share is done. Okay, see, very simple problem. Let's say I'm showing my original coordinate accesses like these. Now you rotate it five by four in an anticlockwise direction. So let's say now along the z, that means z doesn't change. So this is five by four, this is also five by four. Now in this rotated coordinate accesses, these are the components of a vector. Find the components of these vectors in the original system. Now when you rotate a, let's say z axis is not rotated. Let me show you a 2D view. So let's say this was your original x and original y. And let's say you rotate a vector, you rotate these coordinate accesses 45 degrees anticlockwise, okay? 45 degrees anticlockwise. Now any unit vector in this direction, let's say I call it as i cap dash. Can I say i cap dash is nothing but, it is original i plus j by root two. Yes or no? Do you agree with me on this? Similarly, the new y component, let's say j cap dash. j cap dash could be written as minus i cap plus j cap by root two. Do you agree with me on these two fronts? Yes or no? Then only I'll move forward. Yes? Okay. Now, if I want to make a vector, remember k will not change. So k cap dash is same as k cap. So if your vector was in the new coordinate system, two root two i cap dash, three root two j cap dash, and four k cap dash. Then if I convert it back to i and j normal components that we had in the beginning of the system, this i cap dash is nothing but i cap j cap by root two. j cap dash will be minus i plus j by root two. k cap dash is same as k cap, okay? So this root two, this root two cancels, this root two, this root two cancels. So it's two i plus j plus three minus i plus j plus four k. If you expand it, it becomes minus i plus five j plus four k. That means the components, these are the components of the vector in the original coordinate system. That is what we are asked to prove over here. So as you can see, this is what we are asked to prove. Is that fine? So basically here it is the concept of knowing how your unit vector, when you are rotating the x y axis is 45 degrees anticlockwise, how does the unit vector actually gets changed? Here, let's take another one. A man traveling towards east at eight kilometers per hour finds that the wind seems to blow directly from the north. On doubling the speed, he finds that it appears to come from northeast. Find the velocity of the wind. So this north, does it mean east north or the normal? His north. His north and normal north will be the same, right? Turn towards the right and then if it's traveling, then the north will be like the other side. North is fixed. North is a fixed thing. My north and your north will be the same, right? But if they're facing like opposite, then it will be different north. Blowing from the north, north direction doesn't change if we are facing other side. If I turn round, north will remain north, right? It is not from his north or something, it is from the north, that's it. Are you all aware of the concept of relative velocity? When you see this word seems, seems means relative to the man. See this word seems. So whatever is the velocity of the man, the wind relative to man appears to come from north. Okay, this appears to come from north. So let me just make a diagram like this. So let's say the man is moving in this way. Eight I. Okay, so this is your V man. This is what he sees. He sees that the wind is coming from the north. So this is the V man minus, sorry, V wind minus V man. Since it is coming from the north, means it's going towards the negative Y axis. So you can call it as minus P j. So is it A through two? Direction, what is the direction? Sir, is it in one second? Okay, so let's say the wind velocity is Xi plus Yj. So what is given to you here is Vm minus, sorry, Vw minus Vm. Why am I writing M every time there? Sorry. Vw minus Vm is minus Pj. So Xi plus Yj minus eight I is minus Pj. Which is nothing but X minus eight I, Yj is minus Pj. Now guys, we have not yet studied about the concept of linear dependent and linear independent, but here let me tell you, since there is no I component over here, you can safely say that this is zero, which means X is equal to eight. Okay, now second scenario is if he doubles his velocity, let's say he makes it 69. Now this becomes your new velocity of the man. Then the wind appears to come from northeast. So let me say it is minus QI plus J, okay? So this time V wind minus V man is going to be minus QI plus J. So Xi plus Yj minus 16 I is minus QI minus Qj, which makes it X minus 16 I plus Yj is minus QI minus QJ. Now X minus 16 should be equal to minus Q. X is already eight. So minus eight is equal to minus Q. That means Q is equal to eight. And from here Y is equal to minus Q. That means Y is minus eight, okay? So the velocity of the wind becomes eight I minus eight J. That means the wind is appearing to come from the north. This is eight I minus eight I minus eight J, correct? That means the wind is coming from north, west, okay? And the speed of the wind, that is mod of the speed of the wind velocity will be eight root two, whatever units. Clear everyone? Please type CLR if it is clear. Next question. So how important is this chapter like considering the competition? Very important, very important. Bector at least two questions, 3D at least two questions. Vector 3D combined, three to four questions will come. Find a unit vector C if this bisects the angle between C and 3I plus 4J. So basically there is a vector 3I plus 4J, okay? There's another vector C, which is actually a unit vector. So C is a unit vector. And there is a vector which bisects the angle. This vector is your minus I plus J minus K. Find C vector. So here's a very important concept which I want to talk about via this problem is, see if you have any vector A, okay? And you have another vector B. There is a vector which is bisecting the angle between A and B. Okay, let me say C. Then let me tell you that C vector can be expressed as lambda times A by mod A plus B by mod B. Now of course, my question would be how? How is this possible? Can you at least justify this? So what I'm claiming here is that the C will lie along. Lambda is just saying that there is a vector along the direction of this vector. At least somehow you get this vector. Lambda you can always justify. Lambda means any vector which is collinear or lying parallel to that vector. Then we'll come back to this problem. So as of now, part of this problem and let's try to focus on this. Do you want any idea? Aditya, Shreya, Hruthu, Adhra, Samyukta. You can unmute yourself and talk, no worries. Don't know, sir. You have muted yourself to just say, don't know. Sir, could it be like from parallelogram law of vector addition? And if we take the parallelogram, if we take a special case, then we can prove this. But are we allowed to take special cases? Let's see, what is your special case? How special it is? Tell me through special case only, no problem. Technically I'm not sure whether in parallelogram angles are bisected. That's why I'm taking a special case of a square. So then by parallelogram law of vector addition, C vector will be the resultant vector and that. Absolutely, there's nothing wrong in that. See, you actually, you know, you're doing whatever is required. See, if I take a unit vector along A, let's say unit vector is like this. Okay. So let's say I take a unit vector along A cap, let's say. Okay, unit vector along B cap. Can I say, since these two vectors are of the same length, one unit, okay? This parallelogram will have, will be more or less like a rhombus, isn't it? So if I take a bisector over here, can I say this bisector angle will connect this point to the midpoint of these two? Correct? Yes or no? Do you agree that bisector angle will connect the midpoint? So if let's say this is your O and this position vector is A cap, this position vector is B cap, then this position vector would be A cap plus B cap by two. Correct? Can I say any vector in the direction of, let's say I call this point as C. Any vector in the direction of C, can I say it will be some constant, let me call that constant as lambda, not lambda, let me call it as beta, beta times A cap plus B cap by two. Correct? So even C lies in the direction of this. So C is nothing but some constant times A cap plus B cap by two. Can I just pull out a two out and write A cap and B cap as A by mod A and B by mod B and beta by two is just another constant, which we can call it as lambda also. So can I now say that any, this is a very important result guys, let me tell you there is some reason why I am deriving this all because this result is important. This result is directly used in solving many problems. Okay? So any bisector angle vector, which is bisecting the angle between vector A and vector B could be written as lambda times A by mod A plus B by mod B. Getting my point. Now, can you please use this to solve this given problem? Now come back to this question. Now this C is different, okay, don't get confused. This C was my bisector, this C is the one which is getting bisected. So now don't have the same C carry forward, okay? Now solve this. So can I say minus 3 by 5 I cap minus 4 by 5 J? No, no, no, the answer is slightly more complicated. Can I say this vector is lambda times C vector. Let's say C vector is xi plus yj plus zk. So C vector itself is a unit vector. So you don't have to do A by mod A. By the way, let me just make a confirmation here that this will be equal to 1 plus 3i plus 4j by 5. Is that clear? By the same logic that we discussed over here. Okay? This bisector vector is lambda times the unit vector along C, which itself is a unit vector. So I've just written it like this. Plus unit vector along 3i plus 4j, which is 3i plus 4j by 5. Now let us compare the coefficients. Minus 1 is lambda times x plus 3 by 5. J component is 1 over here. J component here is lambda times y plus 4 by 5. K component is minus 1 lambda times z. So now z is minus 1 by lambda. Y is nothing but minus 4 by 5 minus 1 by lambda. No, plus 1 by, sorry. Which you can write it as 5 minus 4 lambda by 5 lambda also. And x is nothing but minus 1 by lambda minus 3 by 5. Which is nothing but minus of, I'm sorry. Yeah? Now how do I get the lambda value by using the fact that x square plus y square plus z square is equal to 1. Because x, y, z are the components of a unit vector. So 5 plus 3 lambda square by 25 lambda square. 5 minus 4 lambda square by 25 lambda square. And 25 by 25 lambda square should be equal to 1. So let's take 25 lambda square on the other side. So it is 5 plus 3 lambda square. 5 minus 4 lambda square plus 25 is 25 lambda square. Let me expand it. If you expand it, 9 plus 25 lambda square here itself I will get. And I will get 30 lambda minus 40 lambda, which is minus 10 lambda. And constant terms would be 25 plus 25 plus 25. That is 75 is equal to 25 lambda square. So 25 lambda square, 25 lambda square gets cancelled off. Giving you lambda value as 75 by 10, which is 15 by 2. Okay? That means x, now put it back over here. Put it back at these values. This value, this value and this value. So x will be nothing but minus 1 by lambda, which is minus 2 by 15, minus 3 by 5. Minus 3 by 5. You can write it as minus 2 by 15, minus 9 by 15. That is minus 11 by 15 is your x. Y will be minus 4 by 5 plus 2 by 15. That is minus 12 by 15 plus 2 by 15, which is actually minus 10 by 15. And z will be minus 1 by lambda, which is minus 2 by 15. So your final answer will be your c vector is minus 11 by 15 i, minus 10 by 15 j, minus 2 by 15 k. Okay? Any questions so far? No sir. No sir. Okay. So this is going to be your answer. Next question. Find the point of intersection of a, b and c, d where a, b, c, d are these given four points. So basically these are the two points and let's say a, b, c and d. You have to find the point of intersection. This is 6 minus 7, 0. This is 16 minus 19 minus 4, 0, 3 minus 6, 2 minus 5, 10. How do you find the point of intersection when you're dealing with 3D points? Could we find the point of cohesion of the unit vectors? How will you find that? Writing one as lambda times another and then finding the unit vectors. Here I will tell you one simple thing. Use the concept of section formula. See, let's say the point p which is the point of intersection. Let me name this point as p. Let's say this p divides this in the ratio of lambda is to 1. Okay? And it divides this in the ratio of let's say beta is to 1. Okay? Now, if you take the lambda is to one fact, the point p will have a position vector of, what will be the position vector? The position vector would be 6i minus 7j times lambda and 16i minus 19j minus 4k by lambda plus 1. My lambda looks like 11. So, okay? And the same point p will, if you take cd into your picture, then the same point p will have a position vector of beta times 3j minus 6k beta plus 2i minus 5j plus 10k by beta plus 1. Since both represent the same thing, can I just compare their i coordinates, j coordinates and k coordinates? So, if you compare their i coordinates, it's 6 lambda plus 16 by lambda plus 1 equal to, this side you'll have 2 by beta plus 1. Okay? If you compare their j coordinates, then minus 19, sorry, minus 7 lambda minus 19 by lambda plus 1 is equal to, j coordinates here would be 3 beta minus 5 by beta plus 1. If you compare the k coordinates, then minus 4 by lambda plus 1 will be equal to 10 beta minus 6, or what am I doing? 6 beta minus 6 beta plus 10 by beta plus 1. Okay? Now, one thing that you would have noticed that, there's a stark resemblance between this term and this term, isn't it? This is just this term multiplied with a minus 2, isn't it? Correct? So, can I say minus 2 times minus 7 lambda minus 19 by lambda plus 1 is actually minus 4 by lambda plus 1? What I did, I multiplied both side with minus 2 and equated their left-hand sides. Can I do that? So, lambda plus 1, lambda plus 1 gone. In fact, this and this will go by a factor of 2. Correct? So, minus 7 lambda minus 19 is 2. That means 7 lambda is minus 21. That means lambda is negative 3. Once you know lambda is negative 3, you can easily get by putting your value of lambda over here. So, coordinates of P would be nothing but minus 3, 6i minus 7j plus 16i minus 19j minus 4k by minus 3 plus 1, which is minus 2. Let's simplify this. So, minus 18 and plus 16 will be minus 2, minus 2 by this will be i. Plus 21 minus 19 is plus 2, plus 2 by minus 2 is minus j. And this will be plus 2k. That means 1 comma minus 1 comma 2 would be the coordinates of this point. Is that fine? Clear how it works? Yes, sir. Okay. The next concept which we are going to talk about is another operation on vector, which we call as scalar multiplication. So, when you multiply a vector A with a scalar quantity lambda, what happens to this vector? Now, if lambda is positive, there is no change in the direction of A. But if lambda is negative, then the direction of A reverses. If mod lambda is greater than 1, that means there is a scale up in the size. And if mod lambda is less than 1, in fact less than 1 greater than 0, then there is a scale down in the magnitude of lambda A by factor of mod lambda. Okay. So, what is the role of lambda? It just changes the direction if lambda is negative. And the magnitude of A gets multiplied with the magnitude of lambda. So, if let's say A is this, then what is minus half A? So, let's say A is this vector. Minus half A would be the same vector, half its length, but in a negative direction, in an opposite direction. Okay. Because minus sign reverses the direction, half means magnitude is halved. This is something which you already know. Okay. The use of this is in understanding a very important concept, what we call as the concept of co-linearity of vectors. Concept of co-linearity of vectors. Okay. But before I start, talk about co-linearity of vectors, there's something which I would like to talk about is linear combination. Very important concept which is very challenging to understand in the first go because many students are not able to relate easily with this concept. So, please all of you listen to this concept very, very carefully. When you write a vector R as m1, m1, m2, a2, m3, a3, etc. till m and an. Okay. Then you say where a1, a2 are vectors, a1, a2, etc., are vectors. Of course, I didn't mention that because I've already shown them with the arrow sign on top. And m1, m2, etc., these are scalar quantities. Okay. Then we say that R has been expressed as a linear combination. R has been expressed as R is a linear combination of, is a linear combination of a1, a2, till an. Is this statement clear? Any doubt till this stage? Are you happy with this statement? Yes, sir. Okay. No doubt regarding what do we call as a linear combination of a certain system of vectors. Clear? Clear? Okay. Now, something which we call as linear dependent and linear independent that evolves from the same concept, linearly independent and linearly dependent. What is linearly dependent? A set of vectors a1, a2, a3, till an is said to be linearly dependent. If a linear combination of these vectors expressed as a null vector implies m1, m2, or if you express this as a linear vector, then your m1, m2, till your mn are all not zero. I'll give you an example. Let's say i, i plus j and let's say 3i minus 2j. My question is, are these vectors linearly dependent? Are these vectors linearly dependent? My question is, are these vectors linearly dependent? Now, in order to know that, we'll take this approach. We'll say m1, m2, m3. Let's say this is a null vector. Now, if I'm somehow able to show that at least one of them is non-zero. That means all are not zero. Not all are zero. Means I just have to show a non-zero case of m1, m2, m3. At least one or two of them. Then my job will be done that they are linearly dependent. If all of them come out to be zero, then that is what we say they are linearly independent, which I'm going to take up next case. But let me just finish off this up. So if you see this, so your i component will be m1, m2 plus 3m3. That's your i. And j component will be m2 minus 2m3. Let's say I call it as j. After equating it to zero vector means it is saying zero i, zero j. So if you compare the coefficients m1 plus m2 plus 3m3 is zero. And m2 minus 2m3 is also zero. So if you take any value of m3, let's say if I take m3 value as k, m2 will be 2k. And here m1 will be minus m2, which is minus 2k, minus 3m3, which is minus 3k, which is actually minus 5k. Now, if I don't choose k value as zero, of course, 0, 0, 0 is a solution, but there can be non-zero values also. So if I don't choose k as zero, neither my m1 nor my m2 nor my m3 will be zero. So there exists not all zero values of m1, m2, m3 such that the linear combination of these three vectors can be expressed as a null vector. If 0, 0, 0 is the only possibility, then that is what we call as linearly independent, which I am going to take up next. But before I go to the next case, is this clear enough? Could you explain that part again? See, in plain and simple language, if one vector could be expressed in terms of the other two, they are linearly dependent, simple as that. For example, if I say 3i minus 2j, I can actually create 3i minus 2j by doing this. 3 times i or let's say 2 times i, let me take minus 2 times i plus j plus 5i. So this vector is generated by some scalar times this vector and some scalar times this vector. That means other two vectors, some scalar you multiply and add them to generate a this vector. That means they are dependent on each other. And this is how they are dependent. Are you getting my point? So we say these three vectors become linearly dependent. I didn't understand how you found that 2 and... How I found this minus 2 and this? Sir. Yeah, I would have taken x and y and solved for x and y, but that would have just taken a bit of time. That's why I... See, what I did was this. X i plus j, y into i. Then I made two equations. 3 is equal to x plus y and minus 2 is equal to x. So x is minus 2 and x is minus 2 means y is 5. That's why I came with minus 2 and 5. This was an easy case. That's why I did not do all these things, but otherwise I would have done it. So the meaning of three vectors being linearly dependent is one of them can be expressed as scalar times the other two. Scalar times let's say this one and scalar times the other one. So if c vector is linearly dependent on a vector and b vector, actually we don't say linearly dependent on, we say all of them are linearly dependent on each other. Let me just correct my this. This, this and this are linearly dependent. Then c could be expressed as some scalar times a plus some scalar times b. Are you getting it? Because if you do this, then at least you have ensured that one of them is a non-zero. For example here m3c, this m3 clearly here is minus 1, isn't it? If you compare this, m3 is clearly your minus 1. So not all of them are 0. Even if these two become 0, at least c component will not be 0. Unless and till, you know, of course two of them have to be non-zero in this case. If a, b, c are non-zero vectors. Are you getting my point? So just remember the simple statement. If you want to find out whether three vectors are more than three vectors, that's the n vectors are linearly dependent. Then a linear combination of this, this is called the linear combination. If it is equated to a null vector, then at least some of this m1, m2, m3 should not be 0. That is the meaning of all not 0. I think this, you will understand this better when I talk about linearly dependent. Sorry, linearly independent. Let's not talk about linearly independent. Now again a set of vectors a1, a2, till an are said to be linearly independent. A linear combination of these vectors, a linear combination of these vectors, if compared to a null vector will only result into these scalar quantities being all 0. Are you getting this point? A typical example of a linearly independent set of vectors is i, j, k. These three guys are linearly independent. That means if you do some m1i, m2j, m3k and equate it to a null vector, this can only be possible when your m1 is 0, m2 is 0 and m3 is 0. There is no other way that this can be equated to a null vector which clearly indicates the fact that these three vectors are independent of each other. In other words, you cannot express one in terms of the other two. That means if you want to express k as let's say some m1i plus m2j, you will not be able to, m1, m2 doesn't exist in this case. This is not possible because they are linearly independent of each other. k doesn't depend on i and j, i doesn't depend on j and k. j doesn't depend on i and k. Are you getting my point? What is the meaning of linearly independent and linearly dependent? Yes, sir. Now, why am I all of a sudden talking about these cases is because you are not going to study collinearity of two vectors. What is the meaning of collinearity? Collinearity means they are parallel. We say vector a and vector b are collinear when they are linearly dependent. Vector a and b are collinear when they are linearly dependent. What is the meaning of linearly dependent? That means if you do m1a plus m2b equal to null, then m1 and m2 not all will be zero. Are you getting my point? In other words, you could express a as minus m2 by m1 times b, which is nothing but you can express a as lambda times b. So if you could express one vector where lambda is some scalar quantity, so if you could express one vector as lambda times the other, then we say that the vector a and b are collinear to each other. In other words, they are linearly dependent on each other. That's why the concept of collinearity evolves with the concept of linearly dependent. If they are non-collinear, my dear friends, let me write the reverse case also. If a and b are non-collinear, then they are linearly independent. In other words, if you try to do this, then it will only result into m1 and m2 both being zero. Now let us try to understand the deeper meaning of this. Let's try to understand this through vector addition. Let's try to have another perspective of it. If let's say this is your vector a and this is your vector b. Are they collinear? According to my diagram, let's say they are parallel. Are they collinear? Yeah. Now if you see this, this is trying to say that if you manipulate these two vectors in such a way that their resultant becomes zero. That means if you do m1 a plus m2 b in such a way that the resultant becomes zero, this is possible without m1 m2 being both zero. How? Let's say this is double the length. What I will do is I will half its length so that it becomes like this. Then I multiply it with a minus one so that it reverses its direction. Now if I add these two, so 1 into a and minus half into b, 1 into a and minus half into b, do you realize that this vector and this vector will close the path and it will become a zero vector? So m1 m2 here both are non-zero. That is what this theorem actually says. It says that it is possible for you to multiply these two vectors with a certain scalar quantities. That means you are fiddling with their direction and magnitude in such a way that they both cancel out each other. That is possible for collinear vectors. And hence these vectors are linearly dependent. Have I made myself clear? Yes, sir. But if you have such vectors, my dear friends, like this, a is like this and b is like this, can you try making the resultant of these two as a null vector? Can you somehow close this path such that this plus this ends up being a closed path? Can you do something like this in this case? No unless and till you make both of them zero, isn't it? There is no way you can close the path. You try fiddling with the direction and length of these two vectors. Do anything you want. Says that they cancel out each other. Can that happen? No, it will not happen. It can only happen when you have multiplied both of them with a zero. That means you have made them both dots. Then addition of two dots will give you a dot. That is a null vector. But if you have any other non-zero quantities for M1, M2, it is not possible to close the path between these two vectors. Or it is not possible to make them cancel each other out. So these cases are called, these vectors would be called linearly independent. So two non-colinear vectors are linearly independent. Two collinear vectors are linearly dependent. Are you getting this point? Now, a same concept can now be applied to collinearity of three position vectors. Collinearity of three position vectors or points you can say. Remember position vectors and points, there is no difference. What is position vectors in vector is points in coordinate geometry. Getting my point. So many points here. So you say position vectors A, B, C. See guys, we never talk about collinearity of two position vectors. Any two position vectors will always be collinear. So it is a futile effort to talk about collinearity of two position vectors. You can always pass a line through two points, isn't it? So in order to make sense, we'll talk about three or more than three points. So three points, let's talk about three points. Let's say point A, point B and point C, whose position vectors are as shown. They will be collinear. That means they will all lie on a line. If. Now there are two ways in which I normally explain it. If A, B, C are collinear, then construct any two vectors by choosing any two of the points. Let's say I take A, B and A, C. Then one can be expressed as lambda times the other. If this condition is satisfied, then point A, B, C will also be collinear. Mind you, you can take any two vectors you like. You can take A, B and B, C also. So even if you can express this as lambda times this, any one of these conditions, then these three points will be collinear. That's one way of understanding it and this is the easiest way. Normally I also choose this method to prove whether my three vectors are collinear or not. Second method is slightly tricky, but you should also understand this. Second vector says that if you express the linear combination of these vectors, if you express the linear combination of these vectors as a null vector, then M1, M2, M3, not all 0, must satisfy M1 plus M2 plus M3 equal to 0. This is slightly challenging way, but this is easy to understand once you know the logic behind it. So what I'm trying to say is that if a linear combination of A, B, C express as a null vector, it should lead to M1 plus M2 plus M3 as 0, but all of them should not be 0. That means of course M1 plus M2 plus M3 will be 0 if all of them are 0. But here my situation is it should be 0 despite M1, M2, M3 not all being 0. Then only point A, B, C would be collinear. How does this actually come about? It's very simple. It actually comes from the section formula itself. Let's say A is this, B is this, and this ratio is let's say X is to Y. So if I use my section formula, B would be nothing but XC plus YA by X plus Y. Now it can be minus also, doesn't matter. If you cross multiply, X plus YB is XC plus YA. If you write this in form of this term, if you express this as this, you'll write it as YA minus X plus YB plus XC equal to null vector. Now I can clearly see here that this is playing the role of M1, this is playing the role of M2, this is playing the role of M3. You can see that M1 plus M2 plus M3 will actually give you Y minus X plus Y plus X, which actually is 0. That's what is this condition saying. But each one of them is not 0. Your M1, M2, M3 may need not be 0 all of them. Are you getting this point? Guys, is this understood? Yes, sir. If it is understood, let me take a problem. I hope you can read this question. The vectors 2I plus 3J, 5I plus 6J, 8I plus lambda J have their initial points at 1 comma 1. Find the value of lambda so that these vectors terminate on one straight line. Done? Do you understand the question first of all? See, there's a starting point 1 comma 1. So one vector is like this. Let's say this vector is 2I plus 3J. One vector is like this. Let's say 5I plus 6J. Another vector is like this. Let's say 8I plus lambda J. All of these three vectors, they end on the same straight line. So this is the terminating point. Let's say ABC are the terminating point. So what should be the lambda values as that? They all terminate at the same point. 11 by 3. 11 by 3 is not the right answer. No problem, Aditya. Let's say the position vector of A is not known. So can I say position vector of A minus I plus J is actually your given vector 2I plus 3J. That means position vector of A is nothing but 3I plus 4J. Similarly, position vector of B would be 6I plus 7J and position vector of C would be 9I plus lambda plus 1J. Now these three vectors, these three points are collinear. That means they are linearly dependent. Take any two of the vectors. The third could be expressed as lambda times the other. So let me take these two vectors and these two vectors. Initial point 1 comma 1. Why does minus 1 comma minus 1? It's 1 comma 1. I think you misread the question. So I'll use the formula A B vector is lambda times A C vector. I'll use this formula or the other way around. A C vector is lambda times A C vector is lambda times A B vector. Anything you can use, not a problem. Now I don't want to use lambda because lambda is already used. Let me use something like beta. So A C vector means C minus A. C minus A would be 6I lambda minus 3J. This is beta times A B. A B will be 3I plus 3J. Now let's compare the coefficients. So 6 is equal to 3 beta. That means beta is equal to 2. And lambda minus 3 is equal to 3 into beta. 3 into beta is 2. And lambda is equal to 6 plus 3, which is 9. Lambda is equal to 9. So one last question we'll take up and then we'll call it a day. If A and B are two non-colinear vectors, then show that these three points, points means position vectors are non-colinear if the determinant given by that expression is 0. Let's say I use the second method this time. So if I say these three vectors are collinear, can I say X times, let's say this vector, Y times or let me call it as, okay, let me use XYZ no problem. Since I was using M1, M2, M3, so I thought I would use the same, but I'm sure you would not get confused even if I replace M1, M2, M3 with XYZ. Let's say I equate this to null vector. Then this should be true for XYZ not all 0. Correct? Yes or no? Now let me first collect the terms A and B separately on my first equation. So if I take A term separately, so XL1, YL2, ZL3, A. Now let's collect the B terms, so XM1, YM2, ZM3. Now please note A and B are non-collinear. Non-collinear means they should be linearly independent. If they're linearly independent and you're equating it to a null vector, that means these two terms individually must be 0, 0 each. That means, let me write it like this, that means L1X, L2Y, L3Z should be 0. M1X, M2Y, M3Z must also be 0. And other than that, we have X plus Y plus Z also equal to 0. Now remember, we have to find a situation where XYZ all are not 0, that means this system of equation, this system of equation must have non-trivial solution. Non-trivial solution means at least XYZ should be non-zero in some cases. Yes or no? Now remember the condition when a system of equation has a non-trivial solution. It can only happen when the determinant formed by their coefficient must be 0. Try to recall your determinants chapter from Cramer's rule for system of homogeneous equations. Is that clear? And this is what we wanted to prove. The other simpler way would have been, if you know these three points, let's say I call this as Li cap, M1J cap, L2I cap, M2J cap, and this point as L3I cap, M3J cap. Now think as if there is a triangle, think as if there is a triangle whose coordinates are known to you, L1M1, L2M2, L3M3. And you want this to be collinear. That means the area of this triangle must vanish. Area should become 0. What is the area expression? Half L1M1, L2M2, L3M3 should be equal to 0. So even if you drop this half, it will be 0. Now you just transpose it. If you transpose it, it becomes this. Another way of understanding the same thing. Is that fine? So next class when we meet, we will be talking about co-planarity of three vectors, etc. And a similar conditions will now be seen in that cases as well. And then after, I'll start with dot product. So today, mostly our session was oriented towards understanding the introduction part of vectors. More importantly, the concept of linear combination, linear independent, linear dependent is what you may have learned new in this part today. Anyways, I'll stop the session here over and out from my side. But yes, all the best for you tomorrow's exam. Do well. Avoid silly mistakes. Okay. Thank you. Thank you, sir. Bye. Have a good day.