 All right. Let's start with the session. Good evening to you. How are you? So as you can see the title of the topic that we are going to study today, vector algebra. Okay. A very simple topic just because you have done it before in physics. Okay. Ironically, we do this topic towards the end of class 12. But this topic is closely related to 3D geometry as well. So it's very important to understand the nuances of these concepts because the same concepts are going to form the formative part of your understanding of 3D geometry. Okay. So vector algebra is what we are going to start with. Now, just a brief introduction because you have all learned vector before. So whatever I'm going to speak in the next 10, 15 minutes, you must be all be aware of it. Okay. It's just a revision for you. So the physical quantities that we see around us. Okay. The physical quantities that we see around us can be broadly categorized as scalars and vectors. Okay. Can be broadly categorized as scalars and vectors. Okay. By the way, many people talk about tensor quantities. Let me tell you scalars and vectors are types of tensor quantity. Okay. So a rank 0 tensor is a scalar. A rank 1 tensor is a vector. Rank 2 tensor is a dyad. Rank 3 tensor is called a triad, etc. Now, we are not going to go into the tensor quantities. Okay. So we are going to keep it simple. Anyways, you're going to study about it later on in undergrads and all. So what are scalar quantities? The quantities which need only magnitude to signify them. Okay. What are some examples? Distance. Right. We only need magnitude to state the distance. Speed. Okay. Temperature, etc. We never mention any kind of direction along with that. So we never say the temperature is 25 degrees Celsius towards north. Okay. There's no direction mentioned. On the other hand, quantities which have magnitude. Okay. Direction. Which also is referred to as sense. Oh, is that left off? Limit of some. Okay. We'll take this up in the last half an hour of today's class. Since we have already started with it, let me just continue with it. Okay. We'll take that up. Just remind me. At 6 o'clock, I'll take that topic. Not a very big one should not take more than half an hour or 45 minutes or so. Okay. So quantities which have magnitude. Direction. Direction is also called sense. Okay. Is this sufficient enough to be? Is these two quantities sufficient enough for a quantity to be called vectors? This is a question to all of you. Any quantity which has magnitude and direction. Is it sufficient for it to be called a vector? No. Then what makes it a vector? Right. It should follow law of vector addition. That's correct. It should either follow the parallelogram law of vector addition or triangle law of vector addition. So whatever are the laws of vector addition that must be followed by that quantity. A classic example is current. Okay. Current has magnitude as well as a direction. Right. Let's say 5 ampere current moves along a wire like this. Okay. Flows along a wire like this. Okay. But it doesn't follow the vector law of addition because let's say if I have another current like this. 3 ampere. Okay. Then let's say the outlet of these two according to Kirchhoff's current law. Here would be 8 amperes. Now this value doesn't change even if your 5 ampere current was like this. 3 ampere current was like this and let's say the outcome of this wire was this. So 5 ampere, 3 ampere, this will still be an 8 ampere current. Okay. However, same will not be true if the quantities were actually vectors. For example, let's say if you talk about forces acting on a body. So if let's say this is a point mass and there's a force acting like this 5 Newton force and this another force 3 Newton force. Then the resultant force is going to depend upon what is the angle between these two acting forces. Correct. So here the resultant which is the sum also called the sum of vectors is also called the resultant that would depend upon the direction between the two initial vectors. Okay. So a quantity to be called a vector should have magnitude direction follows law vector addition. Right. So what are the examples for this? There are a lot of examples in physics. We have come across force. We have come across a spacement. We have come across velocity, acceleration, moment of a force, etc., etc. Correct. So there's nothing new that I'm talking about here. It's just a formal introduction of the topic. Next comes the representation of vectors. Okay. I'll be quite fast over here because you already know all these stuffs. So a vector is represented by a small alphabet with an arrow on top. Okay. If at all in books they don't put an arrow, they should write it in bolds. If you just write a then a should be in bold or just a or you can write just a in bold. In bold. Okay. Sometimes vectors are also represented by stating their initial and terminal points. So let's say AB. AB is a vector whose initial point is A terminal point is B. Okay. And the sense of this vector is from A to B. The sense of this vector is from A to B. Okay. Then we'll write it as AB with an arrow on the top. Is that fine? Okay. Now, normally a vector is categorized by three things. Two of them are already stated. One is the magnitude. Second is the sense or a direction. And third is the support. Now the support is basically the infinite line from which the vector has been carved out. Okay. So let's say AB vector belongs to an infinite line. Okay. It belongs to an infinite line like this. So basically you have carved out this vector AB from this infinite line. Okay. Then this red line will be called the support of that vector. Okay. Just an additional information. It's not like that support is going to be a very important concept for your understanding of vectors. But when we're speaking with each other, we should know what we are referring to. We should be on the same page with respect to the terminologies, with respect to the taxonomy being used. Now, when you talk about terminologies, let us talk also about types of vectors because when I'm talking in terms of the language of vectors, you should know what I'm referring to. Okay. So type of vectors, first we'll start with null vector also called zero vector. Okay. What is a null vector? Null vector is a vector whose magnitude, whose magnitude, remember, magnitude is defined by vector written within two parallel lines. Magnitude is zero. Okay. And what about the direction? Direction of null vector is anybody? What is the direction of null vector? By the way, null vector is written as zero with a arrow sign on top. What is the direction of null vector? Anyone? Direction of null vector is undefined. This is undefined. We don't have a direction for a null vector. Okay. Why do we need this vector is because let's say you go five kilometers towards north. Okay. Five kilometers towards north and you come back five kilometers back. Come back five kilometers back. Five kilometers south. Now, what are the resultant of these two vectors? You will say resultant of these two vectors will be a zero vector. Okay. Now, everybody can know from this that your resultant is zero, but he cannot make what is, what is the direction of this? Because you could have gone five or any unit in any direction and come back in the reverse direction. Okay. It's just like the degree of the polynomial zero. Right. Or it's like the argument of a complex number. Okay. Which is not defined. Though they are complex numbers, though they are polynomials, their degree and argument, etc. Is not defined. Is that fine? Next is unit vector. I'm sure you would have come across i cap, j cap, k cap, etc. They are actually unit vectors. So what is a unit vector? We say a unit vector in the direction of a vector a is given by a cap. Okay. It is nothing but the vector divided by its magnitude. Okay. So if you want to get the unit vector along a vector, you have to divide the vector by its magnitude. Alternatively, you can say if you want to, if you want to generate a vector, you have to multiply the magnitude with a unit vector in that direction. Both are very important. Okay. A few unit vectors have already stated i vector i cap, which is a unit vector along unit vector along positive x axis. Okay. We have j cap, which is the unit vector along positive y axis. And we have k cap, which is nothing but the unit vector along positive z axis. Okay. Negative of a vector. Now this is something which I'm going to deal separately in the scalar multiplication of a vector and a scalar quantity. Negative of a vector, negative of a vector, let's say there's a negative of a vector a minus a. What does it mean? It means a vector which is having the same magnitude as that of a but direction opposite. Okay. So if let's say this is your a vector. Right. Then this vector is minus a. Remember magnitude of a and minus a must be the same only the direction is reverse. Okay. The direction of negative a is reverse of direction of just to formally keep it. I know you're all aware of it. Next is co initial vectors. Co initial vectors as the word itself speaks for itself. Co initial means having the same initial point. Okay. Co initial vectors which emanate from the same point is called the are called the initial vectors. So let's say a and b come from the same point or they are called co initial vectors. I remember for following the parallelogram law of addition, you have to make the two vectors which are supposed to be added as co initial vectors. Okay. Similarly, co terminal vector. Let me move on to the next page. Co terminal or co terminus vectors. What is co terminus vector which terminate at the same point. So let's say a and b both terminate at the same point. P. They are called the co terminus vectors. Okay. Just for your knowledge, it's not like these concepts are very important. But we should know how the naming has been given to these vectors. Next is equal vectors. Equal vectors are two vectors which are exactly in the same direction. Let's say a vector is equal to b. Then direction of a is same as direction of b and magnitude of a is same as the magnitude of b. Okay. These two vectors then will be called as equal vector direction should be same and magnitude should be same. Next is like and unlike vectors. Like and unlike vectors. Two vectors a and b are said to be like. Are said to be like if they have the same direction. If they have the same direction. Okay. That means they are parallel and they have the same direction. For example, let's say a is like this. B is like this. Okay. Then they would be called as like vectors. These two vectors would be called as the like vectors. On the other hand, two vectors a and b are said to be unlike vectors. If they have, in fact, they are parallel. If they are parallel, but they have opposite direction. But they have opposite direction. Okay. So something like this vector a and vector b. Okay. They will be called as unlike vectors. Please note that equal vectors are type of like vectors. Equal vectors are type of like vectors. Now don't get confused with unlike and non-colinear. There is something called collinear and non-colinear, which I'm going to take up next. Okay. Later on, not now. That will be a separate part of your study in vector algebra. Okay. Something very important now is the concept of free and localized vectors. Okay. What are free vectors? Free vectors are one which has got a specific, which has not got a specific initial point. So no specific initial point. Okay. Whereas localized vectors are the ones which has got a fixed or a specific initial point. To give you an example, let's say I'm saying I'm going five kilometers towards north. Okay. So five kilometers towards north is a displacement vector, but it doesn't depend from where you are starting. Okay. Any point in space you start and if you start moving towards north, it would be called the five kilometers towards north vector. Right. But on the other hand, something like the weight force, weight force always starts from the center of mass of the body. Right. You cannot draw a weight force like this. You cannot say, oh, this is the weight force of this body. It is localized. It can't move anywhere. Okay. If you talk about tension, tension on the body, it is always going to start from the body and go opposite to the direction of it. Correct. It is localized. It cannot move. So mostly in physics, you're going to deal with localized vectors. So these are the subject area of physics. Whereas in maths, you're going to talk about free vectors. So in maths, your vectors are free. They have already attained freedom. Okay. Three vectors can be taken anywhere parallel to themselves in space. Okay. So for example, if let's say I have a vector a like this. Okay. I can draw this vector here also and still call it as a, I can draw it here also and still call it as a, please note, I should not change its magnitude and direction. That's everything. If it rest, it is taken anywhere parallel to itself. It will be called as the same vector A itself. Is that fine? No doubt about it. I'm sure you, these concepts are already known to you. That's why I'm rushing through it because you're well aware of these concepts. No doubt. Nevertheless, please stop me and ask me. Next, we'll quickly talk about operations on vectors. Operation on vectors. Normally this part is very big because operation will cover all type of operation like addition, scalar multiplication, dot product, cross product, scalar triple product, vector triple product, et cetera. But I'm going to spend today's time just on talking about addition of vectors. And of course, concepts related to it. I will talk about scalar product and vector product in the subsequent classes. So addition of vectors, the first operation that we are going to learn is how to add vectors. I'm sure you must be already be bored of this particular concept because you have already started adding vectors in class 11th in physics. By the way, when you add two vectors, you obtain something which we call as the resultant. Finding the resultant of vectors. So let's say we have vectors A and B and we want to add it. How does the vector addition happens? Let me show you two ways. In fact, I'll show you three ways to add vectors. The first way that we are going to talk about is called the parallelogram law of addition. In parallelogram law of addition, let's say we have these two vectors, A vector, and let's say we have another vector like this, B vector. Now, when we want to add it, the following steps that we used to find out the parallelogram law of addition. First step is make both the vectors as co-initial. Step number one, make these two vectors as co-initial. Make vectors A and B as co-initial. Please note, these vectors are free vectors, so you can take them parallel to themselves in such a way that they become co-initial. So let me do one thing. Let me maintain the direction of A and let me bring B parallel to itself and fix it over here. So what I have done, I have just taken this B, lifted it and brought it over here like this. Getting my point. So I have made them co-initial. So let's say I call this co-initial point as point O. Now, complete a parallelogram with these two as the adjacent sites. So when you complete a parallelogram with these two at the adjacent sites, the vector connecting the co-initial point O to the opposite vertex, let me call it as B, this vector is your A plus B vector or it is the resultant of A and B. Parallogram law of addition is followed, especially in cases where your vectors are not free to move. Let's say you're talking about calculating the resultant of two forces on a given body. Then probably this law is going to be very, very useful. If I want to subtract two vectors, let's say I want to do A minus B. If I want to do A minus B, you can do it in a similar way by following parallelogram. You just have to do A plus negative B. So I'll show it on the same diagram. So let's say this is your A vector. Instead of B over like this, make a B vector like this. So this is your minus B vector. Now again, complete a parallelogram like this. Complete a parallelogram like this. A vector connecting this point to the opposite vertex, this vector would be called as the A minus B vector. Now remember, this A minus B also directs towards the other diagonal of this parallelogram. Let's say the same parallelogram which I had drawn over here. This particular line is your A plus B line. This is your A plus B vector. The other diagonal over here, if you direct it from this point, let's say I call this as B and A. If you direct it from B to A, this will be called as your A minus B vector. Is that fine? So same parallelogram can be used for subtracting those vectors as well. Is that fine? Any question? Okay. Another way to add vectors. Let me move on to the next page. Another way to add vectors is called the triangle law of addition or triangular law of addition. In this, what do we do? Let's say we want to add vector A to vector B. Step number one, we make the terminal point of A and initial point of B at the same position. That means I connect the terminal point of A with the initial point of B. When I do this, then the side which completes the triangle over here, that's why the name triangle law of addition has been given. This is called the resultant of A and B. Getting the point. So it is like saying that if you want to go from a point, let's say A to B, or let me say O2, let's say P, you take the path from O to A and then go from A to B. So you are trying to say OP is nothing but OA plus AP. Finding the resultant of a vector is nothing but finding a path from one point to another taking the root of those vectors. So if I want to go from O to P, take the root of O to A and then from A to P. So the resultant OP will be called as A plus B vector. So these things are known to you. One important thing that comes out from this particular discussion is the magnitude of A plus B will always be less than equal to the sum of the magnitudes of A and B. And the magnitude of A minus B will always be greater than the difference of the magnitude of A and B. Remember these two? Triangle inequalities or triangular inequalities. Now, before I start talking about anything else, I would be asking you a simple question. Very simple question. Let's say there is a regular hexagon. There is a regular hexagon. Now this regular hexagon, let me call it as A, B, C, D, E, F. We know these two sides represented by vector A and this is represented by vector B. Find in terms of A and B, the vectors CD, D, EF and FA. One minute to solve this. Find in terms of A and B, the vectors CD, D, EF and FA. Remember it's a regular hexagon. It's a regular hexagon. Barring CD, I think D, you can always guess what is D. D is negative. In fact, you can also tell me what is EF. EF will be negative B. How will I get CD? Any idea how to get CD? Now, in order to find CD, I will first talk about very important law which is a generalization of the triangle law of addition. Let me go back to the previous page itself and address that so that it's on the same page. Something which we call as the polygon law of addition. Triangle law is a specific case of something which we call as the polygon law of addition. Polygon law of addition is useful when you are adding, useful for adding more than two vectors. More than two vectors. How does this law work? Let us say we have three vectors A, B, C. If you want to find the resultant of these two vectors, follow the same procedure as what we discussed in the triangle law. Keep connecting the terminal point of one with the initial point of the other. And you can do it in any order. It is not like necessarily it has to start from A only, then B, then C. You can start with B, you can go to A, then you can connect C. Whatever manner you follow, it is not going to affect it. So, let us say I start with A. So, A let me draw it like this. Then B, then C. So, this law says that connect the initial point of the first vector to the terminal point. The initial point of the other vector and you have to show this by a dotted line. Remember, resultant is shown by a dotted line. This vector, which is completing this polygon over here, that would be the resultant of A, B and C. Again, as the definition itself says, you have to go from, if you want to go from P to Q. Let us say this is R and this is S. You have to first go from P to R, R to S and S to Q. A similar thing I will be following in solving the problem, which I just now gave you. Now, if you see, if I connect the A point to D point, would all of you agree with me that this vector is going to be 2B vector? Now, why I am calling it as a 2B vector? Because number one, this vector AD is parallel to BC. You all know that from the symmetry of the figure. And AD length is twice of BC length. You can use your normal geometry to figure out AD length is twice of BC length. So, if this is B vector, I have full right to call this as 2B vector. Now, let us apply a polygon law of addition. So, can I say AD vector is the resultant of AB plus BC plus CD. AD is 2B, AB is A, BC is B, CD, but I don't know what is CD. So, from here, CD vector could be just take it on the left side. So, 2B minus A minus B, that's nothing but B minus A vector. So, 3B was absolutely correct. So, CD is B minus A vector. So, this is your B minus A vector. So, your FA would become opposite or negative of this, which is nothing but A minus B vector. Now, what is important here is something which I am going to spend at least half an hour now is position vector. This is very important concept. Why I call this a concept which is very important because it connects or it relates vectors to coordinate geometry. It relates vectors to coordinate geometry. And all of us have studied coordinate geometry before we were introduced to vectors. Now, why it connects vectors to coordinate geometry will come to know in some time. First of all, what is the definition of a position vector? So, let's say any point in space A. If I say this point A has a position vector of let's say small a, what do I mean by it? What I mean to say is there is some reference point O. Now, where it is I don't know you can choose your reference point anywhere you want. And if you connect that reference point O to this point A, then this vector is your A vector. Alternately, A is the position vector of point A. So, we read this as A is the position vector. A is the position vector of the point A. This is to be read as this. Geometrically speaking, it is a vector connecting the reference point O. Now, you would ask me sir, where is this reference point? See, reference point will never be mentioned to you. It is just like saying that there is a point whose coordinate is 1,2. Now, does the person specify where is origin for you? No, right? He just says there is a point 1,2. It is up to you to choose your origin and from there, relatively you choose your point 1,2. In the same way, if I say the position vector of some point A is A vector. It is up to you to choose your origin. It is up to you to choose your reference point. So, this is not specified. It may not be specified to you. By the way, this reference vector has itself a position vector of a 0 vector. Is that fine? Understood what is the meaning of position vector? So, position vector plays the same role as coordinates play in coordinate geometry. So, it plays the same role as coordinates of a point plays in coordinate geometry. Okay? Now, how can we use position vectors? First of all, position vectors, we need to understand few things about it. Number one, if I say there is a point A and a point B, whose position vectors are A and B respectively. Correct? Then what is your vector A, B in terms of small a and small b? What will be vector A, B in terms of small a and small b? Okay? Now, in order to find this, I am going to use my addition of vectors. So, let's say O is here. Again, O is something which you can choose on your own. It is your call, absolutely your whims and fancies to choose your reference point. So, when I say O is here and I say position vector of A is your vector A, what it means is that this vector is your vector A. Okay? When you say position vector of B is B, what I mean is this vector is your vector A. Okay? Now, can you tell me what would be your AB vector? Okay? If I apply vector addition, can I say O A, O A plus AB is OB. In other words, can I say AB is OB minus OA and OB is nothing but B minus A. This is something which you need to remember as a rule because every time we can't sit and derive this. Remember, if you are connecting a point A to a point B where point A has a position vector of small A and point B has a position vector of small B, then AB vector will always be B minus A. The way to remember this is, let's say I am going from P to Q, then it's always position vector of Q minus position vector of P. Okay? Destination minus source, remember like this, destination minus source. Destination means to where you are going, source means from where you are originating. Okay? So P to Q if you are going, Q is your destination, P is your source. So position vector of destination minus position vector of source. Just remember it like this, right? This is just for your memory aid. Okay? Now, if this is clear to you, I have a question for you all. Let's say I have point A and B whose position vectors are known to us. Can you find the position vector of a point C over here? Okay, let me call it as vector C, which divides the joint of A and B in the ratio of M is to N. Let's say M is to N, we are finding the point which divides this in the ratio of M is to N. Anybody can give me the answer for this? Give me the vector C in terms of A and B. Absolutely. Okay, how do we get this formula? Same logic. Let's say this is my reference vector. Okay? And I start connecting these vectors to this reference position. Now, remember this is your C vector. Okay? This is your A vector. This is your B vector. Okay? Okay? And this is C vector is the resultant of OA and AC. Okay? Now, OA is known to you, that is your vector A. How will I find AC? Remember, AC vector is a like vector of AB vector. Okay? So, AB vector, first of all, we need to find out. AB is, we can say collinear also, but since I have not used the word collinear before, I am not going to use it. Okay? So, AB here is a like vector of AC vector. Now, what is AB vector itself? AB vector is nothing but B minus A. Okay? Now, if you see AC vector, the length of AC vector, okay, is M by M plus N times the length of AB vector. Right? Because the entire ratio is M is to N. N is the total length, then M will be the length of AC. In other words, AC vector is M by M plus N times AB vector. Remember, the direction is also the same. Okay? So, what we can do here is, I can substitute the value of AC as M times AB. AB is B minus A by M plus N. Okay? No, space is gone. Okay? So, let's take M plus N as LCM. So, it becomes M plus N times A M B minus A. Just expand it, it becomes M A plus N A M B minus M A by M plus N. So, M A, M A gets cancelled. So, it becomes M B plus N A by M plus N, absolutely correct. Those who replied to the question, you are absolutely correct. What I want you to appreciate over here is that it resembles the section formula. It resembles the section formula that we have learned in coordinate geometry, right? But instead of using two X and Y, we can use just one position vector. So, it's M into B, N into A by M plus N that is going to be the position vector of C. So, C is going to be this expression. Is that fine? You can also use your midpoint formula if let's say there are two vectors like this. Okay? Sorry, there are two points like this. Whose position vector is A and B and you want to find out the position vector of a point M. Let's say M, then M is nothing but A plus B by 2. Now, based on this, I would like you to solve a few questions. Hope you can read the question. Two forces A, B and A, D are acting on the vertex A of a quadrilateral A, B, C, D. And two forces C, B and C, D at C. Two that the resultant is given by four EF, where E and F are the midpoints of AC and BD respectively. Once you're done, please say done so that I know how many of you have solved it so that we can start the discussion. Let me make the diagram for you. So, let us say this is my A and this is my C, B and D. A, B, C, D. Okay, let me connect them. Let's say E is the midpoint here. F is the midpoint. So, what is this question trying to say is that resultant of, prove that resultant of A, B, A, D, C, B, C, D is four times EF. I'm sure NPS has done this bit of vectors in school. Dealing with position vectors. Okay, NPS. Vector has not been done in school. Okay, we started. So, let me just do this problem now. So, A, B, let me assume A to be the origin. Okay, let me call A to be the origin. Okay, so I can assume that A has the position vector of a null vector. Okay, so what I'm going to do is I'm just going to replace my name A with an O. Okay, so that we can refer to it as O, B instead of O, D. Okay, so all of a sudden this vector has, this problem has become a Bengali problem. Sorry, that was a bad joke. A has become O now. O, B, shake. Okay. Now, O, B, let me call the position vector of B as B, position vector of D as D, position vector of C as C. Okay, so if you look at your left-hand side, if you look at your left-hand side, so if you look at your left-hand side, O, B is nothing but B vector. O, D. O, D is your D vector. C, B. Now, C to B if you want to go, remember, destination minus source, that will be B minus C vector. C, D, C, D will be D minus C vector. Okay, so if you collect all the terms, you get 2B, okay, minus 2C, plus 2D. Okay, so this is your left-hand side of the expression. You can take a 2 common also, so 2B minus C plus D. Let me call it as 1. Now, what is EF vector? EF vector is nothing but position vector of F minus position vector of E, destination minus source. Correct. So what I'm going to do is, first of all, I'm going to figure out what is the position vector of E. Now, E being the midpoint of O and C, can I say E will be nothing but 0 plus C by 2. That's nothing but C by 2. So OE will be C by 2. Now, what is OF? OF is nothing but the position vector of the point F, which is the midpoint of DNB. So you can say B plus D by 2. So position vector of F minus position vector of E will give you the vector EF. Take a 2 common, it becomes B minus C plus D, which is nothing but, yeah. Now, if you do 4EF, if you do 4EF, what do we get? We get twice of B minus C plus D. Let me call this as 2. Now, since 1 and 2 both are same expression, that means your left-hand side is equal to right-hand side and 2. Easy one, just to give you an idea about how position vectors can be used to prove these things. Next question, which is coming your way. So read this question carefully. If O is the ortho-center and O dash is the, sorry, O is the circum-center and O dash is the ortho-center of the triangle ABC, proves that OA plus OB plus OC, that is the resultant of OA, OB and OC, is O, O dash. So there are 3 proofs given to us. Let us prove the first one. Let me make a diagram for you so that we can refer to that diagram. Sorry, this is O dash. So I am perpendicularly bisecting these 2, this is your O. So I have been asked to show that OA plus OB plus OC is O, O dash. This is your O, O dash. How do I prove this? Okay, let me make your life simple. Let me just try explaining this. Let me call this point as OD. Let me call this point as D over here. Okay. Now this is also perpendicular and this is also perpendicular. Let me call this as, okay. So AM and OD both are perpendicular to the very same line BC. So can I say they are parallel actually? So this vector and this vector are parallel to each other. Correct. Okay. Now, how many of you are aware of the distance of the orthocenter from the vertex? If you would have done this chapter of properties of triangle, then that idea comes from there. So let's say I have a triangle like this. Okay. Okay. Let me draw altitudes. What is this? Okay. So let us say this point here is your orthocenter O dash. And this is your vertex A. Okay. How many of you are aware that O A length, sorry, O dash A length is 2R cos A. Not many, I guess, because probably would have not used this. Okay. I think in the revision crash course, most of you would be solving problems. Based on this particular concept also. Okay. Now, now, how do I prove this first of all? How do you approve this? Very simple. The proof is very simple. If you talk about this triangle, let me call this point as M. If I talk about this triangle, A, B, M, right? Okay. So can I say this angle, since it is angle B, this angle will be 90 minus B. Okay. Correct. Let's say O dash A length is X. Okay. Let's say I call this length as X. So can I say this length, let me call it as AP. So can I say AP length would be X sin B? Yes or no? Yes or no? Now, all of us know the naming convention of a triangle. This is called B. This entire thing is called A, etc. Now look at the triangle. APC. Okay. Since this is angle A completely, this entire angle is angle A. And this side is B. Can I say AP will also be B cos A? Yes or no? AP will be B cos A. Since both of them represent the same thing, can I equate them? So can I say X sin B is equal to B cos A? And from sin rule, we know that. From sin rule, we know that A by sin A is equal to B by sin B is equal to C by sin C. And this ratio is to R. What is R over here? R is the circum-circle radius. Circum-circle? Radius. Okay. So from here, I can say small a is to R sin A. Okay. And small b is to R sin B and small c is to R sin C. Okay. This formula, any way you should remember. Okay. So in this expression, can I put this small b as to R sin B? Okay. Sine B, sine B gets cancelled off. What is the value of X? 2R cos A. So that's what I was talking about. Here. That the length of the orthocentre from vertex A is to R cos A. Similarly, what is the length of orthocentre from vertex B? You will say simple 2R cos B. Okay. So in general, in general, O dash B is equal to 2R cos B. And O dash C is 2R cos C. Just remember these formulas. They're important. Okay. Now, how is it going to help me solve this question? Okay. All of you get back to the question now. So this derivation is done. Let's get back to the question. Let me draw a line over. Now, do you all agree with me that since OD is perpendicular to BC and this length is R, OD length is going to be R cos A, right? Because this is R and this is, remember this entire angle. Okay. A will be this angle because, you know, the angle subtended at the center is double the angle subtended at the vertex. So both will be A each. Okay. So I'm not drawing the circle. There is a circle which is passing through ABC whose center is at O. So this angle is also A. So OD is R cos A. Now, since OD is parallel to, since OD is parallel to O dash A and O dash A is double of OD, can I make this statement that vector O dash A is to OD? Sorry. Vector A O dash, not O dash. Vector A O dash will be to OD. Do you agree with me on this? Okay. Now, what I'm going to do next is since D is the midpoint, since D is the midpoint. Okay. Let me take the reference at O. Remember, I can take my reference at any point I wish to. Okay. So can I say OD vector, OD vector is the position vector of D and position vector of D will be position vector of B plus position vector of C by 2. See what I'm doing is something like this. Let's say this is my O, this is my B and this is my D and this is my, okay. So what I'm saying is that let this be the position vector B. Let this be the position vector of C, then position vector of D will be B plus C by 2, right? Which is nothing but your OD vector. So OD vector is OB plus OC by 2, which means OB plus OC is equal to 2 OD, right? Now, let us look into the question. This is asking us to prove that OA plus OB plus OC is OO dash. Am I correct? This is what the question is asking me. Yes. Now, instead of, instead of OB plus OC, I'm writing 2 OD, okay. And left O is OA as such. Okay. Number 2 OD is from here, 2 OD is AO dash. 2 OD, I can replace it with AO dash, okay. Now, all of us can take a guess. If I'm going from O to A and A to O dash, let me show you the path. I'm going from O to A and A to O dash indirectly from where to where I'm going. I'm going from O to O dash. That is nothing but this expression is O to O dash, which is your RHS. So this problem would have been very simple had you already known the result of the distance of the ortho center from the vertex. But nevertheless, it's never too late to learn anything new. Now, can you all try to prove the second one? The first one I've already done for you. Check. Second one. Can you prove on your own? O dash A plus O dash B plus O dash C is 2 O dash O. Done anyone? Okay. Let me help you out with this also. Let me erase all unwanted things. Okay. Let me do a small construction. Let me join O dash to D. Let me join O dash to D. Now, will you all agree when I say that O dash D, O dash D. Let me write it down over here. O dash D is nothing but O dash B plus O dash C by 2. Simple because again, what I'm doing is I'm shifting my reference point to O dash position. Okay. So again, D is the midpoint of BC. So this relation will hold true for that also. That means O dash B plus O dash C is twice of O dash D. Okay. That relation will be still valid. Okay. Now, O dash A. O dash A, can I say it is twice of DO vector? Correct? We have just now proved little while ago. O dash A is twice of DO vector. Right. So what I'm going to do is in the left-hand side of the expression, which is nothing but O dash A plus O dash B plus O dash C, I'm going to put this as 2 O dash D. Sorry, not O dash D, 2 DO vector. Anybody saying something? Let me just check. 3D we're saying something. Got it? Okay. And your O dash B and O dash C is nothing but 2 O dash D. Right? So I can write it as 2 O dash D plus DO. O dash D plus DO is nothing but O dash O vector, which is nothing but your right-hand side. This is what we wanted to prove. Okay. So see how position vector is helping to make our life easy. So second part of the question also done. Third part. Again, I'll give you a minute to solve the third part. A hint, you can use your second part result for your third part. Just try it done. Or you can just unmute yourself and speak also. No problem. We don't have got many people over here. Guys, one more minute I'll give you. Okay. Simple problem. What I'll do is I'll write it as 2 AO dash. Minus AO dash plus O dash B plus O dash C. Okay. So what I've done. So what I've done. I've just written AO dash as 2 AO dash minus AO dash. Correct? Now, if you switch the position of A and O dash, it becomes 2 AO dash plus O dash A plus O dash B plus O dash C. Okay. This is already known to us. This is already known to us and that is 2 O dash. So this is to this. If you combine this, it becomes 2 AO dash plus O dash O. That is nothing but twice AO. That's what we wanted to prove over here. And they're calling this twice AO as AP vector where P is the. Where AP is the diameter through it. That's fine. This is something which is optional. So in fact, I wanted to prove that it is twice AO vector, which I've already done, which is your RHS. Okay. Is that clear? So see how important is your concept of position vectors to deal with such questions? Any doubt? Please type CLR on your screen if it is clear to you. CLR. CLR. Okay. A simple NCRT question. Just on the use of position vectors as finding the internal and external bisectors. Find the position vector of a point R, which divides the line joining points P and Q whose position vectors are given to you in the ratio 2 is to 1. Internally, number 1 externally, number 2. Quick. You should not take more than a minute to solve this. It's a conventional question. This is internal division. This is external division. Okay. So let's say position vector of P is known to us. This is your position vector of P. This is your position vector of Q. Done. Okay. Very simple. So position vector of R will be 1 into position vector of P, which is nothing but I plus 2J minus K. This into 1 plus 2 into position vector of Q, which is minus I plus J plus K by 1 plus 2, which is 3. Okay. Let me quickly simplify this. So this is going to be minus I plus 4. J plus K whole divided by 3. This should be your position vector of R. Done. Okay. Similarly, when you're talking about external division, all you need to do is to change the sign in between. So 1 into I plus 2J minus K minus 2 into minus I plus J plus K whole divided by 1 minus 2, which is negative 1. Okay. So let me simplify this as well. So you have 2I plus I3I minus 3I. Okay. J, I think will be getting cancelled. So 0J. K will be minus K minus 2K, which is minus 3K minus 3K by minus 1 is plus 3K. This is going to be your position vector of R. In case of external division, that's going to be your answer. Is that fine? Very simple question. Now, guys, without wasting much time, I'm going to talk about resolution of vectors. That means resolving vectors along two perpendicular directions. And those two directions are normally chosen as your X and the Y axis. The reason why I'm slightly fast right now is because you are already aware of this concept. Okay. So if you want to resolve a vector A, let's say like this. Okay. Let's say this is your O point and this is your A point. Okay. What do we mean by resolving this vector? Means finding the vectors, components along the X and the Y axis whose resultant would be your A vector. Correct. So in order to do this, what I'm going to do is I'm going to drop a perpendicular from A on to X axis. Let me call it as M. Let me call this angle as theta. Okay. Now, if this is your vector A, its length will be mod A. Right. So OM length will be mod A cos theta. And AM length would be mod A sin theta. We all know that. Correct. Now, if I know the length and I know the position vector, remember when I was talking about unit vectors. So if I know the magnitude of a vector and I know a unit vector along that vector, I can always generate a vector in the direction of that vector of that given magnitude. So I can generate OM vector. Okay. Let me show you on the diagram. I can generate OM vector by multiplying the magnitude of OM vector, which is mod A cos theta with i cap. Remember along positive direction, the unit vector along the positive X direction, the unit vector is i cap. Similarly, MA vector is nothing but mod A sin theta. And this direction is along the direction of positive y axis. So it is j cap. Now, I can say from vector addition, OA vector is the sum of OM vector and MA vector. That means nothing but mod A cos theta i mod A sin theta j. Mod A sin theta j. Okay. Let me tell you this point mod A cos theta is nothing but the x coordinate of point A. And this is nothing but y coordinate of point A. So you can write it as xi plus yj. And if O is the origin, we can all see that this represents the position vector of A. Okay. And this is how you can actually relate position vector to coordinate geometry. So I was telling you earlier also the coordinates of a point in coordinate geometry is synonymous to the position vector of that point A. So let's say A is a point. So this is just attach a i and j along with the coordinates. It starts playing the role of position vector. Okay. One more important thing is that the magnitude of any vector which has been resolved along the x and the y axis is nothing but under root of x square plus y square. How do we know that? Very simple. You know that by Pythagoras theorem. O is the hypotenuse. Correct. So hypotenuse square is OM square plus MA square, which is nothing but x square plus y square. So if you know the components of a vector along the x and the y, you just have to square those components and add them up and take the under root to get the magnitude of that vector. Okay. Now something very important over here is if you do A by mod A in this expression, this is your vector A. Correct. So if you divide the A vector by mod A. Okay. This is something which we call as A cap, right? That will actually become cos theta i plus sin theta j. Okay. Now normally this sin theta, we write it as cos 90 degree minus theta. Okay. Now remember here, theta is the angle which A makes with, theta is the angle which A makes with x axis and 90 degree minus theta is the angle which A makes with the y axis. Okay. If you take the cos of these two angles. Okay. If you take the cos of these two angles and attach a i and j next to it, it actually becomes the unit vector along A. Now this concept I'm going to scale up to 3D vectors and I'm going to explain you the concept of something which we call as direction cosines. Okay. So here cos theta and cos 90 minus theta are the direction cosines of a 2D vector. Okay. How many of you have heard this word direction cosines in school? Any one of you? Have you heard the word direction cosines in school? Anyone? Just type yes or no? No. Okay. You may not understood. Okay. Now I'm sure you understood what do we call as direction cosines. In fact, I'm going to take up this concept next with you when we are talking about resolution of a vector in 3D. So this was the concept of resolving a vector in 2D, in two dimensions. Next we'll talk about resolution of vectors in three dimensions. Just one minute. Yeah. Now when you have a vector in three dimensions, first of all, let me make a right-handed coordinate system. Remember, we always follow a right-handed coordinate system while representing any coordinates or position vectors. Right-handed coordinate system means any coordinate system where if you stretch your fingers along the x-axis and naturally curl it towards y-axis, naturally curl it. When I say natural curl, you know how naturally the fingers curl. Then your direction of the thumb will be your direction of the z-axis. Don't take any line as x, y, and z. Please note we have to follow a right-handed coordinate system. Now, let's say I have a vector. Let's say I have a vector like this. Okay. Vector OA. Okay. Now first of all, let us understand what are direction cosines. You already understood what are direction cosines in 2D. It is cos of the angle made by the vector with the x-axis and cos of the angle made by the vector with the y-axis. In the same way, let's say this vector OA makes an angle of alpha with the x-axis. Now, many people are confused because they are not able to imagine how do I imagine the angle between the vector and the x-axis look like. See, very simple. Take a plane which is passing through the vector and the x-axis. Okay. So think as if you are taking a plane and you are passing it right through the vector OA and the x-axis. Now, on that plane, the angle between the OA and the x-axis that represents your angle alpha. Similarly, let's say the angle made by the vector with the y-axis is beta and the angle made by the vector OA with the positive z-axis is gamma. Okay. So I will write it down. Alpha is the angle between OA vector and let's say i cap. Okay. Beta is the angle between OA vector and let's say j cap and gamma is the angle between OA vector and k cap. By the way, all of you know the angle. How do we define an angle between two vectors? Everybody knows that. I'm sure in physics you must have done this. See, just to create, just to be on the same page. When we say what is the angle between these two vectors? Okay. Normally, two things come in our mind. Okay. One is this angle. Let's say I call it as alpha. Other is let's say beta. Correct. Yes or no? Which is the answer over here? Alpha or beta? If somebody says, what is the angle between the vector A and vector B here? What will you state? Alpha or beta? Aariman says beta. Beta. Is there anybody who is thinking of alpha also as the answer? If you are, let me tell you your own. Okay. Beta is the correct answer. The angle between two vectors is defined as the shortest angle between the two vectors when they are assumed to be co-initial. So A and B, when you make them co-initial, the shortest angle between them, okay, not the reflex one. This angle will be your angle between the two vectors. Now, what are direction cosines? Direction cosines are defined as the cos of the angle that the vector makes with the x-axis, y-axis, and z-axis respectively. So this triad, which we normally nickname it as L, M, N, just to make our life easy. These three parameters together constitute something which we call as the direction cosine of that vector. Are you getting my point? Is that clear? Okay. Later on, this concept is going to be used for expressing the equation of a line in three-dimension. Okay. We'll be also using this for the concept of equation of plane as well. Okay. So let me just summarize quickly. Direction cosines are nothing, but it is the cos of the angle. Or it is the triad, which is made by taking the cos of the respective angles made between the vector and the respective axes. Okay. Later on, we'll go on to prove that L square plus M square plus M square will always be equal to one. We'll prove this in some time. But before that, the idea itself is pending, which is the resolution of this vector along the X, Y and Z axes. So let me do that. Now, let me just drop a perpendicular from A onto the XY plane. Let me call this point as P. Okay. And again, let me drop a perpendicular from P onto the X axis. Okay. Let me call that as Q. Okay. Now, how many of you would agree to the fact that OQ length, OQ length, not OQ vector, OQ length would be the length of OA, which is mod A, into cos of the angle alpha? See, basically, I'm trying to take the projection of OA on the X axis. Would you agree with me on that? Yes, sir. Okay. Would you agree with me further if I say QP length would be mod A cos beta? Because QP length is just like taking the projection of the vector OA along the Y axis. In the same way, you will also agree when I say AP is nothing but mod A cos gamma, that is the projection of the vector OA along the Z axis. So, if I want to reach from O to A, can I take a path like O to Q, Q to P, P to A? All right. O to Q. O to Q is nothing but mod A cos alpha times I cap, because you are going in the direction of positive X axis, so magnitude into unit vector along that. QP will be mod A cos beta J cap and PA would be mod A cos gamma K cap. Okay. This finally is your A vector, which has been resolved in I, J and K components. So, this is called the resolution of vectors in three-dimension. Let me show you something very interesting. First of all, if the coordinates of A is XYZ, you would agree that the vector A could be actually written as XIYJZK, which actually forms the position vector of the point A. Okay. Second thing, if you look at this expression again, A by mod A is cos alpha I cap cos beta J cap cos gamma K cap. And we very well know that this is nothing but L, M and N. So guys, one thing I want to highlight over here is this fact, which most of us would be needing in our 3D geometry is that the direction cosines of a given vector are nothing but the I, J and K components of a unit vector along that vector. If you want, you can write this down. The statement is going to help you out throughout the 3D geometry chapter. So, L, M, N, these are I, J and K components respectively of a unit vector along that vector. So, these are I, J and K components, components of a unit vector. In fact, there's always one unit vector along a given vector. I should say the unit vector, not a unit vector of the unit vector along A. Getting my point. Now, this concept gives me the fact that if this is the unit vector, the magnitude of this vector, which is given by under root of L squared plus M squared plus N squared, just like we did it in two dimensions, should be equal to 1, which proves the fact that L squared plus M squared plus N squared is equal to 1, as I promised you earlier that I would prove. And here is the proof for it. Are you getting my point? Is the concept of position vectors clear in mind of everybody over here? If it is, then I have a question for you all. No, no, no, no, it's always plus 1, 3D. Why minus 1? How can it be minus 1? L squared plus M squared plus N squared, they are three positive quantities. How can the sum be minus 1? It will always be 1. Okay, so I have a question in front of you. I hope you can all see it. 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. When I say velocity, give me the magnitude and direction. Two and a half minutes to solve this. Time starts now. Anyone ready with the answer? Two and a half minutes are over. Come on, a simple... Okay, we have an answer from Tudi. Tudi, we are absolutely correct in terms of magnitude. May I have the direction, please? And please send me the answer in private so that nobody is influenced by your answer. Everybody, please send your answers in private to me. Okay? In subsequent questions, please send it in private. Minus J direction. That is also correct. Well done. Time up for the others who are trying to solve this. Now, all of us know the concept of relative velocity. When you say the wind velocity with respect to man, what does it mean? It means the wind velocity minus the man's velocity. This is what we call as how the wind will appear. What is the wind velocity? This is called wind's velocity with respect to man. Let us say the wind's velocity was xi plus yj. Okay? So when the man was moving with a speed of 8 km towards east, let me call it as 8i velocity. This is the velocity of the man. The wind appears to come from northeast... sorry, the wind appears to come from north to him. Okay? This is the velocity of wind with respect to man. This is something which we are given that... can I call this as minus pj? So let me call it as minus pj. Why minus pj? Because let's say p is the magnitude, then minus j signifies that it is going towards the negative y direction. So wind velocity minus man's velocity is your... let's say minus pj. That's equation number one. Let me first group the terms together. 8 minus i plus yj is minus pj. Now remember there is no i component on the right-hand side. So x minus 8 must be 0. So x is equal to 8. So at least now I know the i component or the x-axis component of the wind's velocity. Next equation is given to me as when he doubles up his velocity to 16i, then the wind appears to come to him from northeast. That means let's say minus qi plus j. Okay? This is i plus j direction. This is i plus j. So negative of that, that's why minus qi plus j. So second equation that I can formulate is xi plus yj minus 16i is equal to minus qi plus j. Correct? Okay? If you group the i terms together, okay? Now x is already 8. So can I say 8 minus 16 is minus q? That is q is 8. If q is 8, y is minus q. That means y is also minus 8. So I got the y component of the velocity of the wind. x component I had already got as 8. So velocity of the wind would be 8i minus 8j. Correct? So modulus of the velocity, that is the speed of the wind will be under root of a square plus 8 square which is 8 root 2 kilometers per hour. And it is coming from minus i plus j means the wind is coming from northwest. Coming from northwest. Simple. Is that fine? Any question with respect to this? Please type noq if there is no question. Next question. Again, three minutes to solve this, time starts now. Just type done if you are done. Read the question carefully. The question states the axis of, the axes of coordinates are rotated about the z axis through an angle of 5 by 4 in an anticlockwise direction. And the components of a given vector is 2 root 2, 3 root 2 and 4 prove that the components of the same vector in the original coordinate system would be minus 1, 5 and 4. Done. Okay, Vidush is done. I need at least two more response. Okay, see it. Let's say this was your original coordinate axes. Now let's say you rotated your coordinate axes and rotated it about the z axis only. So basically now this becomes your new one. Remember, they are still perpendicular. Okay, now, if you rotate everything 5 by 4 anticlockwise, your basis of your x axis and y axis and z axis itself would change. Your unit vectors themselves would change. So initially let's say for your initial coordinate axes, let's say i cap, j cap and k cap where the unit vectors along the x, y and z axes respectively, then for the rotated one, then for the rotated one, I will now convert to let's say i cap dash, j will convert to j cap dash and k will convert to k cap dash. Remember, since you're rotating about the z axis, k cap dash and k cap dash would be the same. But your i cap dash would be nothing but do you all agree that it is going to be i plus j root 2? See it's just like you are finding a unit vector in this direction. What is the unit vector in this direction which is making 45 degrees? You yourself will say i plus j direction lambda. Since it is a unit vector, lambda square plus lambda square under root should be 1. That means lambda should be root 2. Sorry, 1 by root 2. Okay. Getting over. In a similar way, j cap dash would be minus i plus j by root 2 because this unit vector you are finding out. So its direction will be minus i plus j. It will be along the direction of minus i plus j. But you want a unit vector, so you have to divide by root 2. Now, if a vector has its component as 2 root 2 along i, that means it is 2 root 2 i cap dash, 3 root 2 j cap dash and 4 k cap dash. What is the vector along the original coordinate axis? All you need to do is replace your i cap dash in terms of i cap and j cap. Is that fine? Now just take care of the root 2s. 2i minus 3i is minus i. 2i plus 3j is going to be 5j and 4k. So its component along the old coordinate axis will be minus 1, 5 and 4 respectively. That is what we are required to replace. See, the vector along this direction. See, you have rotated it. What is the unit vector along this direction? You tell me. Treat this as your old y and x axis. So if you are taking 45 degree turn, what will be the unit vector along that direction? i plus j. See, this vector will be in the direction of i plus j. But since I want a unit vector, I have to divide it by the modulus of i plus j, which is root 2. That is why i plus j by root 2. Clear? 3d, if it is clear? Great. One question I wanted to ask about the position vectors. But I forgot to ask you earlier. Now I am going to ask you. Yeah, this question. If you are talking about section formula and position vectors, let me ask you this question. Find the point of intersection of a, b and cd, where a, b, c and d have been specified to you on your screen. Three minutes to solve this, time starts now. After this question, I am going to do your limit of a sum as definite integrals. As I promised you. Because only one hour remains after this. Note that it's 3d dimension. It's not a two-dimension that you'll find the equation of a line and simultaneously solve it to get the point of intersection. It's the coordinate mention in three dimensions. Are we classy? Yeah. Okay. Anybody, any success? Which Nikola, Kuchaya? Minus one, one, two. Okay. Okay, fine. I'll give one more minute to everyone. Okay. Time up. Other than Akash. Okay. Let me solve this now. So let us say that the point of intersection is p. Okay. And p divides internally the joint of a and b in the terms in ratios of 1 is to lambda and joint of c and d in the ratio of 1 is to beta. Correct. So considering that p is internally dividing the joint of a and b in the ratio of 1 is to lambda. Okay. Vidush has also given the same answer. Okay. So now let's check what would be the coordinates of point or what would be the position vector of point p in terms of lambda? Okay. So from, from AB, from AB, we can say that position vector of p would be lambda six I minus seven J plus one 16 I minus 19 J minus four K by lambda plus one from CD. The same point p will now have a position vector of beta times three J minus six K plus one time. Two I minus five J plus 10 K by beta plus one since both represent the same points. Can I compare? Can I compare their I J and K components? So if I compare their I components, can I say six lambda plus 16 by lambda plus one is going to be two by beta plus one number one. Number two, comparing the J components minus seven lambda minus 19 by lambda plus one will be equal to three beta minus five by beta plus one. That's number two. And third equation comparing their K components is minus four by lambda plus one is minus six beta plus 10 by beta plus one. Okay. If you see this term very carefully, it's minus two times three beta minus five by beta plus one. Which clearly resembles this expression over here. Getting my point. Yeah. So I can use this as a very important tool to get the value of lambda and beta. One of them is what is required. We don't require both of them. Lambda is also fine. Just beta is also fine. So I can say minus four by lambda plus one is minus two times this term, which is actually minus seven lambda minus 19 by lambda plus one. This cancel out a factor of two over here. Lambda plus one, lambda plus one gone. So I can say seven lambda is minus 21. Lambda is going to be negative three moment. You get the value of lambda job is done. All you need to do is put in this expression. Lambda is minus three to put lambda is minus three. So your P will be minus three six. I minus seven J plus 16. I minus 19 J minus four K by four. Oh, sorry. By minus two. I'm so sorry. So sorry. Minus two. Correct. So your answer will be minus 18 plus 16 by minus two, which is one I plus 21 minus 19 is two two by minus two is minus J and K will be plus two K. That means the point will be one minus one two. This is your answer. I think those who responded. I'm so sorry. Okay. Those who responded, you were all correct. You are correct. You were slightly off in your sign. Is that fine? Okay. So guys, I'm going to put this to a hold and I'm going to talk about quickly limit of a some definite integrals. So there are some questions in limits which are of the form zero into infinity or convertible to zero into infinity, which can be very easily solved by the use of definite integrals. Okay. Now I'm going to explain this concept through an example. Okay. I'm going to explain this concept to an example. Let us say I ask you this question. Find the limit and tending to infinity. One by n, one by n plus one, one by n plus two, all the way till one by let's say two n. Okay. Now many of us make this mistake. We think that since n is infinity, each one of these terms are tending to zero. Correct. And we say the answer is going to be zero. Please let me tell you this is wrong. What is the mistake over here? The mistake here is not a simple one. It's a deep conceptual mistake. The concept is you have mistaken tending to zero quantities which are present infinitely many number to be zero, which is not the case. Something which is very small tending to zero, but present in infinitely lot of quantities may give a substantially big quantities also. For example, a drop of water. Right. Volume wise, it is tending to zero. But when present in infinitely many number of quantities, infinitely many quantity, it can become an ocean of water also. Correct. So this approach must never be taken if you realize your problem is of the nature tending to zero into infinity. Now the problem is how do we solve this because our summation process also doesn't seem to work on this. Right. Because your denominator is constantly changing. Correct. So how do I do this? Now, in order to do this, we follow a mechanism which we called as converting limit of a sum as definite integer. Just follow the steps of this mechanism very, very carefully and you'll get all the problems correct on this particular concept. Step number one. Okay. Just follow my algorithm. So whatever algorithm which I'm going to talk about in the next few minutes, that's going to help you solve all the problems of this type. Step number one. Write down the rth term of the series. Remember, this is a series that you are summing up, right? So write down the rth term of this series. Okay. Many of you. Okay. Many of you will say n plus r minus one n plus one by n plus r, you will say, okay. And of course you're trying to sum up from r equal to zero to r equal to n. Right. So this is your rth term. Okay. So you must be thinking that how come zero's term has been. Okay. Doesn't matter. You can take r plus one also or r minus one also and start from r equal to one doesn't make a difference to my answer. Okay. So step number one is done. Okay. Step number two. From your rth term, you pull out a one by n default as out. If you put one by default, one by n by default out, your expression will look like this now. Correct. It's obvious that if I'm taking one by n common, I have to compensate by putting an n on top. Correct. Now, this part of this expression, step number three, let me call this as a function. Okay. Try to convert this function as a function of r by n. Okay. That is step number three. Let me not write it over here. Let me write it in this third step. Okay. So whatever is left apart from one by n, convert it as function of r by n. Now, when I say function of r by n, let me tell you the meaning of what I want to say over it. You must not see anything other than r by n and constant in that. I must be wondering how do I do that? See, very simple. If you divide by n throughout, it becomes this. This becomes a function of r by n. Getting it? So there are only constants and r by n expression, nothing else. I don't want any stray r terms or any stray n terms. Correct. Now comes the crucial step, step number four. Write one by n as dx and write r by n as x. Okay. So do the following replacements. Replacement number one. Put one by n as dx, put r by n as x. Okay. That means this expression of tr would now start looking like this. One by one plus x dx. Getting the point. Now, most of the structure is ready. Now, what is the limit? So if you're summing it up, it is as good as integrating this up. Correct. Now, I need to know what is the limit of integration. That can be found very easily. See, you are summing from r equal to 0 to r equal to n. Correct. Now, x which is r by n will now become 0 by n to n by n. Yes or no? See, you are calling x as r by n. So when r is 0, x will be 0 by n. Simple as that. And when r is n, x will be n by n. That means you're summing from 0. You're summing from 0 to 1. So when you integrate it from 0 to n, where n is very, very large, you are as good as integrating it from 0 to 1. So the answer for this question that I gave you was just an integral 0 to 1 dx by 1 plus x, which is nothing but ln 1 plus x put the limits of integration. That is nothing but ln 2 is your answer. Are these steps clear? I want you to have a quick look at these four or five steps which I have written over here. Step number one, write down the RS term. Step number two, pull out 1 by n from the RS term. Step number three, whatever is left after pulling out 1 by n in that, you try to make a function of r by n. Step number four, replace 1 by n with dx. Now depending upon from where to where you are summing up on r, your x will change and those will form your upper and the lower limit of your integration. Just perform the integration. Your job is done. Is it clear? Yes or no? Please type CLR on the screen if it is clear. Well guys, good to know that. Let's have some questions. This is also called Riemann sums. Riemann sums. Riemann sum. Okay. It has no correlation with Riemann by the way. Okay, questions. The first question that I would like you to solve is this question. Limit, n tending to infinity. 1 by n square. 1 by n square. 1 by n square. Sorry, sorry, sorry. 2 by n square. 2 by n square. 4 by n square. Then 3 by n square. 2 by n square. 9 by n square. And all the way till 1 by n. 2 by n square. 1. Find the limit of this sum as n tends to infinity. And tends to infinity. Exactly two minutes for this. Shall I give you two and a half? Time starts now. Are we done? Are we done? Sorry, I'm not giving you any break because we just have a very short duration class today of three hours. Guys, let's follow the steps that I discussed. Step number one. Write down the rth term. Undoubtedly rth term would be r by n square. 2 by n square. There's no doubt about this. If it is, let me know. What is the doubt? To be more precise. Step number two. Take a 1 by n common. When I take 1 by n out. This is what I see. Remember what I told you in step number three. Try to see this as a function of r by n. Which I'm sure you must be able to see it quite clearly because there's only r by n terms and seek term over here. So step number four. I can directly go to step number four. Step number four. You have to just replace your... I'm sorry. You just have to replace your 1 by n with replace your 1 by n with dx and r by n with x. So it looks like this. All you need to do is sum this up from r equal to 1 till n is as good as integrate this from. Now you will tell me from where to where you have to integrate. Remember you are calling x to be r by n. So when r is 1, x is 1 by n which is as good as 0 because n is tending to infinity. And when r is n, x is n by n which is 1. So this is your lower limit. This is your upper limit. So you have to integrate it from 0 to 1. So if you say it's pi by 8, let's check. Let's check. Now here all I need to do is put x square as t first. So x dx will become half dt. So limits of integration would be unaffected. It becomes secant squared t half dt. Secant squared t integration is tan t. So half tan t is 0 to 1. So when you put 1 in place of that tan of 1 minus tan of 0. Okay. So your answer is half tan 1. See the tan 1 is not pi by 4 da. Is that fine? Any question guys? Any concerns? As I told you follow these steps. It simply will take you to the answer. Okay. Let's have few more questions. Find summation of first of all limit tending to infinity. Summation from 1 to n of 1 by under root 4n square minus r square. Pi by 6 is what most of you say. Let's check it out. So one thing that I don't have to do is write down the rth term because the problem itself has stated that. So my life becomes easy there. Step number 1. Step number 2. I need to pull out a 1 by n out. So when I pull out a 1 by n immediately I get a n on the top. Correct? Step number 3. Arrange these terms in such a way that apart from 1 by n the rest of the term becomes exactly the function of r by n. So I can do that by dividing both numerator and denominator by n so it becomes 4 by r by n the whole square. Is that fine? Step number 5. I'm sorry. Step number 4. Replace 1 by n with dx r by n with x. When you do that you end up seeing something like this under root of 4 minus x square dx. Correct? Now the limits of integration would be decided by from where to where you are integrating in terms of r. You are integrating in terms of r from 0 I'm sorry. 1 to n. So x will be from 1 by n to n by n which is nothing but 0 to 1. We are integrating this from 0 to 1. So this term becomes By the way this is the integration of sin inverse x by 2. Sin inverse x by 2 is sin inverse half that's going to be pi by 6. Absolutely correct guys. Those who replied pi by 6. For others I have few more questions for you. Let's try to solve this. Hope the question is clear. 3 minutes to do this time starts now. Anyone? Any success so far? First of all limit is you know the algebra of limits right? It can apply individually to all the terms. So you can read this to be like this. So this can be your sum. Sure. Take your time. So you can see here that we have limit n tending to infinity summation of r square limit n tending r equal to 1 till n limit n tending to infinity summation of n cube by limit n tending to infinity summation of r to the power 6. Right? Let us do one thing. Let us take make it as n r by n and multiply with an n square. Let us make this as r by n multiply with an n cube. Here also I take n to the power 6 multiply this. Is that fine? Now let us divide both numerator and denominator by n to the power 7. When I do that something very interesting happens. I get 1 by n square over here which I can distribute as 1 by n 1 by n each to these two terms. So I can say 1 by n r by n the whole square summation. Now individually treat this as if we have three definite integral problems. So here as you can see its integration from 0 to 1 x square dx I need not tell you how and why because you know what to do 1 by n is replaced with dx r by n is replaced with x since you are integrating from 1 to n it is as good as the limits of integration being 0 to 1. The other term here would be 0 to 1 x cube dx the term in the denominator is 0 to 1 x to the power 6 dx and I think this is quite solvable. Integration of x square from 0 to 1 is x cube by 3 which is 1 third this is x4 by 4 which is 1 fourth and this is 1 seventh so the answer is 7 by 12 correct fiduci absolutely correct 7 by 12 is the right answer is that clear everyone please type CLR if it is clear let's solve this limit question I think this limit question you would have seen n number of times n tending to infinity right and same as the question also so limit n tending to infinity n factorial by n to the power n 1 by n solve this 1 by e is what fiduci says anybody else okay let's discuss time to discuss now let me call this limit as L let's take log on both the sides so it becomes 1 by n ln of n factorial by n to the power n remember 1 factorial n factorial can be written as 1,2,3 all the way till n and this n to the power n can be split up between these terms like this so this becomes ln 1 by n plus ln 2 by n plus ln 3 by n all the way till ln n by n now from here your problem starts you can write this as a series infinite series okay and you are trying to find out the limit of this sum correct so if you write down the rth term it is going to be 1 by n ln r by n now everything is so beautifully coming up here that you don't have to apply that conversion of the function to r by n function is already evident from here so it becomes ln x dx limits of integration will be 0 to n okay so this will be your summation of tr from r equal to 1 till n and limit on n is tending to infinity so ln x integration we all know is ln x ln x minus 1 okay 0 to 1 when you put a 1 you get this and when you put a 0 you get a 0 so your answer is finally negative 1 but remember this is the answer to log l to the base e so l is e to the power minus 1 which is 1 by e absolutely correct Vidushi she was the first one to get this problem correct let's have next one 3 minutes to solve this time starts now so if you are wondering how is the rth term in this case it is n 1 by n plus r n plus 2 r okay we are going to sum this up from r equal to 1 till r equal to n so we have an answer over here Vidushi wants one more minute so this says ln of 3 by 2 okay okay let's solve this second step is you pull out a 1 by n common so it becomes n square by n plus r n plus 2 r right now whatever is 1 by n I have to make it a function of r by n so it becomes 1 plus r by n 1 plus 2 r by n okay now make these replacements put r by n as x put 1 by n as dx okay limit of integration will of course be again from 0 to 1 guys please be careful it is not always 0 to 1 so it is a matter of the coincidence that we are getting 0 to 1 every time in the last few problems but be very careful by choosing the limits of integration okay now here we can apply I think we can apply partial fractions for this can we? so partial fractions are related as minus 1 by 1 plus x and 2 by 1 plus 2x okay so we are integrating this from 0 to 1 this is your limit of integration so this becomes minus ln 1 plus x this becomes ln 1 plus 2x okay so when you put a 1 it becomes minus ln 2 and this becomes ln 3 when you put a 0 it becomes 0 anyways so the answer is ln 3 by 2 that is correct well done 3div first one to answer this correctly let's take another one how do you do partial fractions so fast see I had to have a factor of x and 2x getting cancelled from each other right so one has to be 2 other has to be minus 1 all was required was to cancel off that x right so one had 1 plus x other had 1 plus 2x so if I multiply 1 with 2 other with a minus 1 they will get cancelled each other and luckily all the constants became 1 so I didn't have to do anything outside okay let's do this question 3 minutes to solve this time starts now okay 2 ln 2 2 ln 2 is not correct check it out check it again no 1 by is also not correct this is of a similar nature as what we did earlier that is n factorial by n to the power n whole to the power 1 by n it is of the same nature ah no rmr it's not 1 okay let's look into this no it's not 2 either so first of all introduce this n inside that power so if you introduce this n inside the power it will get introduced as n to the power n right let me call this limit as l take log on both the sides so limit n tending to infinity 1 by n log off now n plus 1 n plus 2 etc till n plus n divide by n individually correct so limit n tending to infinity 1 by n log if you see this is 1 by 1 by n then you have log 1 plus 2 by n all the way till log 1 plus n by n correct now treat this as a series pull out the rth term rth term will be 1 by n log 1 plus r by n this term is self-explanatory put r by n as x 1 by n as dx since we are integrating from r equal to 1 till n limits of integration will be from 0 to 1 so this will be your answer to log of l okay yeah so log 1 plus x expansion is very simple log 1 plus x integration is very simple take 1 plus x as t okay so dx is equal to dt so log of t dt integrated from 1 to 2 okay that's going to be your answer for log l that's nothing but t log t minus 1 limits of integration will be from 1 to 2 so when you put a 2 you get 2 ln 2 minus 1 and when you put a 1 it becomes 0 minus okay so answer is 2 ln 2 minus 2 plus 1 that is 2 ln 2 minus 1 that is ln 4 minus ln e that is ln 4 by e okay so remember this is nothing but your log of l which is log of 4 by e so l has to be 4 by e so it's 4 times the previous answer no problem idushi is this clear guys so well I'm going to stop over here we have officially completed definite integers in vectors let me tell you a lot of complicated things are awaiting we have to understand linear combination of vectors that is a very critical aspect of the introduction of vector part where I will be talking about linear independent linearly dependent set of vectors and that is going to pave the path for co-linearity and co-planarity of vectors post that I'm going to do scalar product which is the dot product a dot product and vector product will be taking one class okay then scalar triple product and vector triple product will be taking one more class so I think two more classes would be required in fact three more classes would be required to finish vectors 3d will take another two classes I guess so I think one more month of continuous completion is required meanwhile I will also send you some videos on vectors so do watch it before coming for the next class that will make your understanding much easier fine then over and out from my side all the best for your Tuesday's exam I think that is chemistry which exam is Tuesday physics I'm sorry physics and Friday is your English right okay all the best thank you over and out from my side have a good night