 So three-dimensional lines or three-dimensional geometry, we are officially starting with the topic today. Last class, we ended up on vectors. So vector triple product and vector equations were the last thing that we did under vectors. Yes, yes, yes, you'll be having more tests for neat as well. Yes, in fact, 490 tests will be happening for neat. So that is very much in our plans. Yeah, so under three-dimensional geometry, we are going to talk about 3D lines. Okay, so today our topic will be 3D lines. And what is the overview of the topic? What are we going to study under this? We are first going to talk about the concept of direction, cosines and ratios. Okay, so direction cosines and ratios will be the first topic under lines. Then we'll be talking about the Cartesian form and the vector form and the vector form of equations of 3D lines. Equations of 3D lines. Okay, so Cartesian and vector would be studied hand in hand. So the same problem which I'll be giving you in Cartesian form can also be done in vector form and vice versa. Okay, so many of the many of the concepts you realize that the vector forms. In fact, most of the concepts you realize that the vector forms are more convenient way or efficient way to solve the given problem. So vector will be actually adding like acting as a help for us. So it'll be as a aid to us for understanding our equations and concepts in three-dimension geometry. So we'll be talking about angle between lines, angle between lines, followed by distance of a point from a line, point from a line. And finally we'll be talking about distance between, distance between lines. Okay. So under distance between lines, we'll be talking about three cases. One is the distance between parallel lines. And we'll be also talking about distance between skew lines. And if the lines are intersecting, that means the distance between them is zero. How do we talk about those cases? Or we can say how do we talk about the condition of coplanarity in that case. So we'll be discussing that in the fifth subtopic over here. Okay. I think we have enough time to finish all these concepts in one class itself. Yes. So just to tell you about the crash course, which is going to start from December 27. You're going to have a three hour session almost every day, six days a week, and the seventh day will be a testing day. Okay. Meanwhile, the neat aspirants, you will be continuing with your regular topics because in bio, I believe the syllabus has not been completed. So you will be having regular classes going on. In the evening, you can actually attend the crash course as well. Okay. So as far as possible, we have tried not to have any clashes between your math class and the bio class so that you can attend both of them in case you want to write both JEE and neat exams. However, this combination is very rare. So our testing will happen almost every week on Sunday. That will be full-length tests. No, no, it includes everybody. It includes old and new students, everybody. Okay. And we'll be starting all the way from class 11 topics. Full syllabus tests. Yes, yes. So now you should start writing full-length tests because almost everything is over. Okay. All right. So let's get started with our direction coscience and direction ratios concept. So our current students are our old students and there are some students who will join in just for the crash course. They are new students. Okay. So direction coscience. First of all, let us talk about direction coscience. So from our introduction to 3D geometry, we have learned that we are going to talk about a right-handed coordinate system. So let us say this is a right-handed coordinate system. Okay. So this is your x-axis, y-axis and z-axis. Right-handed coordinate system, as we all know, is a coordinate system which basically, yeah, which basically is based on a simple rule, which is called the right-hand thumb rule. That means if you stretch all your fingers in the direction of the x-axis and curl them naturally towards y, your thumb should point towards the positive z-axis. This is called a right-handed coordinate system. And as I already told you, you may also be asked to solve a question using left-handed coordinate system, but that depends upon the instruction provided in the question. Okay. Now let us say there's a vector in 3D. There's a vector in 3D. Let's say this is a vector. Okay. I'm just making the tail of the vector or the initial point of the vector coincide with origin. In 2D, it is very easy to talk about the direction of a 2D vector. So we basically mentioned its angle with the x-axis or the y-axis. But in 3D, it is not sufficient to just mention the angle from one of the axes. Right. Because it involves a three-dimensional figure. Okay. So just giving you angle with the x-axis is not sufficient. Right. Because there can be several vectors making the same angle with the x-axis. That means you can get an entire generator of a cone making the same angle with the x-axis. Right. So there can be plethora of vectors which make a particular angle. Let's say alpha, whatever you want to state with the x-axis. So it is not sufficient that we state the angle with one of the axes. We require at least two angles. Okay. Now let me just discuss this out first. Let's say OP is a vector. Okay. Let's say the direction of OP makes an angle alpha with the direction of the positive x-axis. Okay. And let's say direction of OP makes an angle beta with the direction of the y-axis. And direction of OP makes an angle of gamma with the positive z-axis. Okay. Then remember the cosine of these angles. The cosine of these angles, they are called the direction cosines of the given vector. So they are called the direction cosines of the OP vector. Okay. No problem. We had just started with the first topic, reaction cosine purple. You have not missed anything. Okay. Now, why did I tell you that, you know, two of the angles is sufficient because there is a relationship connecting cos alpha cos beta and cos gamma, which I will be deriving in some time. But again, for the benefit of those who have joined a little late direction cosine of a vector is nothing but cosine of the angle that the vector makes with the positive x-axis, positive y-axis and positive z-axis. This is also called as LMN for brevity so that we can write them pretty quickly. Now, many times students ask this question, why cosine? Why cosine of these angles, right? You could have taken sign also. But remember, sign does not make any distinction between an angle and its supplementary angle. For example, let's say 45 degree and 135 degree, right? Both are different angles. So these, these angles could be different in case of a vectors. So sign of 45 and sign of 135, both are one by root two. So they doesn't make any distinction between the difference of the angles while cosine, you know, performs this particular distinction very, very. You can say, you know, meticulously. So if an angle goes in the second quadrant, it will automatically change its sign so that you can know that it is not the, you know, angle in the first quadrant. It is an angle in the second quadrant. Now, let us say this vector that you have over here. I'm just calling it as x i plus y j plus z k. Okay. Or let me call it as x 1, y 1, z 1. That's to be. Okay. I don't want to use the variables which we normally use for the axes or a general point. So let us call this as x 1, i cap, y 1, z, j cap and z 1, k cap. In other words, this point has a coordinate of x 1, y 1, z 1. Correct. Now, how is this x 1, y 1 and z 1 and this cos alpha cos beta cos gamma related to each other. Let's try to figure that out. Now, let us imagine that there is a plane. There is a plane which contains the OP vector and the x axis. So imagine that there's a plane sheet which is passing through OP and the x axis. Okay. So on that sheet, this angle is alpha. Okay. So can I say this length, let me call it as OM. Can I say this length is x 1? Yes or no? So x 1 divided by modulus of this vector will be cos of alpha. Okay. In other words, x 1 is nothing but modulus of OP vector times cos of alpha. Okay. Similarly, if I take a plane which passes through OP vector and the y axis, that means that plane contains OP vector and the y axis and this angle beta is on that plane. Okay. So let's say I call this point as n. So OM is your y 1. So this whole distance is your y 1. Correct. So by the same logic, I can say y 1 by modulus of or you can say magnitude of OP vector will be cos of beta. Okay. Similarly, if I take, if I take a plane which contains OZ and basically a drop-up perpendicular from P on to the Z axis and there's a plane containing OZ and OP, then this is an angle on the same plane, gamma. So this length is going to be gamma. Oh, sorry. This length is going to be Z 1. So can I say Z 1 by magnitude of OP, that would be cos gamma. Okay. In short, y 1 is magnitude of OP cos beta and Z 1 is magnitude of OP cos gamma. Correct. Shouldn't X 1. Okay. You got your mistake. Fine. Now OP vector itself, OP vector itself was X 1 i cap, y 1 j cap, Z 1 k cap. Right. Instead of X 1, if I use modulus of OP cos alpha, instead of y 1, I use modulus of OP cos beta and instead of Z 1, I use modulus of OP cos gamma. You would realize something very interesting here. You realize that OP vector is nothing but modulus of OP times cos alpha i cos beta j cos gamma k. Okay. In short, OP by modulus of OP or magnitude of OP is nothing but cos alpha i cos beta j cos gamma k. That is nothing but li plus mj plus nk. Now, when you write a vector divided by its magnitude, what does it actually represent? It represents a unit vector in the direction of that vector. Okay. So that li mj nk vector is nothing but it's a unit vector in the direction of the given vector. So direction cosines are nothing but please note this down. In fact, I'll write it down here for the benefit of creating a note. So please note this. Direction cosines. Direction cosines. We normally write it as lmn and we use an angular bracket. We don't use a round bracket because if I use a round bracket, you will get confused that whether it's a coordinate or it's a direction cosine. So direction cosines are nothing but they are the. They are the components of the unit vector in the direction of the given vector. Okay. So direction cosines lmn of let's say a vector, let me write a vector over here are nothing but they are the x, y, z components. So l is the x component, m is the y component and n is the z component of the unit vector, which is in the direction of that given vector. And remember, direction cosines, because of this property, follow this particular nature that their magnitude, anyways, unit vector magnitude, we all of you know that it's going to be a one. So because of that, they follow this important property. l squared per cent square per cent square will be equal to one. So as to say that the square of the direction cosines will add up to give you a one. Okay. So many times when we know at least alpha or beta gamma value can be found out. Of course, two values will come out from there, but it can be found out. Okay. So direction cosines are unique for a given vector. We cannot have more than one direction cosine. So there's only a one direction cosine possible because they can only be one unit vector in the direction of a vector. We cannot have multiple unit vectors for the same vector. So since there is a unique unit vector for a vector, hence there is a unique direction cosine. You can say a triad. So this is a triad. So a unique direction cosine triad will be there for a given vector. Okay. Is this fine? Any questions with respect to what's the direction cosine? Okay. Let's take a basic question on this particular property or this particular concept. Let me just go to the next slide. Okay. So let's start with this simple question. If a line makes alpha beta gamma with the coordinate axes, prove that sine square alpha plus sine square beta plus sine square gamma is equal to two. Now this is a very obvious question. Do let me know with the done once you're done. Should not take you more than 10, 15 seconds. Great. So if alpha beta gamma are the angles made by a line. In fact, line vector will talk about the difference later on. So cos square alpha cos square beta cos square gamma. We all know that it's going to be one. So we know this fact. Now as per the left hand side, we have sine square alpha, sine square beta, sine square gamma, sine square alpha can be written as one minus cos square. Sine square beta can be written as one minus cos square beta and sine square gamma can be written as one minus cos square gamma. So that gives us three minus cos square alpha cos square beta cos square gamma. Okay. So that means it is three minus one. That's nothing but two, which is equal to RHS hence proved. Is it fine? Any questions? Easy. Maybe it can come as a one marker for you all. Not that useful. Like to solve questions as in solving questions based on this topic. Okay. Let's take, no, no, no, no, not an important property that you would require it to solve some questions. This was a one standalone question on itself. Okay. Not a useful result that you may use in some different question. Nothing like that. Okay. Copy through origin. O is inclined at 30 degree and 45 degree to O X and O Y. That is your X axis and Y axis respectively. Find the angle at which it will be inclined to O Z or the Z axis. Okay. Okay. I'm going different types of answers. Okay. So basically you're trying to say that this condition is not possible. Okay. Okay. So, okay. So I think I have given out the response already. Let's try to look into the situation. So it has been given that alpha is 30 degrees. Okay. Beta is 45 degrees and gamma is not known to me right now. And we already know that cost square alpha cost square beta and cost square gamma would sum up to give you a one for sure. Right. So cost 30 degrees square cost 30 degrees root three by two root three by two square is three by four cost beta square will be one by four. Okay. Sorry. One by two and cost square gamma is unknown. Okay. So from here you get cost square gamma as half minus three by four, half minus three by four means minus one by four. Right. And this is not possible because square of a number here is a negative answer, which is not possible. Okay. So such a situation is not going to arise where you are going to see a line which makes 30 degree with the X axis 45 degree with the Y axis. So such situation is not going to arise because this is not possible. Okay. This is not possible. So no such situation will arise. What if we use the property we got in the previous question. Okay. Yeah. So what do you get from there? A different result. Oh, even that is coming out to be wrong. Right. So yeah. So it's not going to work out. Right. All right. So having done the concept of direction cosines will come back again to this concept. But before that we will try to discuss quickly the concept of direction ratios. Okay. Also called direction numbers. So what are direction ratios? So direction ratios. Okay. Okay. Is the direction ratio. Or is one of the direction ratios. Let me write it down. Is one of. The direction ratios. Off a line. If. The ratio of. A to L is same as ratio of B to M is same as the ratio of C to where L M N are the direction cosines. Where L M N are the direction cosines. Where L M N. Are the direction cosines. So direction ratios are simply three numbers which are in the same proportionality constant. As the direction cosines are. Okay. So direction ratios are much more simpler figures. And they don't have to satisfy the fact that else. They don't have to satisfy anything like, you know, the direction cosines were satisfying that is L squared percent square percent square equal to one. That restriction is not put on direction ratios. So direction ratios. We will see in some time. They are nothing but they are just corresponding to any vector which is collinear to that given vector whose direction cosines are L M N. Okay. Let me explain this. See, let us say there are, there's a vector. There's a vector OP. Okay. And this vector has direction cosines of L M N. What does this mean? It means it shows you a unit vector. Let me just draw a small unit vector. Okay. This unit vector OP cap. This unit vector is your li plus mj plus nk. Correct. Now direction ratios are nothing but three numbers which are proportional to the direction cosines. That means you can say from this, let's say I call this as Lambda. That means a is Lambda L B is Lambda M and C is Lambda M. Correct. So if you create a vector like this AI plus BJ plus CK, it becomes something like Lambda times li mj nk. Correct. Yes, sir. In other words, this is Lambda times a unit vector in the direction of OP. So which basically gives you a vector which is collinear to vector OP. Okay. So direction ratios are nothing but components. Components of any vector which is collinear to OP. Okay. And you would realize that later on that we will use direction ratios more commonly vis-a-vis direction cosines. So direction ratios are more useful for us vis-a-vis direction cosines. Okay. So please note this down that if I take any vector, let me make it in gray color. Okay. So if I take any vector in the direction of a collinear to OP, I should not say in the direction collinear collinear means it could be in the reverse direction also. So either this way or this way doesn't make any difference. So this would be your vector made from the direction ratios. Okay. So no doubt direction ratios of a vector direction ratios of a vector is or are or are the components of any vector collinear to the given vector. Any vector collinear to the given vector. Yeah. Collinear to the given vector. Okay. So there can be infinitely many they can be infinitely many vectors which I can write which is collinear to a given vector OP. Okay. So direction ratios are also infinitely many in number unlike direction cosines which is only one. Please note there can only be one direction cosine for a vector because they can only be one unit vector in the direction of that given vector. Okay. So the components of that unit vector are unique for a given vector. So direction cosines can only be one for a vector but direction ratios could be any set of numbers which are proportional to the direction cosines. In fact, it could be the components of any vector which is collinear to that given vector. So since we can make millions and trillions and zillions of you know vectors which are collinear to a given vector in the because of that we will be having infinitely many direction ratios possible. That is why I wrote this fact here. ABC is one of the direction ratios because they can be seven direction ratios possible. Okay. So any such numbers which are proportional to the direction cosines will become the direction ratios. I'll give a simple example. Let us say that in cosines for a given vector is one by root three minus one by root three one by root three. Please note I cannot write any three numbers as a direction cosines. They must satisfy the fact that L squared plus M squared plus N squared. So let's say this is your L. This is your M. This is your N. L squared plus M squared plus N squared should add up to one which is happening in this case. Okay. So one of the direction ratios which I can suggest could be something like one minus one one. It could be like minus one one minus one. It could be two minus two two. It could be minus two two two. It could be hundred minus hundred hundred. Okay. So there's no depth of direction ratios. You can keep writing infinitely many direction ratios for the same vector. Okay. We will come to the lines later on for a line. Yes. There are two direction cosines, but for a vector, there's only one direction ratio cosines, but they can be infinitely many direction ratios. Fine. So now let us talk about, let us talk about if the direction cosines are given, how do we find direction ratios? And if the direction ratios are given, how do we find direction cosines? So that activity is very simple. If direction ratios are given. Okay. Then how to find direction cosines. So if let's say L M N are given to you. Okay. So this is given to you. So all you need to do, take a suitable proportionality constant, multiply it with L M N. That will give you, that will give you one of the direction cosines, ratios, one of the direction ratios. Okay. If direction ratios are given to you, then how do we find? Sorry. I think I should say direction cosines are given to you. Yeah. Now if the direction ratios are given to you. Okay. Then how do you find the direction cosines? Let us look into that. So let's say somebody has given you ABC as the, as the direction ratios. Okay. So what I'm going to do is, in order to find direction cosines, I will use the fact that A by L is equal to B by M. And let's say I call it as a lambda. So L is A by lambda. M is B by lambda. And N is C by lambda. Okay. And use the property with the direction cosines satisfy. L squared plus M squared plus N squared equal to one. So a square by lambda squared, B squared by lambda squared, C squared by lambda squared equal to one, which means lambda squared is A squared plus B squared plus C squared. So lambda could be plus minus under root of A squared plus B squared plus C squared. Okay. Now all of you please pay attention. So having found the value of lambda, you can have A by, I mean, let's say plus minus under root of A squared plus B squared plus C squared. Again, plus minus B by under root of A squared plus B squared plus C squared. Okay. And plus minus C by under root of A squared plus B squared plus C squared. Is it fine? Now, if the direction ratios are given, you will end up getting two answers for the direction cosines. Why? Because direction ratios, they don't actually give you the, they don't actually give you a vector in the direction of the vector always. They could give you a vector opposite to the direction of that given vector. Isn't it? So if I give, if I mentioned the fact that let's say a vector p is collinear to vector q, right? Does it make it evident that p is in the direction of q? No, right? It could be in the direction. It could be opposite in the direction also. They are collinear. Collinear means do not, collinear doesn't mean having the same direction. So collinear means they have the same support or parallel supports. Okay, direction can be opposite. Now many people ask me here, sir, you told direction cosines is unique. How come I am getting two possible answers? By the way, this is only two possible answers. Either you take with a plus sign or you take with a minus sign. Okay. So if you take a plus, plus, plus, that is one set of direction cosines. If you take a minus, minus, minus, that is another set of direction cosines. So how come two answers are coming? One answer should be coming. So two answers are coming because if somebody gives me a vector, collinear to a vector, then I will not be able to accurately find out whether the direction is the same as that of the given vector or opposite to that. And hence two possibilities of the actual vectors can be possible. And because two possibilities are there, two unit vectors will be possible. And if two unit vectors will be possible, two set of direction cosines will be possible. Are you getting my point? What I'm trying to say? So here, two direction cosines are obtained. Okay. Or are possible. Two direction cosines are possible. Is it fine? Any questions, any concerns anybody has? Clear everyone? Okay. So let me ask a simple question to all of you. If the direction ratio is given, even the vector it corresponds to can have two possible directions. No, no, no, no. Sorry, if direction ratio is given, yes, the vector that it corresponds to can have two possible directions. Hence two possible direction cosines. No. Why did I write two direction cosines here? Because if I give you a vector which is made by using the direction ratios component, the actual vector could be either in the direction of it or opposite to the direction of it. And hence, if it is in the direction, it will have a direction cosine, one particular direction cosine. And if it is opposite in the direction, it will have another possible direction cosine. So if direction ratios are given, two direction cosines will come up because of that. Got it? Okay. All right. So I have a question for all of you here. Let us say, let us say these are the, these are two vectors. Okay. These are two vectors. I have purposely made them intersect here. This vector has direction cosine of L1, M1, L1. Okay. And this vector has direction cosine of L2, M2, M2. Okay. I would like to know what is the direction cosine of a vector which is bisecting the angle like this. So the blue one and the green one. So the blue vector is dividing this angle. And let's say this is my green vector, which is dividing this angle. You can use, you can use theta two in your answer. You can use theta in your answer. Two vectors are always coplanar. So there's a vector which is coplanar with the two and bisecting the angle. Let's say this whole angle was theta. This whole angle was theta. So what is the direction cosines of this vector and what is the direction cosines of this vector? You can use L1, M1, M1, L2, M2, M2 along with theta in your answer. Yeah. But we are trying to find out nothing related to the vector. They're just finding out the direction cosine related to the bisector. Yes, they are bisectors, but we're trying to find out the... No, I don't think so. Locust will help. In fact, you can use your vector skills. Vector is going to help you out. See, just the understanding that direction cosines are nothing but they are components of a unit vector in the direction of a vector. Okay. That is more than enough to solve this question. Okay. So let's see a small segment over here of this entire scenario. Okay. What I've done here is I have cut a unit vector along the given two vectors. Okay. So let's say this point is O. Okay. And what I've done, I have cut a unit vector. So this length, let me call it as PQ. Okay. So OP is a unit vector along the first line. Let me call it as L1. Okay. So OP is a unit vector along L1. OQ is the unit vector along L2. And this was the angle between them. Let's say theta. Okay. Now, all of you please pay attention. Can I say, can I say the bisector would be meeting the midpoint of the line, joining these two. Okay. Let's say I join the endpoints, endpoints of P and Q. The bisector vector is going to meet the midpoint of this. Let's say I call this as OB. So OB will be the midpoint of P and Q. Okay. And since this is a unit vector, I can use the direction cosines to frame a unit vector like this. So L1, I, M1, J, N1, K. This is your OP vector. L2, I, M2, J, plus N2, K. Okay. This is your OQ vector. Okay. Now, let me just ask you a very simple question. What would be the vector corresponding to OB? What will be the vector corresponding to OB? Now, of course, your response will be. So P, Q position vectors are these given vectors. Right. So this is your OP vector and this is your OQ vector. So these are the position vectors of the point P and Q. Correct. So B vector will be nothing but OP plus OQ by 2. In other words, it will be L1 plus L2 by 2 ICAP, M1 plus M2 by 2 JCAP, and N1 plus N2 by 2 KCAP. But is this my direction cosines of the bisector vectors? Answer with a yes or a no. Can I say this number, this number, this number? They form the direction cosines of the given vector. The answer to that is no, because this is not a unit vector. Please note, in order to become direction cosines, they must be components of a unit vector. Please note, if this length is 1, this length cannot be 1. This length will be smaller than 1, isn't it? Yes or no? So OB length is what? OB length is nothing but it is cos of theta by 2. Are you getting my point? So to get a unit vector, you need to divide OB by its magnitude. Then only you'll get OB, I mean a unit vector along that OB vector. So I need what I'm looking for. I'm looking for the direction cosines of OB. So I need to know what is a unit vector along OB. Once I know the unit vector along OB, then only its components will be called the direction cosines. But OB right now in the present shape, OB is not a unit vector. OB is a vector smaller than unity. So what I have to do, I have to divide by its magnitude, which is in this case cos theta by 2, to make it a unit vector. So if I do that, I will end up getting something like this. M1 plus M2 by 2 cos theta by 2j and N1 plus N2 by 2 cos theta by 2 k cap. Now I can claim that the direction cosines of the internal angle bisector. Let me call it as internal angle bisector. Internal angle bisector vector. That will be L1 plus L2 by 2 cos theta by 2. Sorry for that. M1 plus M2 by 2 cos theta by 2 and N1 plus N2 by 2 cos theta by 2. Is it fine? Can you explain why its magnitude is cos theta by 2? That's a good question. There is a triangle like this. This angle is theta by 2. This is length 1. So what is the base of this triangle? Let's say this is OB and Q. So what is this length? If I ask you, what will your answer be? Harsha. OB by OQ is cos of this angle. OB by OQ is cos of this angle. So OB is cos theta by 2. Got it now? Is it fine? Can we also write down the direction cos of the external angle bisector which I have shown with the green? It is not about origin. I have just written O doesn't mean it is an origin. I can place my reference point anywhere. If I am making a vector, first of all that vector is free to translate anywhere in space. So I picked up a vector over here. I picked up a vector over here and I brought them to origin. What is wrong in that? Are you getting my point? So vectors are free to translate themselves parallel to themselves anywhere in space. So what is wrong in this fact that I make their tails to be at origin? I make their initial points to be at the origin. Nothing will go wrong. So that is what I did. Yes. How would you find out the direction cos of this green vector? Okay. A couple of things that you can do here without redoing the whole process. You can consider that there is a vector in the opposite direction whose direction cos is negative L2, negative M2, negative N2. Because the unit vector in the direction of L2 is in the direction of line L2 is L2M2N2. Then the unit vector in the direction opposite to that line would be negative L2, negative M2, negative N2. And if this whole angle is theta, this angle is pi minus theta. So each one will be how much pi by 2 minus theta by 2. Okay. So all you need to do is in order to know, in order to know the direction cosines of the external angle bisector, external angle bisector. I'll write for that. Two things you have to do. One change L2 to minus L2 and change your theta with pi by, change your theta by 2 with pi by 2 minus theta by 2. Getting my point. So you'll end up getting something like this. L1 minus L2 by 2 cos pi by 2 minus theta by 2, comma M1 minus M2 by 2 cos pi by 2 minus theta by 2, and N1 minus N2 by 2 cos pi by 2 minus theta by 2. Sorry for missing out at minus in between. Okay. So in short, it will become by the way, let me put a barricade over here so that you don't get confused. So this will become L1 minus L2 by 2 sin theta by 2. M1 minus M2 by 2 sin theta by 2 and N1 minus N2 by 2 sin theta by 2. This is your answer to the second one. So these two results which I have given you, they have come as direct questions in several combative exams. Is it fine? Any questions, any concerns, anybody has to let me know. All right, so should we move on to the next question? Okay, so we'll take up the next question quickly. Find the, okay, actually this concept requires concept of angle between the two vectors. So let me just discuss that first. How do we find the angle between two vectors whose direction cosines or direction ratios are known? So let us say I have two vectors. I'm making them co-initial, doesn't make a difference. Let's say this vector, I know the direction cosines are L1, M1, N1. And this vector direction cosines are L2, M2, N2. Okay. How do I find the angle between them? Let's say alpha. Very simple. See, if you know the direction cosines, you indirectly know the unit vectors along those vectors. So whatever is the angle between the unit vectors, the same will be the angle between the given vectors also, isn't it? Okay. So in short, what do we know? We know the unit vectors along OP and we know the unit vectors along OQ. Yes or no? So the angle between them will be same as the angle between these two unit vectors. Okay. So cos of alpha would be nothing but OP.OQ by modulus of OP, modulus of OQ. OP.OQ unit vectors will give you nothing but L1, L2, M1, M2, N1, N2. And these are all anyways one one each. So that will be one all itself. Now please note that if the question setter is asking you for the angle between the two vectors, please do not take any kind of a modulus over here. Are you getting my point? There is nothing like acute angle between the vectors or obtuse angle between the vectors. The angle between the vectors is only one angle. That is the shortest angle between them when they are when they are co-initial. That means they are all emanating from the same point. So please do not try to take a modulus here that we do in case of a 3D line. So people who have already done 3D line, don't get confused between the two fact. So this quantity, let it be with whatever algebraic sign it has, whether it is positive, negative, fine. If it is positive, then the angle is automatically acute. If it is negative, then the angle is obtuse. So don't take a modulus and try to make it acute. Are you getting my point? So in case of two vectors, the angle between them is just given by cos inverse L1, L2, M1, M2, N1, N2. Sorry, I missed out that question that you asked. You had some question if the third vector was not in plane with the second vector means I didn't get you. If it is not in the plane, then how come it becomes a bisector? All right. Now, all of you please pay attention. If the direction cosines are given, then this is the answer. But if the direction ratios are given, then what do you do? So in the case of a direction ratios, please note that there could be two possible answers. As I already told you, direction ratios are nothing but components of any vector collinear to that vector. So if you are using direction cosines, then one answer will come out. But if you're using direction ratios, then please note because of that, there will be two answers coming up. Now, how are those two answers coming up? Very simple. See, let's say these are your two original vectors. Okay. Now, if somebody gives me that the direction ratios of this vector is A1, B1, C1. Okay. Now, whether A1, I, B1, I, sorry, B1, J and C1, K is in the direction of this vector or opposite to the direction of this vector, I cannot say. Okay. Are you getting my point what I'm trying to say? Direction ratios, in direction ratios, we don't get to know what is the actual direction of that vector. Right. So if you're using, let's say I use any two vectors, let's say I call it as OP and OQ. Okay. So I'll call as OP1 to be any vector in the direction of this or opposite to the direction of it. And I call OQ1 to be any vector in the direction of OQ or opposite to the direction of OQ. Sorry, I should write direction ratios only. So this could be your possibility. Correct. So let's say this is angle alpha, then please note that cos alpha. Okay. That would be A1, A2, B1. Let me write down the formula. OP1 dot OQ1 by modulus of OP1, modulus of OQ1. Okay. Now, please note. It could also result into the obtuse angle. Right. So it could also be the cos of pi minus alpha between the two angles. Right. So it's very difficult to guess. In other words, what I'm trying to say is that this expression that you get. Okay. This expression could be cosine of the angle between them or cosine of the supplementary of the angle between them. You will not be able to know that. Is it clear? So in other words, cos of alpha is plus minus whatever quantity that you will obtain from it. Hence, knowing the angle between the two vectors when their direction ratios are given will lead to multiple answers. That is why such questions are made in lines, not in vectors. So there would be, I don't think so. It's a rare situation to ask a question for the angle between two vectors when the direction ratios are given. These type of questions are mostly asked in 3D lines, which I will come to in some time, not to worry about it. But direction, cosines, this is the only formula that will give you the angle between the two given vectors. Is it clear? Any questions, any concerns? Okay. Now maybe I'll go to the question that I had started with, but I could not do it. If you want to copy anything down, please do so. Now let's take that question, which I was about to give, but I did not give. Find the angle. Let's make it. Okay. Let it be lines, no issues. Find the angle between the lines whose direction cosines are satisfying these two conditions. You can treat this to be vectors as of now. I don't want to use lines without talking about it. So let it be vectors. Find the angle between the vectors whose direction cosines satisfy these two equations. Note that two vectors, two set of direction cosines will be there and both set of direction cosines satisfy this condition. Give me a response on the chat box. Is the question clear to everybody? What I'm trying to say. See, the question is, let's say there are two vectors. One is this white vector. Another is let's say this yellow vector. Okay. Let's say this white vector has direction. Cosines has L1, M1, N1. Okay. And this yellow vector has the direction cosines of L2, M2, N2. Now both these direction cosines satisfy these two equations. That means L1, M1, N1, they will add up to zero. So will L2, M2, N2 will also add up to give you zero. Okay. And two L1 square, two M1 square, minus N1 square will add up to give zero. And so will two L2 square plus two M2 square minus N2 square will add up to give you zero. So they're finally asking you what is the angle between the two given vectors? What is this angle? That is what they're asking you. Okay. I'm getting all different types of answers. Some of you are saying zero degrees. Some of you are saying 180 degrees. Some of you are saying 90 degrees. Okay. Any further more answers? Okay. Kinshuk has a very different answer. Fine. Anybody else? Anybody else? Okay. Let's look into, let's look into the two given equations. I would start with the last equation that is given to me. That is two L square. Okay. Let me take a two common. Okay. Now we already know that any set of direction cosines will always satisfy this criteria. Right. That means L square plus M square is one minus N square. Okay. So if I put it over here, I get two one minus N square minus N square equal to zero. That means two is equal to three N square. That means N is plus minus under root of two by three. So two values of N already have come out. Okay. So two values of N have already come out. Let me call N1 as root two by three and N2 as negative root two by three. Okay. Now, again, let us write, come back to this equation once again. So two L square plus M square is equal to N square. And what is N from the first equation? From the first equation, from this equation, N is negative of L plus M. Correct. So can I not use that over here? So let's use. Let us use negative of L plus M the whole square over here. Okay. So two times L square plus M square is this, that means twice of L square plus M square is equal to L plus M the whole square. Let us simplify this. So this will be L square M square plus two L M. I think one set of L square M square will get cancelled off. Oh, wonderful. So this gives us L minus M the whole square equal to zero. So L and M values are actually same. Okay. So one fact that I came to know here is that L and M value are actually same. Now, again, going back to L plus M plus N is zero. If M is made to be L, so that means two L plus N. Now, let me use one of the values root two by three over here first. So L value from here will come out to be minus one by root six. That means M value. Let me call it M one. That is M one value will also come out to be this and N one. I have already fixed root two by three. So this is one set of values. And if I take the second situation where my N value was negative root of two by three, then from here L one is one by root six. Sorry. M two will also be one by root six. And N two as I already took negative of this. Right. So here comes my three set of value two set of direction cosines. Okay. So once I know them, I already know alpha is cos inverse of L one L two M one M two N one N two. Right. So that is going to give me L one L two is going to give me minus one by six. This is also going to give me minus one by six. This is going to give me minus of two by three. Yes. I hope I'm not missing out anything. So this is cos inverse of minus one by three. Yeah. Hope everything is fine. Minus one by three. Yeah. And this is also minus two by three. Right. So that gives me cos inverse of minus one. So that's a pie. Is it fine? Any, any questions? Anything that I'm missing out here, please do highlight. And if you have any questions, please do get it clarified. All good. No issues. No worries. All right. So now with this, we are, you know, almost done with the concept of direction cosines and ratios. Now we are slowly going to move towards the similar concept applied to 3d lines. Okay. So first of all, I would like to know from you. What do you think is the difference between a 3d line and a 3d vector. So how a vector in three dimensions and how a line in three dimensions differ or resemble each other. Let's talk about that first. So I want you to give me some pointers. How is a vector in three dimension and a line in three dimensions different from each other? Let's make a table if you want to. Let me make a table. Let's see this and let's make a small table. Yeah. So at one, we have a 3d vector and 3d line. Yeah. So what are the first difference that you all, you know, have to say? Yes. In a line, there is no direction. Okay. So it has direction cosines to direction cosines, but there is no direction assigned to it. Okay. So this has a direction. Okay. But there is no direction assigned to a line. Of course, a vector is has a finite magnitude. Okay. So it has a magnitude. Okay. This guy doesn't have magnitude because it's a infinitely extending geometrical figure. Okay. Main differences. Main differences. A vector is free to translate parallel to itself, translate parallel to itself in space. But a line is a fixed in space. Okay. So a vector can move parallel to itself anywhere in space. Okay. There's no restriction on its movement unless until. Okay. It changes its direction or it's changes its magnitude. So the same length, the same direction, if it follows, it can be placed anywhere in space. But a line. If a line is there like this, and if I make a line, another line exactly parallel to it, they would be considered to be two separate lines. Okay. So these are the major differences that you will find in case of a vector and a 3D line. But having said that there are a lot of concepts which a 3D line will borrow from vectors. Okay. So it will be, it will be following many things which, you know, a vector also follows. The first thing that it follows is the concept of direction cosines itself that we were talking about. And of course the action ratios. So in case of a line, we have no direction. Correct. So a line has two possible direction cosines because there could be a unit vector in this direction. Sorry. Let me make it in my. So it could have a unit vector in this direction. Or it could have a unit vector in this direction. Let me assign the direction to it. I have not written any direction next to it. Okay. So there are two possible. Let's say these are all unit vectors. Okay. So a direct, a line can have two possible set of direction cosines, but those two direction cosines will be exactly opposite in sign because these two vectors are exactly opposite in direction. Okay. So a line has got a 3D line has has two direction ratios. Sorry. Direction cosines. Okay. And they are opposite in sign. That means if this is one by root three minus one by root three, one by root three, I'm just taking an example for a purpose. Let's say this is then the other one could be negative of the same thing that is minus one by root three plus one by root three. So each of them will become a negative. Negative here doesn't mean all of them will be negative. No. This means whatever was LMN, the other direction cosine will have a negative sign or opposite sign to what the other has. Is it fine? So a line has two direction cosines unlike a vector which has only one direction cosine because you can assign two unit vectors which have the same support as that given line. Okay. Is it fine? And the concept of direction ratios doesn't change. So a line can have infinitely many direction ratios. Okay. As long as the direction ratios, I'm sorry, as long as the direction ratios are proportional to any of these two sets, they will be considered to be one of the direction ratios of that given line. So no, no change in the direction ratio concept. That is why direction ratios are more, you can say, you know, utilized. They don't have a lot of restrictions like direction cosines have. Okay. So they are mostly used in our dealing with lines. In fact, we will see it in some time. Now, the first and the most important thing that comes in our mind is, okay, so I know the direction is the ratio of a line or cosines of a line. Can this help me to get the equation of a line? Okay. So the next thing that we are basically going to talk about is in terms of equations. Okay. So let us go to the next slide because I think the space is not good enough for me to, okay. So let us go and find out equations of equations of 3D lines of a 3D line. So I'll be first talking about the vector form. Okay. Vector form and Cartesian form. They are the two forms that we'll be discussing more frequently, but I'll be coming from vector point of view. Okay. By the way, when you want the equation of a unique line, is it just enough to know the direction cosines or ratios of that line? Or you need some extra information also. Recall your 2D lines. When you are finding the equation of the 2D lines, how many set of information was always given to you? Either you know two points or you knew the slope and one of the point or you knew the X and the Y intercept or you knew the distance of the perpendicular from the origin onto the line and the angle made by the perpendicular. So these are the different forms of 2D lines which you did in class 11. So in all these we basically required two set of information. Okay. So I'll be also talking about the equation of 3D lines when we have been provided with two set of information. The first set of information that I'll be providing here is I know the direction cosines or the direction ratios and one point on the line is given to me. Okay. So let's say the question center has provided me with one point. Okay. Let's say that point is X1, Y1, Z1. You can say one point or one position vector. Both of them are in the same thing. So if the position vector on that given line has been given to you and you have been also provided with the direction ratios or cosines of that line, how do we get the equation of a line? So we will talk about vector form first of all. Okay. Let's say vector form. Then I will talk about Cartesian form a little later on. So let's say this is a line and on this I know a position vector. Let me call this point to be point A. Okay. So point A has position vector. So what's the position vector? Position vector is nothing but a vector connecting origin to that point. You all know position vector from our vectors chapter. Okay. And I also know the direction. Let's say this is my direction cosine or somebody has mentioned me or somebody has mentioned me the direction ratios, one of the direction ratios, A1, B1, C1. Okay. Then how do I get the equation of this line? So what is equation first of all? Equation is nothing but it's a relationship between the x, y and z coordinates of any point lying on that line. Please note it is equally valid to any curve existing. What's the equation of any curve? When somebody says what do you mean by equation? So if you say x plus y equal to five, what should I understand from it? What message does it convey to me that a line's equation is x plus y equal to five? Let's say a 2D line equation. What do you understand by it? It's nothing but it just says that any set of points whose x and the y coordinate add up to give you five, that will lie on that line. Okay. So it's a relation between the abscissa and the ordinate of any point that happens to lie on that line. Okay. In the same way, in order to get a equation of a 3D line, I need to strike a relation which connects x, y and z of any random point which I take. Let's say I take any random point p here. Okay. This point p has, let's say, position vector x i, y j, z k. Can I strike a relationship between x, y and z? Or can I strike a relationship between this entire vector itself with the given position vector a and the direction ratios or cosines? That is what is called the equation. Okay. So normally what do we do? When we write such, you know, generic point, we normally use the word r for it. Okay. So r is basically reserved in vector equations for representing any generic point lying on that curve. In this case, a line. Okay. So we write r for it. Okay. So r is, I'll just write it next to it. r is any generic or general position vector on that line. Okay. We don't write x i, y j, z k, but of course, internally, you know that r is actually nothing but a position vector made by x, y, z of any random point taken on that particular line. Okay. Now, can I say that this PA vector? Okay. PA vector or AP vector, whatever you want to call it. Okay. This PA vector is going to be collinear to a vector which is made form, or you can say it's a collinear to l 1 i m 1 j n 1 k. Correct. Because you know that the unit vector lying on this line, let me make it in yellow, a unit vector lying on this line is your l 1 i m 1 j n 1 k. Okay. So PA vector will be collinear to this vector, unit vector. Okay. So if I use this fact, can I not say that. Can I not say that r minus. Okay. Let's let's call this also as a simple expression, maybe let's say a. Okay. Can I say r minus a vector is collinear to is collinear to this vector. Okay. In other words, can I say r minus a is some lambda times m 1 minus take out m 2. Okay. And many times we give this also a very simple name. Let's say I call this this unit vector as B cap. Okay. So I can say r minus a is lambda times B cap. So r is equal to a plus lambda B cap. So please note that this becomes the vector equation. This becomes the vector equation of a line, which happens to pass through of your position vector a and has a direction cosine, which is given by your, which is given as components of this B cap vector. Okay. So if I have to write it in a, you know, expanded version, it is nothing but something like this. Plus lambda times the B cap vector. This is your B cap vector. Okay. So this, this equation is what we call as the vector form of the equation. Okay. So let me write it down. This is the vector parametric form. How many people say why a vector parametric form? Okay. Here also I can, you know, right vector parametric form because the parameter, if there is a parameter involved, let me raise this what required. Yeah. So here there is a parameter involved, which is lambda. What is this parameter actually doing? This parameter is actually a quantity which keeps changing. And because it keeps changing, you keep getting different, different hours. Right. So if this changes, this will also change. Yes or no. But as your lambda change, you will start getting different, different points which lie on the very same line. Okay. For example, if I put lambda as zero, then the R will correspond to that a point given to me in the, in the question itself that it is, it'll give, it'll become this point itself. So this lambda is like, you know, quantity which changes and as it changes, it starts giving you different points lying on that line and thereby giving you the equation actually. So many people say, sir, I say every point will have a different lambda. Yes. So lambda will be like the, you can say the adhara card number of every point lying on that line. Right. Just like we all have different adhara. Correct. So if I put one adhara number, I'll get a person corresponding to it, which is your R. So R is a person. Lambda is the adhara card number of that person. So if I put lambda as one, I get different R. So which is a point on that line. If I put lambda as 100, I get another R, which is another point lying on that line. So every point on that line or every R lying on that line will have a different, different, different, different lambda values for it. Okay. So that depends on the point. So lambda can only be found out when I know the point on that line. For example, for X1, Y1, Z1, lambda is zero. Are you getting my point? So there's some other point. If you give me a point, I can find the lambda corresponding to it, but it has to be provided. It should lie on that line. Is that fine? So what is A? A is nothing, but I have called this as your A vector, okay? Just to call it, just to keep it brief, that's it, okay. Is it fine? Any questions, any concerns here? Alright. Now one important thing that I mean, it's not going to be that important, but you should be aware of it. So if, let's say there is, this is my position vector A. This point is a position vector A. If I take an R over here and this let's say R1, let's say this R1 has got its lambda value as positive. Let's say lambda 1. Then if I take another point which is on the opposite side of A, let's say this is R2, then for this point your lambda value will be negative. Please note that. And for A itself, lambda value is 0. Are you getting my point? There is nothing like zoom out, Harsha. You're probably attending the class after a very long time. So there is no feature to zoom out. I can drag the screen here and there, but there is no feature to zoom out. So what I was trying to say is that if there is a point, let's say in one direction of A, let's say here, all the points in this direction will have positive lambdas. So all the point that lie on this direction, let us say, I'm just assuming, then if they have a positive lambdas, then all the point which lie on this direction, the lambda will be negative. So it cannot happen that lambda changes its value, positive, negative, positive, negative on the same side of A. So let's say you are going in one direction, the lambda will continue to be either positive, positive, positive, positive, positive for all those points or negative, negative if you go to the other direction. Just a good fact to know, but I've not seen its utility in any question yet. Yes. So what do you want to copy here, Harsha? Which part do you want me to stop at? Got it? Great. So can I do the same activity even if I was using the direction ratios? Yes, why not? So I can also say that PA or AP, whatever you want to call it, is collinear to, is collinear to your a1i, b1j and c1k. That means you can say r minus a vector, which was a given point on that line. This is some lambda times. You can change the parameter here. You can call t also. You need not always call the parameters to be lambda. You can call it as st, uv, whatever you want to call it. Okay. So we end up getting almost the same type of an expression for the equation here. So it will be something like this. Why did I write a2? Yeah, it's a1 only. So this is another, this is when you have been provided with the direction ratios. Okay. Is it fine? Now, many people say, sir, can I also do this fact that if PA is collinear to, let me write it down here. Please note, if PA vector is collinear to, let's say the unit vector, which is made of the direction cosines, can I also say that r minus whatever was your position vector, this cross product with l1i, m1j, n1k equal to a null vector. This can also be said. Yes or no? If two vectors are collinear, their cross product will be a null vector. Correct. In other words, I can say x1, I can say x, let me write it down, r minus x1i, y1j, z1k cross product with, this is a null vector. This type of equation is called the vector nonparametric form, vector nonparametric form, simply because there is no parameter involved here, unlike in the previous one where we had a lambda and a t, etc. But this is rarely going to be used. Okay. This is rarely used at least for board exams. It is not going to be used. Your teachers in the school are not going to teach you this fact. But for competitive exams, this may be given to you as one of the options. Okay. So please note this down as well. Please note this down as well. Why? There can be so many still vectors which are satisfying the cross product as 0. So all those vectors, all those position vectors, which satisfy this condition, there will be different, different points on the line. So when you say cross product is a 0, see if I say i cross i is a 0 vector, correct. 2i cross i is also a 0 vector. 5i cross i will also be a 0 vector. So there are so many, there are so many such vectors which satisfy the same criteria. So all such vectors whose cross product with this is giving you a null vector, those all will be in my answer. And there can be so many such vectors. All right. So having seen this vector form, let's go back to, let's go to the next slide and do a quick final this thing. So the point here is that, please remember this, the point here is that if somebody provides you with a position vector, let's say a, and mentions the direction ratio or cosines. I'm calling, I'm giving them a single term, let's say b vector. So you know a vector which is either in this direction or this direction, doesn't matter. Line doesn't care about the direction of b. So as long as this b vector is collinear to the line, the equation of, the equation of such a line is given by this. This is called the vector parametric form. So this is in a parametric form. Or this, which is actually a less known equation, this is called the non-parametric form. Non-parametric form. Is it fine? Any questions? R is any position vector, just like we know, just like when we write a equation of a curve, we put x, y and all. So x, y is any point on that curve. Same way, R is any generic position vector on that line. So any generic position vector on that line should satisfy this and this equations. One is called the parametric form, other is called the non-parametric form. Fine. Any questions? All right. So this was related to the vector equation. Now let us talk about the Cartesian form of the equation. Now Cartesian form will be derived from your vector form only. That is why you should never shy away from using your vector concepts. Many people I have seen, they have some kind of a repulsion towards vector method. Vector is there for your help. Vector is not there to make your life difficult. In fact, vector expressions are much more simpler as compared to coordinate geometry expressions. In coordinate geometry, you need to mention your x, y, z separately. A vector doesn't require that. So vector is always a simpler version of the coordinate geometry forms. So never shy away from using vector forms in order to get to a result if at all it is helping you. So what is Cartesian form of the equation of a 3D line? So let us take the same example where we had a point x1, y1, z1 and where somebody had mentioned me the direction cosines of this line or direction ratios of this line. So I will use my vector form only, the vector form as you saw in the previous slide. This was the vector form. Let us use direction cosines as of now, lambda times l1, i, m1, j, n1, k. Now what I am going to do is, as I already told you that r, the position vector, it is nothing but r is nothing but it is an expression for xi, y, j, z, k. x, y, z being any generic point. So x, y, z being any generic point. It can be anywhere on the line. So I am just replacing this r with xi, y, j, z, k. Let us copy the right hand side as it is. What is wrong with me? Why I am writing m2? Sorry. Let us take this to the other side. You have something like x minus x1, i, y minus y1, j, z minus z1, k equal to lambda times l1, i, m1, j, n1, k. Now if these two vectors are equal to each other and we know that i, j, k are linearly independent vectors, it means x minus x1 is lambda l, y minus y1 is lambda, sorry, lambda l1. y minus y1 is lambda m1 and z minus z1 is lambda n1. In other words, x minus x1 by l1 equal to y minus y1 by m1 is equal to z minus z1 by n1 are equal to lambda each but we will not write that lambda and we will stop over here and we will claim that this becomes your Cartesian form of the equation of a line. So unlike in 2D vectors where there was 1 equal to, in case of a 3D vector, you have 2 equal tos. Are you getting my point? So please make a note of this. Now, if you analyze this equation carefully, let us say this is equal to lambda. Then what does this lambda actually represent? This lambda actually represents the distance between the generic point x, y, z and the given point x1, y1, z1. How? See, very simple. See here. Let us say this is your x1, y1, z1 and this is your x, y, z. Can I say the vector? Let us say I connect this to, let us say I make a vector like this. This vector is x minus x1i. Let me call it as p and let me call it as q. So pq vector is x minus x1i, y minus y1j and z minus z1k. This pq vector is nothing but the magnitude of pq into the unit vector along pq. As per our basic understanding of vectors, any vector can be generated by the product of its modulus with the unit vector along it. In other words, I am trying to say the length pq multiplied with l1i m1j n1k. So this is your given vector, z minus z1k. Now again, if you compare, you realize x minus x1 is mod pq into l1. So this divided by l1 is mod pq. So this term that you see over here, which I am circling out, this is actually mod pq. Similarly, this will also be y minus y1 by m1 and z minus z1 by n1. So this lambda, which I had skipped writing, that lambda actually has a meaning. That lambda actually shows, in fact, if you take a mod of this, it actually shows the distance between the given point and the generic point that you are taking on that line. That is why many a times we call this as the distance form also and many times we call it as a parametric form like I already told you in case of vectors also. So this is what we start calling this line as also in addition to calling it as a Cartesian form. So this is one of the several Cartesian forms that you can see. So this is also called the distance form or the parametric form. Okay, so please note this down. Another version of the same equation is something like this. We call x as l1 lambda plus x1. Let me write it like this. x1 plus l1 lambda and y as y1 plus m1 lambda or lambda m1 and z as z1 plus lambda n1. Okay, this is another way of writing the same thing. That means you can always assume a point on this line to be like this. So this equation has a lot of importance because it helps you to assume a point on that line provided you should know one of the points x1, y1, z1 and you should be aware of the direction cosines or ratios. Ratios also will work. We will see in some time. Is this fine? Any questions? So if I have to assume a point on this line, let's say if I have to assume a point on this line, I can assume this point to be x1 plus some lambda l1 comma y1 plus lambda m1 comma z1 plus lambda n1. Of course, lambda value has to be figured out because that is the Aadhaar card number for this unknown point. Let's say call it as r. Are you getting my point? So model of the story is that we had derived the Cartesian form from the vector form itself. Vector form itself is acting like the core principle on which we have got the Cartesian form. And in this Cartesian form, when you are using the direction cosines here, this lambda, which is the proportionality constant here, which we don't write normally, but I'm just giving you some inside information about that lambda. That lambda would become the distance between the generic point x, y, z that you have used in this equation and the given point x1, y1, z1, which was provided to you in the equation. This entire equation can be also written like this and this is a form which facilitates us to choose a point on the curve. Is it okay? Any questions? Any concerns so far? Now, how would this equation change if I had used direction ratios in place of direction cosines? Not much, but of course, there would be a small change, which I'll be telling you right now. If you want to use your direction ratios instead of direction cosines, then let me just write that same equation once again. So in place of lambda1, sorry, in place of l1, you can use a1 divided by lambda, whatever you want to call it, divided by some constant, let's not call lambda, let's call it as t, maybe. Similarly, y minus y1 and m1 can be written as b1 by t and z minus z1 will be called as c1 by t. So one by t, one by t, you can just score off from the three sides and once you score off, you'll end up getting something like this. Now, you must be thinking, sir, there is no difference, it's just that you have replaced l1, m1, n1 with a1, b1, c1. There is no visible difference, but there is an internal difference which is going to happen. In this case, the proportionality constant, let's say I call it as some value, maybe let's call it as beta or something. This beta will not be the distance, this is not the distance between the generic point and the given point. So this is the only difference. So please note the difference here. Please note the difference. So this is the only difference between this equation and this equation. So you can see that you can easily replace your direction or signs with direction ratios. Nothing will happen to the equation, but internally what will happen, this ratio where earlier it used to represent the distance between the given point x1, y1, z1 and the generic point that you are involving in that equation, that will not be present while you are expressing it like this. Is it clear? Any questions here? But anyways, we don't show lambda and beta in the while we are writing the equation. So lambda and beta are not written, they are kept anonymous. But internally you know that this equation when you are using the direction cosines, this ratio that you are trying to show over here, these ratios are these ratios are actually equivalent to the distance between the given point and the generic point. Here it doesn't represent the distance. Is it clear? All right. So I think we have discussed enough related to it. Now time has come that we start talking about equations. Okay. So before I start talking about equations, one small thing I would like to discuss, I'll go back to this angle between the two lines, angle between two vectors. In fact, I have not discussed that with you. Yeah. So when I was discussing the concept of angle between the two vectors, there is a very important thing that I missed out that that concept was condition for parallelism and condition for perpendicularity. So when are two vectors perpendicular and when are two vectors parallel to each other, that information can also be obtained from their direction cosines and ratios. Okay. So if two vectors are parallel to each other, please note that their direction cosines will exactly match. L1, L2 will be equal, M1, M2 will be equal and N1, N2 will be equal. That means to say that L1 by L2 equal to M1 by M2 is equal to N1 by N2 and each one of them will be equal to 1. Okay. If they have provided you with the direction ratios, then we can say A1 by A2, sorry for writing B1 here, A1 by A2 will be equal to B1 by B2 will be equal to C1 by C2. Okay. So please remember these two things when they are parallel. Okay. When they are perpendicular, when two vectors are perpendicular. So what is the condition for perpendicularity? Condition for perpendicularity. So as you can see in this formula, cos 90 degree will become a zero, right? So condition for perpendicularity will be L1, L2, M1, M2 and N1, N2 will add up to give you zero and same will be true even if you are using direction ratios. So if your alpha is 90 degree, whether you use in this formula or in this formula, this is anyways going to become a zero, right? So your A1, A2, B1, B2, C1, C2, that is going to become a zero in case of direction ratios provided to you. So this also please note it down. So these are the conditions that you should keep in mind because they are very commonly used in solving problems also. Okay. So the moral of the story here is that if you use the same principle for your lines as well, they are going to work fine. So in lines, I will just tell you what are the minor differences. Of course, many of the things will be same. What are the minor differences when you take the same, when you borrow the same concept for a line? Oh yes, yes, why not? By the way, many people ask me, sir, can I write this as a one? No. It would be some value but need not be one. Okay. Some value need not be one. Need not be one. Okay. But here it will be one only because the two vectors should have the same unit vector if they are parallel. I'm not saying collinear, they're parallel. Parallel means same direction. Done? Shadda? Done. Okay. Now how do these conditions change for a line? Let's talk about that. Okay. Angle between two lines. Okay. So when two lines are provided to you and they have given you their direction, ratios or cosines, then how do I work with them? So let's say L1 is your white line. L2 is your yellow line. Now, before I start talking about angle between the lines, do the lines have to intersect for this concept to hold true? Can I find the angle between lines only if they intersect? What do you think? Do they have to intersect? Or even if they don't intersect, I can still talk about angle between them. Right. So please understand this fact. See, in vectors, you can make the two vectors come in the same plane. Okay. And you can just talk about their angle between them. But lines. Okay. When you're talking about two lines, okay. One line can be like this. One line can be like this. Okay. We'll talk about such cases a little later on. So when we talk about the angle between the two lines, even for such cases, we can talk about the angle even though they are not intersecting. Okay. So please do not have this wrong notion. I've seen people having this wrong notion, sir. They're not intersecting. How can you talk about angle between them? So they need to not intersect. Fine. But still I can talk about the angle between them. Okay. So when I talk about angle between two lines, L1 and L2, when their direction cosines are given, let's say L1, M1, N1 are the direction cosines for, why am I writing in yellow? I'm showing the line in white and I'm writing the direction cosines in yellow. I mean, doesn't matter, but still I should maintain the sign. I should maintain the color coding. Okay. Let's say these are your direction cosines. Okay. See here, the angle between the lines could be two angles. It could be the acute angle between them or it could be the not so acute angle between them or obtuse angle between them. Okay. So here, please understand the formula that you learned in your vectors. If you're talking about alpha being acute, then you need to take the modulus here. Okay. Please get this. Okay. And if you're not taking the modulus and if the answer to the right is a negative quantity, then if you're calculating alpha from there, that means you are giving the examiner the obtuse angle between the lines. Okay. So please read the question. What does the question actually ask you? Does the does the question setter ask you the acute angle between the lines or does the question setter ask you the obtuse angle between the lines or does the question setter doesn't mention anything? You can, you must mention both the answers. Okay. Are you getting my point here? And if somebody gives you the direction ratios of these lines, let's say a1, b1, c1 and let's say a2, b2, c2. Okay. And use the same formula mod a1, a2, b1, b2, c1, c2 by under root of a1 square, b1 square, c1 square that we had learned in our vectors. So this, the same formula I used in vectors. Okay. You can check, go and check back your notes. Okay. So this can also be used to get the acute angle between them. Is this fine? Any question? Any concerns? Here also the same condition for parallelism, similar condition, I should say not the same. So condition for parallelism here would be, please note down l1 by l2 will be equal to m1 by m2 will be equal to n1 by n2. The only change that it could be both plus minus one. Why? Because if the lines are parallel and you're mentioning the direction cosines of those two lines, they can be in the same direction or in the opposite direction. But for line, their direction doesn't matter. See, we know that direction ratio cosines give you the unit vector on lying on that line, lying on that line as a support. Okay. So there could be two directions of the unit vector, isn't it? Right? So if they are parallel and you have considered on one line the unit vector like this, let's say this is one line and this is another line. Okay. They are parallel, right? But if you have considered a unit vector like this on this line, okay? And you have considered a unit vector like this on the other line. So in that case, your ratio will be minus one. And if you're considering in the same direction, it will be plus one. But for lines, it doesn't matter. Line will say, I don't care whether they are in the same direction or opposite direction. Okay. Vectors do care. And of course, same will be true even for their direction ratios. Okay. And this could be anything. Let's say this could be anything need not be plus minus one. Condition for perpendicularity, that would remain exactly the same, no change. Condition for perpendicularity of two lines, l1, l2, m1, m2, n1, n2 will be equal to zero. And same will be holding true for a1, a2, b1, b2, c1, c2. This will also be there. So please note the minor differences, okay, between vector and line. Of course, they are different objects, they're different geometrical objects. So there have to be some difference. But of course, in many cases, you can borrow concepts from vectors to address your 3D lines problem. Okay. Now let's start solving some questions. I think we have discussed enough theory to get us started with some questions at least or maybe one or two questions we'll take and then we can go for a break, small break. Okay, let's take this one first. Find the vector equation of a line passing through two minus one one and parallel to the line whose equation is this. All of you notice this fact that they have given you, they have given you a line already. And let's try to do very basic exercise over here. Okay, let's take this line itself. Okay, this is the Cartesian equation of some line. Okay, now tell me a few things that I want to, I want you to answer. Are these numbers sitting in the denominator? Are they direction ratios of the line or are they direction cosines of the line? Tell me that first. Let's do some basic question-answer session. Right, they are direction ratios. They cannot be cosines because cosines must always satisfy which relation? L squared plus m squared plus m squared should add up to one, but they are not going to add up to one. They're going to be I think beyond 50. Okay. So these are direction, they are part of direction ratios. So two seven minus three is the direction ratios of this given line. Okay. Now, if I want to write a line equation in Cartesian form, let's say Cartesian form first, passing through, passing through, passing through two minus one one and having the same direction ratios and having the same direction ratios two seven minus three. Right. How will you write the equation? Very simple. You'll say x minus two by two. So x minus this, this value divided by this value equal to y minus minus one, which is y plus one divided by seven and then z minus one divided by minus three. Okay. In short, this becomes your equation of a line passing through this point and having a direction ratio same as this. Okay. Now, even if you have a direction ratio proportional to this, the answer is still not going to change. You can still maintain the same answer. For example, even if I had let's say a direction ratio, let's say I take another set of direction ratios to be let's say four, 14 minus six and passing through the same point. Will my equation change? You will say no, because even if you're writing four, 14 and minus six, you're still going to cancel out a factor of two from everywhere and come back to the very same equation. Okay. So when we were talking about the condition of parallelism, I told you that direction ratios of a line parallel to a given line, they are proportional to the given direction ratios. Right? So a one by a two equal to b one by b two equal to c one by c two. So two lines if they are parallel, their direction ratios will be proportional. Okay. So that doesn't actually impact the denominator part of our equation because that proportionality constant can be always be cancelled. In other words, if somebody asking you for a line which is passing through two minus one one and parallel to this line, you can just change the numerator part to x minus two, y minus minus one, z minus one. But the denominators of each of these terms, you can take these numbers themselves. Okay. You need not put something always proportional to it. Even if you write, nobody's going to deduct marks. Okay. Right. Now, please understand this fact, we have been asked to write the vector form. This is the Cartesian form. So vector form, if you remember, sorry, vector form, if you remember, it is r equal to the position vector lying on that line, which is two i minus j plus k plus some proportionality constant, you can use any alphabet that comes in your mind, mu, lambda, alpha, beta, t, s, are these frequently used ones? Okay. You can use any alphabet that comes in your mind. Don't use x, y, z, but because you'll get confused with the variables involved. And just write down the direction ratios in vector form. That's it. Got the point? Any questions? Okay. So this becomes your answer to this question. Okay. Let's take another one. The Cartesian equation of a line is given to you like this. Find its direction ratios and also the vector equation of the line. So first, give me the direction ratios. Everybody, by the way, your answers need not be all the same, right? Note that direction ratios can be infinitely many. So don't expect that what answer you got your friend would also have written the same answer in the exam, not necessarily. So direction ratios is nothing but the component of any vector collinear to that vector. Or in this case, it will be any line direction collinear to that. Okay. So it can have infinitely many answers for it. Done? Okay. Our NPS Raja Ji Nagar has done this chapter and spoon. Just start it. Okay. And what about HSR, NPS HSR and what about other schools, GEAR, DPS are not done. We have online tomorrow, full day school starting from third. Yeah, I think that's the situation everywhere now. But I really doubt this will continue for very long because the cases are on rise to be very frank. Okay. See, we are, you know, as soon as we get an approval, we'll be there in the school. Okay. But we are going to be in talking terms with the school management to allow us inside the school. Okay. What one class? We want to take several classes. Okay. So we'll keep you updated related to that. Don't worry. So if you have full day of offline classes, we will be definitely there in the school premises to take the class. Not to worry. But I think we are going to start with the crash course this week itself. So crash course will be online. Okay. Anyways, let's look into this. So first of all, by the way, this equation that is provided to you is not in the form that we would like to see it in because in my equation which I gave you the coefficients of x, y, z were one each. But as you can see here, they have actually kept six, three and two. So what do we do in such cases? We normally take the coefficient common out something like this. And then we put that six as a one by six down something like this. Okay. And I mean, you could just write one by six, one by three and half also, but you know, it would be advisable that you multiply the denominators with six. So you can say that the equation which was given to you could be easily rewritten like this. So you can say one of the direction ratios or possible direction ratios could be one, two, three. Okay. So this could be a possible answer. If somebody writes two, four, six, he is still right. If somebody writes minus one, minus two, minus three, he is still right. Okay. But of course, as you know, we don't waste time writing unnecessary in the formats of the same answer. It is just going to waste your inviting. Is this okay? Now vector equation is what they're asking us. Okay. Now everybody please understand this fact. When you're writing the vector equation, I have seen many people, they only use the point that they have, they have got while they have written it like this. So as you can see, one by three, minus one by three and one is a point. And of course, lambda times one, two, three with ijk. Okay. Please note that even this can change from one person to another. You did not always use that one by three, minus one by three and one. Are you getting my point? However, many people will use it because they don't want to apply their mind in finding some other point. Okay. But trust me, you could use other points also. Now what other points you can use? Now all of you please pay attention here. See, let's say I call this as lambda. So x is one-third plus lambda, y is two lambda minus one-third and z is three lambda plus one. Okay. Now if somebody who has used this point, he has actually used lambda value as zero. But I can use any lambda I want, lambda is a parameter. I can use lambda as one also. So if I use lambda as one, I could get a point as four by three. I could get a point as five by three. And I could get a point as let's say four. So I could also write down the equation like this, four by three i minus, sorry, plus five by three j plus four k plus some other, let's say I am just using a different parameter to distinguish between the given equation and the previous one. Okay. So this is also correct. This is also correct. Nobody can nobody can mark this wrong. Are you getting what I'm trying to say? So please note that this is something that they will do in their options also. Okay. They will say the line which does so and so and they will just change the point. And what people will think, they will think, oh, this one, two, three is matching. But this should be one by three minus one by three and one. That is not matching. And they will not mark that particular option. I'm talking about a situation where there could be more than one also correct. So please don't be under the impression that it's mandatory to use the same point in the equation. No, not mandatory. Right. You can keep any lambda and get us, you know, point on that line and use that in your equation. Are you getting what I'm trying to say? Clear? No. Okay. So this can change depending upon the person who is writing it. Okay. You can't say boss, you're wrong. You can't say that. Okay. The only thing that he cannot mess with, the person cannot change. He can change this also. But whatever he changes, he should keep it proportional. For example, here I can say two, four, six, even this is fine. I can keep three, six, nine, even this is fine. But whatever I keep, I should keep it proportional to one, two, three. That thing I cannot randomly change. But I can take a random point on the line. Is it clear? Any questions? All right. So I think we deserve a break now. I will not give you a long break. We'll meet at 6.30 because I have a lot of things to cover. Okay. Let's meet after a 13 minutes short break. Now another equation, another, I can say conditions given to us while forming the equation of a line is they may give you two points on that line. Okay. For example, let's say somebody mentioned that there are two position vectors. Okay. So when two points or position vectors are given to me on that line or two position vectors are given in the previous situation, we had taken one of the point. So one point was given to me and we knew the direction ratios or cosines of that line. Now another situation when you can ask, when you can be asked the equation of a line is when they provide you with two points or two position vectors on that given line. Okay. So let's say as a coordinate, they are x1, y1, z1 and x2, y2, z2. Okay. So how do you find the equation of line in such case? Let us start with the vector form first. In vector form again, we use the same logic. Okay. So let's say I call this point p and I call this point q. Okay. If I take any generic position vector, let's say r here. Okay. And let me call this as r. Can I say rp, rp is collinear to, is collinear to pq vector. Right. In other words, I can say r minus p. Okay. By the way, this is pr, but it doesn't matter because I've used the word collinearity. So this is collinear to let's say q minus p or p minus q actually doesn't matter. Okay. So better to use proper representations so that you are also not confused. Okay. That means r minus p is lambda times q minus p. In other words, r is equal to p plus lambda times q minus p. By the way, there are various facets to it. You can either write it like this or you can write it like this. It doesn't make any difference to your answer. You can also write it like this. Okay. Or you can always write it like this. Okay. They all mean the same things. There is no difference between these four representations. Okay. Nobody's going to deduct marks for writing any, you know, anyone out of these four. Are you getting my point? Now, if you look at this expression, basically it's trying to say that you have written r as 1 minus lambda times p vector plus lambda times q divided by 1 minus lambda plus lambda. What does this remind you of? What does this remind you of? What does this remind you of? What thoughts come in your mind? Section formula. Okay. Now actually section formula is happening, right? Because if you take any point on that given line, it will divide the joint of p and q in certain ratio, isn't it? Isn't it? So that itself is a relationship between any generic position vector and the two given points p and q, right? Yeah. So this form is basically inspired from your section formula itself. Okay. So don't be confused if somebody writes the same equation like this. That is also acceptable. Is it fine? So this is the vector form. We can call it as a vector parametric form. So let me call it as a vector parametric form. Okay. You can also write like this. The same thing you can also write that r minus p cross q minus p is equal to a null vector. Okay. So if you're writing like this, you are basically writing the vector non-parametric form. So for competitive exams, be prepared with both the versions. Okay. Anyone can be asked to you. Anyone can be given to you. Oh, I'm sorry. I meant to write it like this. I think I've written p, q, p minus q. Thanks. Sorry. I wrote the same thing once again. Is it fine? Any questions? Any concerns? Could you scroll up one second? Yes, why not? Are done Shraddha. I'm going to show you. Take your time. Okay. Now the same thing in the Cartesian form, the same thing in the Cartesian form. So I'll be taking a clue from the vector form itself. Okay. So r is p plus lambda times p minus q. This was your, or you can say lambda times q minus p, whatever. Doesn't actually matter. Okay. So here, what I'm going to do, I'm going to call my r as x i, y j, z k. As I told you, r is any generic position vector. That means it is less a vector form of x i plus y j plus z k. This is your x 1 i, y 1 j, z 1 k. Okay. And lambda times the difference between the two vectors that will lead to I think x 2 minus x 1 i, y 2 minus y 1 j and z 2 minus z 1 k. Okay. Let's bring this vector to the left hand side. So x minus x 1 i cap, y minus y 1 j cap, z minus z 1 k cap is x 2 minus x 1, you can say lambda times x 2 minus x 1 i cap, y 2 minus y 1 z cap, z 2 minus z 1 k cap. So since i j k are set of linearly independent vectors, the coefficient of i cap and i cap on both the sides will be same. So x 1, x minus x 1 is lambda x 2 minus x 1, y minus y 1 is lambda y 2 minus y 1 and z minus z 1 is lambda z 2 minus z 1, which clearly implies, which clearly implies. Okay. So from this, we can clearly say x minus x 1 by x 2 minus x 1 is equal to y minus y 1 by y 2 minus y 1 and z minus z 1 by z 2 minus z 1. Okay. Please note, it doesn't matter even if you flip the positions of 1 and 2. For example, you can write x 1 minus x 2 here, y 1 minus y 2 here, z 1 minus z 2 here. Both are acceptable. So let me write that down. So it doesn't matter whether you're writing it like this. Okay. Okay. Neither does it matter to change x 1, y 1, z 1 to x 2, y 2, z 2. Okay. So you can also write it like this, x minus x 2 by x 1 minus x 2, y minus y 2 by y 1 minus y 2, z minus z 2 by z 1 minus z 2. But please maintain the order which you are following. Okay. Don't start doing x minus x 1 by x 1 minus x 2 and the other one you write it as y minus y 1 by let's say y 2 minus y 1. That will lead to erroneous results. Okay. So all these forms are, all these forms are the same thing. No difference. Please note it down. Okay. So they all mean the same. They all mean the same. Okay. Fine. Could you scroll up for a minute? How up? This, this is good enough or? Yeah, this is the beginning. Okay. One thing is very obvious from here is that if you, if you collect, if you recollect your previous form that we had seen when I was provided with the point and the direction ratios or cosines, it basically is a very interesting fact that you should all note down if you have been given two points on a given line. Okay. Let's say these are the two points on that given line. Right. Then remember the direction ratios. Let's say if A, B, C are the direction ratios are one of the direction ratios are one of the direction ratios of the line. Then please note that your A1 is proportional to the difference of the x coordinate. B1 is proportional to the difference of the y coordinates and C1 is proportional to the difference of the z coordinates. Okay. So please understand that your direction cosines themselves are or you can treat the difference of their coordinates themselves as one of the direction cosines. Okay. So when two points are given, you can actually take the difference of the respective x, y, z coordinates themselves as your one of the direction cosines, direction ratios. Okay. And then you can start using it in your equation as well. Okay. So this is a very, very important, you know, observation over here. Is it fine? That is your DRs could be your difference in the coordinates like this. In any order, it doesn't matter. Even if you want to take x2 minus x1, y2 minus y1, z2 minus z1, both are fine. Both will give you direction ratios only. Is it okay? Hariharan, do you still want me to scroll anywhere? Left top. Left top, left top, we are here. Okay. Let's take a question based on the same. Find the vector equation and let me add one more thing here. Find the vector as well as Cartesian equation, both also Cartesian equation. Of the line passing through these two points. Oh, Cartesian equation. They've already asked. Sorry for that. Didn't read the question completely. They've also asked you the Cartesian equation and the vector equation. Let's do it. You can just type a done on the chat box if you're done. No need to type your response. By the way, it's my prediction. I'm not commenting anything. UP election is going to happen in April. Okay. So I think your exam dates will come early in the month of March. Yeah, because exam should go on for 15, 20 days at least. So it cannot happen in the last week of March because UP elections are there. And you know how important is the elections of UP for the government because it has the largest population. Very much related. UP election is going to be very, very critical because teachers are going to be involved in the elections. Yeah. So teachers are normally they play a lot of role in election duty. So in order to keep them available, exams will be done much before the at least one week or two weeks before the UP election, which happens to be in April. So expect your second semester date sheet to come very early in March. May start in the first week of March. Why not after because all the competitive exams are there, they will clash. J main, day one, CT, they'll all start clashing. Okay. Yes. So that is why we rush to be completed the syllabus, you know, before your semesters are one only, of course, one or these two topics are left. And yes, when is the first attempt of J main? Now again, this is a speculation which people are making. They're saying that there will be still four attempts and first attempt is going to happen in February. Again, speculation. Nothing is fixed till the notification comes up. So hence we started, we want to start our crash course so that by February we are done with at least, I don't know, our crash course and you are ready for the exam. No, not yet. Registrations have not opened it. Okay, let's discuss it. So your schedule is going to be very hectic. So semester two topics are going to be covered. Then your, we have crash course, then your J main first attempt will happen, then immediately second semester is going to happen. So be prepared guys and girls, lab records, project, class works, absolutely hurry up. Yes, that's a very recent question you asked me. I'm not in the board of NTA. But if there are two attempts, then it will happen post UP election. One should happen in February anyhow. And one should happen post. But again, people who are banking on second, third, fourth attempt, etc. Please don't bank on it. Your first attempt should be the best one. You know why? You know why? Because everybody will be less prepared. Okay, so let's write down the vector equation. Now see vector equation I can use. I can use any one of the two points. Any one of the two points, let's use three, four and minus seven. Okay, plus lambda times the difference of the two. And you can take the difference in any order also. So let's say I subtract the second one from the first one. So 2i 5j minus 13. And again, this answer is look and feel can change. Somebody can use the second point as this point. And he can take the difference in other way also. So if somebody writes down the answer like this, he or she is still right. Nobody is going to deduct any marks for it. Okay, so this is also correct. This is also right. Now let us write down the Cartesian form. Yeah, let us write down the Cartesian form. So these are your vector forms. So Cartesian form as you already know how to write a Cartesian form from a vector form and vice versa. Let's say I want to write down the Cartesian form for this one. So you can write it as x minus 3 by 2, y minus 4 by 5, z plus 7 by minus 30. Okay, this is your answer. That's why prakul, all your copying work should be done when your brain is not working. I mean, you have already studied very hard that day, you have solved so many Jee main, Jee advanced questions, you're tired. That time you should sit and copy. That time you will realize that okay, it's worth giving this time. If you ask me how I used to do it, I used to write it late in the night. That's a 10 o'clock, 10 30 after my dinner when I'm sleepy and all then I used to copy stuff. Of course, I didn't used to make any mistake while copying. But when your mind is fresh, you should keep it for productive work. Okay, let's take a question also on finding the angle. Okay, find the angle between the pair of lines. Let's say I want to, I want you to find out the acute angle, acute angle. Prakul is saying there is a lot of project reports to be copied in school and also that seems unproductive. So I was answering to that. Done. So when you're finding the angle between the lines, please note these guys, they don't have any, the points, they don't have any role. Okay, what is important are these guys, this and this, they basically carry the information. So they carry the info related to the directions. Okay, so the points, they don't have any role to play while finding the angle. Okay, so right now they are the direction ratios because they are not unit vectors, isn't it? So you can use the formula cos of alpha, let's say alpha is the angle, you can use the formula a1, a2 or you can just use the dot product of these two vectors and divide by the modulus. Okay, so y2, okay, just take dot product of these two. Okay, and I don't think so, it'll give you a negative answer. So modulus is of no significance divided by their magnitude, divided by their magnitude. Oh, sorry, this is 6 square. So this will become 3 plus 4 plus 12, okay, 3 plus 4 plus 12 by 3 and this will be 7. So this becomes 19 by 21, which is nothing but alpha is cos inverse of 19 by 21. Is it fine, any questions? Okay, so coming to the slightly difficult part of this chapter where we are going to learn how to find the foot of the perpendicular and distance of a point from a line. Let's take that part. Let me just start with the question, by the way, the topic name is how to find the foot of the perpendicular, you can say foot and length of the perpendicular, length of perpendicular drawn from a point onto a line, from a point onto a line. Okay, now let me discuss this concept through a problem because then you will understand it better rather than me giving you a generic theory, you will not understand that much as what you will understand from this particular example. And I will first take this concept in the vector form itself, which is as given to you in the question. So you can see that there is a, the question says find the foot of the perpendicular, draw from a point, let's check all this point as a point A, which is given to you as 2i minus j plus 5k. And you have been given a line whose equation is also mentioned to you. Okay, 11i minus 2j minus 8k plus lambda times 10i minus 4j minus 11k. Okay, so we have to not only find the foot of the perpendicular, let's say m, but also we have to find out what is this length. Preq will I know that everybody is on the same page. I think that the, I mean, it is up to the teacher, but you can request for some extra time. I should not be saying this on records. Anyways, so how do you find the foot of the perpendicular? Let's try to address this fact. Any suggestions from your side, which can help me to get m? Any suggestions? You can unmute yourself and talk. Because if I get m, then, I mean, I know A and I know m. I can use my distance formula, which I learned in my class 11th 3D geometry to get the distance A. But how will I get m? That is the main thing. Right, Aditya. So Aditya has a very, very, you know, you can say interesting suggestion. He says that anyways, this expression is actually representing any position vector on this line. So let this be the position vector. Let, let this be the position vector of m. That is to say that, that is to say that let position vector of mb, mb, 11 plus 10 lambda i, okay, minus 2 minus 4 lambda j, and minus 8 minus 11 lambda k. Okay, so this point itself has this position vector. Remember, lambda is the only thing which is stopping me from getting the coordinates. Rest, everything is like, you know, settled here. The only thing that is stopping me from getting my final answer is this lambda. Had I known, if I know the Adhaar card number for m, I will be able to get what is my m. It's the same way. If I know somebody's Adhaar card, I can easily go to UIDA office and know who is that person. Isn't it? So it's like, the only thing that I need to know here is lambda. If I know that, m will be exposed, my m will be known. Correct? So how do I get m now? Sorry, how do I get lambda now? Very simple. I can use the fact that am vector or ma vector is perpendicular to the direction of that given line, which is 10i minus 4j minus 11k. Yes or no? That is to say am vector dot product with 10i minus 4j minus 11k, which happens to be the direction of the line or one of the directions. It can be this way, this way. Doesn't matter. It is still going to be perpendicular, so it is going to be zero. So what is am vector? Am vector is nothing but position vector of m minus position vector of a. So position vector, any question anybody has? For cool, I will address your concerns a little later on. I understand your situation. I was also in the same situation 20 years back. So am vector will be this minus 2, by the way, let me just directly write it. So it will be 9 plus 10 lambda. And this minus j will be nothing but minus 1 minus 4 lambda. Correct me if I am wrong? And this minus this will be minus 13 minus 11 lambda k. So this dot product with 10 minus 4j minus 11k, this is equal to zero. Yes or no? Correct me if I have missed out anything, do let me know. All set? I hope I have not missed out anything. So that means 10 times 9 plus 10 lambda minus 4 times minus 1 minus 4 lambda and minus 11 times minus 13 minus 11 lambda is equal to zero. So in this case, you end up getting 90 plus 4 plus 143. So what is 90 plus 4 plus 143? 143 and 94. That's going to give me, I think 237. And I think this is 100 plus 116, 121, 116, sorry, 116, not 116, 116 and 121. That also gives you 237. Oh, good. So plus 237 lambda, 237 lambda equal to zero. So lambda is negative 1. Now I know what is the Adhaar card number for M. So M is exposed. So if I put my lambda as minus 1 here, which I have obtained a little while ago, I get the point to be, correct me if I am wrong, I get the point to be I and this is going to be 2j and this is going to be I think plus 3k. So this is your position vector of, this is your position vector of M. So in short, I ended up knowing the vector M. Now, foot of the perpendicular is found out, no problem with that. Now what is the length AM? Length AM or D is nothing but the modulus of AM vector. So AM vector will be the difference of the vectors here, which is minus i, destination minus source, I think you already know it. So you have to find the modulus of this vector to get your distance, which happens to be under root of 1 square plus 3 square plus minus 2 square or 2 square, you can call it. That gives you root 14. So this distance is root of 14 units. Is it fine? Any questions, any concerns in the approach? This approach is important. Don't try to memorize anything here. So the approach is, again I'll repeat, the foot of the perpendicular, since it lies on this line, you can take this expression of R itself to be your M because lambda is anyways not known to us. Whether you call lambda, whether you call T, whether you call S for that particular point, it is something which is supposed to be found out anyhow. So take it to be the same expression. So take it to be like this and then to get the lambda, you use the fact that AM is perpendicular to the direction of the line and this guy contains the information about the direction. Okay, direction information is hidden in this particular vector. So take the dot product with this, put it to zero, you get lambda value. Thankfully our lambda came out to be very simple. Normally it comes out to be simple only unless until you are unlucky and then put that value back in your position vector of M, you get the point. Once you get the point, you know the vector AM or MA. Find the modulus, that will be the length D. Is it fine? Any questions? Okay. Now we'll try to do a similar question, not the same question, a similar question, but this time not in vectors, but using Cartesian form. Okay. So one important thing I would like to tell everybody that if a question is, let's say if you get a subjective question in your next semester and you're supposed to solve it and let's say the question is in vector form, try to solve it in vector only. Don't try to convert it to Cartesian. So if you convert it to Cartesian, it gives a very long notion to the examiner that you're weak in vectors. Okay. So if a question is in vectors, solve it in vectors. If a question is in Cartesian, solve it in Cartesian. Okay. Don't try to convert vector to Cartesian in order to solve it. I've seen many people doing that by the way. Converting vector to Cartesian. Let's take this question. Find the coordinates of the foot of the perpendicular drawn from one zero three to the joint of B and C. Okay. So this is your A, this is your, let's say B and C. So you want to know what is the foot of the perpendicular here? Okay. How would I solve this question? Everybody please attempt this question and give me your response for M coordinate on the chat box. Quick. We don't have much time. Done. Okay, Aditya. Anybody else? Okay. Let's discuss this quickly. Let us first try to figure out. This is an extra work that we are doing. Equation of BC line. Okay. Equation of the line BC. So as I already told you, you can use X minus four, Y minus seven, Z minus one. And you can take the difference of these in any fashion you want. Four minus three, one, five, seven minus five, two. And this is minus two. Okay. Let's say for a particular lambda, you end up getting the coordinates of M. Okay. So let M coordinate, sorry. Let M coordinate be lambda plus four comma two, lambda plus seven minus two, lambda plus one. Now, can I say direction ratios of AM? I can take it as a difference of this point and the coordinates of M. So can I say it is lambda plus three comma two, lambda plus seven comma minus two, lambda minus two. So basically remember in one of the initial slides, I told you that the difference in the point itself acts like a direction ratio. The difference in the coordinates of the two given points on the line that itself acts like a direction ratio. Okay. So I've just taken the difference of A and M. So lambda plus four minus one, that is lambda plus three, two lambda plus seven minus zero, that is two lambda plus seven and minus two lambda plus one minus three. That gives me this. Now, this is the direction ratios of the line, which is nothing but the difference of the coordinates, which you have already written over it. One, two, minus two. Okay. So think as if this is like A1, B1, C1, A2, B2, C2, and you want to make the two lines perpendicular. So remember the condition of perpendicularity. Condition for perpendicularity is that A1, A2, B1, B2, C1, C2 should add up to give you zero. Right? This is your A1, B1, C1. This is your A2, B2, C2. Correct? So A1, A2 means lambda plus three into one. B1, B2 means two lambda plus seven into two. C1, C2 is minus two lambda minus two into minus two. This should be zero. So let's try to simplify this. So lambda plus three, four lambda plus 14. This is again four lambda plus four equal to zero. I think this will give you nine lambda plus 21 equal to zero. So lambda value comes out to be minus seven by three. Slightly ugly value. But yes, it's possible to get such ugly values. I'm not surprised. Okay. So now put that in your position vector of M. So put this guy over here. So it'll become minus seven by three plus four. Minus seven by three plus four will be five by three. And minus 14 by three plus seven. That is going to be seven by three if I'm not mistaken. No, minus 14 and seven. Yeah, seven by three. Correct? And here it'll give you 14 by three plus one, 14 plus three is 17, 17 by three. So this is going to be your answer. Aditya, what did you do? Your answer looks so nice, but it is not right. Is it fine? Any questions anybody has? Okay. Now an additional question may be asked to you sometimes to find the image of a point in a line. Okay. So for finding the image of a point in a line, you can use the same logic that we use in case our 2D lines. Right? So let's say if I extend this question and I also say find, let me use a different color, find image of A in that line. Okay. Please note M is not called the image. M is called the projection of A on the line. Please be very careful about the words. Right? So if I take a mirror image, it will be here. A dash, let's say. Okay. So how do you get A dash? Very easy. Let's say A dash is alpha, beta, gamma. Okay. So you know that M, let me just put a barricade here. You know that M is the midpoint of A and A dash. Okay. So you can use this fact. So alpha plus 1 by 2 is equal to your M's X coordinate. What was your M's X coordinate? 5 by 3. Okay. Similarly, beta plus 0 by 2 is Y coordinate of M, which was 7 by 3. And gamma, what was it? Gamma plus 3 by 2 is going to be your Z coordinate, which I think was 17 by 3. So get your alpha, beta, gamma. Let me write it down directly here without much waste of time. So this will be 10 by 3 minus 1, which is 7 by 3. 10 by 3 minus 1, 7 by 3. This is going to be 14 by 3. And this is going to be, if I'm not mistaken, 34 by 3 minus 9. 34 by 3 minus 3, which is 9, which is going to be 25 by 3. Okay. Is there any questions with respect to the image? So now you know how to find out foot, perpendicular length, and image all the three. Any concerns? Any questions? Please, I might. Okay. The last concept that we are going to talk about is distance between two lines. Okay. Now, when I say distance between two lines, even though I have not used the word shortest, distance between two lines is always the shortest distance. Okay. Whenever you say distance of a point from a line, you mean the shortest distance, isn't it? So the word shortest may not always be erratic. Okay. But it is implying the same thing. Now, two lines, they can be in the following manner. They can either be intersecting like this. Let me make a different, use a different color. So let's say a white line and yellow line, they can either be intersecting like this. Okay. Or they could be parallel like this. Okay. Or they could be skew like this. Okay. What is a skew line? A skew line is something like a flyover and a road. Okay. So they're not parallel. Let's say the road is, let's say my hand is the road and this marker that I'm showing you, that is a flyover. Okay. So this flyover and this road are not parallel. Right. But they're not intersecting anyways. Okay. So that is called the condition, that is called the skew lines. Okay. So let me write it down here. This situation one, we call them as intersecting lines. Okay. Situation two is called parallel lines and let me add one more thing. Both together are called coplanar lines. Of course, they're coplanar. Correct. You can always pass a plane through the first case and the second case. Okay. The third one is called skew lines. And this is a case where these two lines are non-coplanar. Is it fine? Any questions? Any concerns? Okay. Now, the question is, first of all, if the two equations are given to me, how would I come to know which of the three cases is happening? And if I come to know which of the three cases is happening, then how do I find the shortest distance? Okay. So that is the problem that we need to address. So all of you listen to this carefully. Very, very important thing I'm going to tell you. Let us say the two lines given to you have the equation. So let the lines be, let the lines be r equal to, let's say a1 plus lambda b1 and let's say r equal to a2 plus, let's say, tb2. Okay. I'm purposely taking a1, b1, a2, b2 to keep it as generic as possible. Okay. So let the two lines given to you are like this. Now, actually looking at them, you cannot distinguish between the first and the third case, but you can definitely figure out whether it's the second case or not. How can anybody tell me that? How would you know looking at these two equations that they are parallel or they are not parallel? Not parallel means it could be intersecting or it could be skewed. So at least you can know whether it is the second case or not the second case. How would you come to know that? Right, b1, b2 will be proportional in that case. Please understand this fact. Note this down. Very, very important. If b1 and b2 are proportional, that means the direction of these two lines are such that they are proportional and you can easily make them out. That means the coefficient of i, j, k, you know, components of b1 and b2 are proportional. Then it's the case of parallel lines. Okay. Then situation number two is holding true. Okay. Correct. And if they're case of parallel lines, then remember, then see what do we do normally? See. So if they're parallel lines, that means you know that these two lines are parallel. How do I get the distance between them? Okay. So let's say this is r equal to a1 plus lambda b. Let me write b only. Okay. And let's call this as r as a2 plus let's say tb. Now, you must be wondering, sir, you gave b1, b2, and now you made them bb. Why? The answer to that is very simple. If the two lines are parallel, I know they're parallel, I can actually make their direction same also. Just like what we had in case of 2D lines. In 2D lines, if the lines are parallel, we could make the coefficient of x and y as same in the general form of the line. So here also, I have a liberty of making them the same. Correct. Am I right? Correct. So basically what I'm trying to say that both these lines, let's say they are parallel to a given vector b. Okay. And let's say a1 is this point. Okay. And I'm just taking a2 as this point. I'm just taking some, you know, orbit positions. Okay. Now, what I need, I need the distance between the two parallel lines. So what I'm going to do is I'm going to just make a small diagram over here. Okay. So let's say this is the distance between the two lines. Okay. This length is what? This length is what? Or this vector, let's say I call this as a vector. This vector is what? a1 minus a2 vector. Correct. And this vector is b vector as I've already shown you on the diagram. Correct. Now, all of you please pay attention. I need a d and d is supposed to be a1 minus a2 this length into sine of this angle, isn't it? Am I right? Everybody agrees? Everybody's happy with this? Yes or no? Say something. Correct. Okay. Now, all of you please pay attention. I'm not going to take the cross product of this with b and take a modulus of this. When I do that, I know that this is going to give me a1 minus a2 mod into b mod sine of theta. Correct. So we know that the modulus of a1 minus a2 cross b is going to give me this. Okay. Which means, let me write it here. Which means mod a1 minus a2 cross b is nothing but mod b and what is a1 minus a2 mod sine theta? That is actually your d value. Correct. In short, you can get d value by using this simple technique. Okay. Please note this down. This is going to be a very useful formula for you, not only in competitive exam but also in school exams. So this is how we find out the distance between two parallel lines. Okay. a1 minus a2 cross b whatever vector comes out, take its magnitude, divide by the magnitude of b that will give you the distance between two parallel lines. Okay. Done. Any questions in this? Please do let me know. That's a very late realization. a1 a2 are yes, two position vectors on those lines. What were you doing when I was talking about vector form of a line? So when you write an equation of a line, this guy is a specific point on that line and this guy is the direction which is parallel to that line or collinear to that line. Is this fine? Any questions? Okay. Now, if b1, b2 are not proportional, if b1 and b2 are not proportional, then what to do? If b1 and b2 are not proportional, then it could be either of the two case. It could be a intersecting line. Okay. Or they could be skew lines. Okay. So what do we do in such cases? Please understand this. Okay. I have a very interesting thing. Let us say I take a very, very generic case where the two lines are skew actually. I mean, I'm just assuming it to be skew. Let the time tell whether they are skew or not. So let's say this is one of our lines, r1 a1 plus lambda b1 and this is your another line r equal to a2 plus tb2. Okay. And I want to know the perpendicular distance between them. Okay. So please note the distance which will be perpendicular or the shortest that will be any that distance will be along the perpendicular to both the lines. Isn't it? Right. This is something which everybody understands. It's a basic fact that the shortest distance between the two lines is always along the mutually perpendicular to both the lines. Isn't it? Okay. Now, let me ask you a simple question. If let's say this vector is b1 and this vector is b2. Can you tell me a vector which is perpendicular to both of them? Can you tell me a vector which is perpendicular to both of them? Can you tell me this vector b1 cross b2? Absolutely. So can I say b1 cross b2 will be perpendicular to both the given lines? Right. Now, how does this help us to get this distance d? Let us, let us try to understand that as well. Now let's say this point a1 is here. Okay. Let's say this point a1 is here anywhere you can take actually and this point a2. This point a2 is somewhere over here. Okay. Can I say d and correct me if I'm wrong. Can I say d is the projection of a1 minus a2 vector in the direction of b1 cross b2. Can I say this? Right. Let's understand it from a simple figure. Let's say this is b1 cross b2 and this is your this is your let's say this vector is let's say a1 minus a2. Okay. These are the position vectors. So this vector is a1 minus a2 vector. Correct. So what I'm trying to say is that if you take the projection of this vector on this b1 cross b2 vector then this distance will actually match with your d that you are trying to find out. So this d will be nothing but it will be the projection of this guy in the direction of b1 cross b2. Agreed. Now many people say, sir, why not take b1 cross b2 magnitude only? See, b1 cross b2 is a vector which is perpendicular to both the lines but that doesn't give me the length or that doesn't give me the shortest distance between them. So many people wrongly think that the magnitude of b1 cross b2 should be the answer. No, that is not the case. Okay. So here I have to come from our scalar product concept that if I want to know this distance I have to project a known value or a known vector in the direction of that given vector. Are you getting my point here? So this point is the projection of a1 on the perpendicular. This point is the projection of a2 on this perpendicular b1 cross b2 and then the distance here will give me the shortest distance between them. Correct. Now let us recall our scalar product, a small recall I would do here. What is the projection of a vector p? What is the projection of p on q? Let me see how many of you remember vectors? I know you have been busy with your board exams. What is the projection of p on q? Correct. p dot q cap. Or you can say p dot q by mod q. Yeah, both are the same thing. Okay. So in the same way projection of p in the direction of q, can I say that will be nothing but a1 minus a2 dot b1 cross b2 divided by a modulus of b1 cross b2. Yes, I know. But normally after doing this we take a magnitude of it because it may happen that dot product can give you negative, but negative as distance cannot be negative. So many times we take the positive value from there. So please note this down. This is the shortest distance between the two lines. Okay. Now how? Now please remember this is just a case where my b1 and b2 are not proportional. But how does it still tell me whether they are intersecting or not? Now here is the game. If this distance d becomes zero, that is if a1 minus a2 dot b1 cross b2 becomes a zero, that means the lines were intersecting. Of course, isn't it? If the shortest distance is zero means the lines are intersecting. So this expression actually becomes a zero if they were actually intersecting. Okay. And if d is not equal to zero, if d is not equal to zero, that means if a1 minus a2 dot b1 cross b2, this is not equal to zero, then the lines are skew lines. Then the lines are skew lines. Please note this down. Very, very important. So this single formula basically clarifies whether, see, when they are parallel, you can know easily because b1 and b2 will be proportional. When they are not parallel, then there would be a doubt. Are you, this is intersecting or this is skew? I don't know. Okay. So what I'll do, I'll use this formula. Okay. And on using this formula, I realize, oh, I'm getting d as zero. That means the lines were intersecting, not the point. If d is non-zero, that means the lines were skew. Now clear? So this formula helps you to distinguish between this dilemma, whether intersecting or skew. Yes or no? Okay. Now one very interesting fact evolves from this also. First note this down, then I'll tell you that interesting fact. So we are almost done with the straight line spot. Tomorrow is going to be a lengthy day. Maybe tomorrow, I may have to take the class till eight o'clock. A year, sir. We have a lot of project work to write. If you're not able to finish tomorrow, then one of the Sundays of the crash course I will call you while the crash course is on. Note it down. Anything that you would like to ask me here? Okay. So when very important point, a1 minus a2 dot b1 cross b2 equal to zero is called condition for coplanarity of two lines. Okay. So if two lines are coplanar, please note a1 minus a2 dot b1 cross b2 will always be zero. Now, of course, you already have seen that if it is zero, the shortest distance between them is zero. So they intersect each other. But many people ask me, sir, if they are parallel also, they are coplanar. Correct. So this formula is going to work in that case also. So this formula works also for parallel lines which happen to be coplanar. Parallel lines are coplanar. Isn't it? So when two lines are parallel, you can draw a plane through them. So it's a coplanar. So it also works in, also works for parallel lines also. How? See, b1 cross b2 in that case will become zero because b1 will be lambda b2. Isn't it? So that will make b1 cross b2 as a null vector. So this condition will work for that case as well. Are you getting my point? So this is the condition for coplanarity of two lines, whether they intersect or they are parallel, this condition is going to work fine in either of the two cases. Are you getting my point? So let us try to summarize everything. So in this last part of the exercise, we were talking about a distance between two lines, a shortest distance between two lines, whatever you want to call it. If the two lines are parallel, you know that b1 and b2 will be proportional, which is our this case. Second case. Correct. b1 will be proportional to b2. If b1 is proportional to b2, we get the distance between the two lines by using this formula. Cross product of a1 minus a2 with b, take the magnitude of it, divide by the magnitude of the b vector. If the two lines are such that b1 and b2 are not proportional, then it could either mean that they are skew lines or they are intersecting lines. So nevertheless, we found out this expression. If they are intersecting, the numerator part will become 0. And if they are not intersecting, means they are actually skew lines, then the numerator part will be a non-zero value. By the way, this is modulus. And from this exercise, we also learned that a1 minus a2 dot b1 plus b2 equal to 0 is the condition for coplanarity of two lines. Because if it is 0, either the lines are intersecting hence coplanar or the lines are parallel and hence coplanar. Are you getting my point? So coplanarity can mean they are intersecting or they are parallel. And this works in both the situations. Getting my point. Okay. So we will wind up this with a simple question. Sorry for extending this class by a few more seconds. Okay, let's take this question. Find the shortest distance between these two lines. Okay. Now, let me solve this question in the interest of time. Are these two lines parallel first of all? What's your answer? Yes or no? They are not. Why they are not? Because you can clearly see that these two are not proportional. This is 1 is to 2, 1 is to 2, but this is 3 is to 5. So not proportional. So because of this, they are not parallel. So I would not use the first formula. I will use the second formula. What was the second formula? Initially, it will take some time for you to remember. But just remember this thing that you are actually taking the projection of a1 minus a2 along b1 cross b2. So when I was a student, I didn't remember this formula, but I remembered a statement that in order to find the shortest distance, I have to take the projection of a1 minus a2 or a2 minus a1 doesn't actually matter because you are modding at the end of the day. So it's the projection of a1 minus a2 in the direction of b1 cross b2. Okay. So first, you have to find b1 cross b2. b1 cross b2, you can use your determinant method. Okay. So let's evaluate it. How much does it come out to be? i times minus 10 plus 12, right? So 2. And this is minus 5 plus 6, which is 1. And this is minus 5 plus 6, which is again a1. Okay. Correct me if I'm wrong. I hope these figures are correct. Please check somebody. Thanks. Okay. So it's 2i minus j plus 0k. Okay. Thanks, Aitu. Thanks, Aditya. Yeah. And what is a1 minus a2? Just take a difference between these two. You can take it in any order. It doesn't matter. So a1 minus a2, it'll be 3i, 0j minus 2k. Okay. So you have to take a projection of this guy in the direction of this. So as we already have written the formula, we'll follow the formula. Okay. So modulus of dot product of these two, which is 6 by magnitude of this vector. Magnitude of this vector will be under root of 2 square plus 1 square. So it's actually 6 by root 5 units. Okay. So these two lines were actually, what is the conclusion? These two lines were actually, what type of lines? SQ lines. Okay. Is it fine? So I have completed this topic, but if I get more time, I will be doing more problems. Maybe in one of the extra Sundays, which I'll call you, I'll just call you for more and more questions. Meanwhile, tomorrow, I will get started with planes. Okay. Thanks a lot. All right. Take care. See you tomorrow again. I'll be putting the notification for tomorrow's session in a short while. Thank you. Goodnight.