 So planes, so planes is yet another three dimensional figure actually it's a figure which is going to be very, very useful for you in your undergrad courses also because you'll be coming across three dimensional figures in your undergrad like ellipsoid hyperboloid cone, etc. And you realize that team plays a very, very important role in those kind of three dimensional figures as well. So what is a definition of a plane? No, you can take one week to complete the assignment, not an issue. Okay. Don't worry about assignments. You can just take one week of time to finish it off. So yes, so what is the definition of a plane? Even though this definition is not of much significance but still officially writing the definition of a plane. So a plane is a surface. So a plane is basically a surface is a surface such that such that if any two points are taken on it. If any two points are taken on it. Two points are taken on it. Then the line segment joining them. Lies completely on the surface. Okay. Joining them lies completely on that surface. Okay. So as I told you this definition doesn't have much mathematical significance is just that we are defined a plane like this. Now please try to understand in order to get a unique plane. In order to get a unique plane. We need to have an information. Which basically tells us. Let me just draw a structure of a plane here. So in order to get a unique plane. We need to have an information about two things. I should either know directly or indirectly a point on the plane. Okay, let's say position vector a and I should also know the direction ratio or cosines of the normal to the plane. Let me make it in white color. Okay, so this line that you see this line is a normal to the plane. Okay, so in order to know a unique plane. For a unique plane, you should have two things given to you. You should know one of the points lying on the plane. And the normal to the plane. And the normal to the plane. This information can be given to you directly also or it could be given to you indirectly also. Right, so we need two things. These are the two necessary things that we need. We need at least one point on the plane. At least what I'm writing at least one point on the plane. Okay. And second thing is we need the direction ratios. Or direction cosines of the normal to the plane. Of the normal to the plane. Okay, so if somebody just gives me a point. It is not possible for me to get a unique plane out of it. Because passing through one point they can be infinitely many planes. Or if somebody gives me the normal to a plane and asked me to get the plane there could be many planes, which have the same line as a normal to it. Okay, so both the conditions individually will not help us to get the unique plane. So I need to have either direct or indirect information about a point lying on the plane. Or, sorry, and I must have a direct or indirect information about the normal to the plane without these two information, we cannot find a unique plane. Is it fine. Any question with respect to what is the two necessary requirements for us to get a unique plane. Okay. Now, let us try to get the equation of the plane using these two fact in vector form so let us say the question setter has provided me with a position vector on the plane. And he has also mentioned this normal vector in. Okay, let us try to find out what should be the vector form of the equation of a plane. So here I would use a concept which you have already learned in your dot product. So when I take a generic point, let's say our position vector. Can I say a line connecting. Let's say a to R this line will be perpendicular to the normal. It doesn't matter whichever R you take on the plane every R on this plane is going to meet the fact that the given point, if you make a vector or from that given are if you make a vector towards a that vector would be perpendicular to this normal. Okay, so please everybody imagine this fact, or please realize this fact that any line line on a plane is always perpendicular to the normal to the plane. So any, any line, for example, let's say I draw a plane like this. Okay, if I take any line on this plane, I take it like this I take it like this I take it like this I take it like this I take it like this I take it like this. Each of these lines would be at a 90 degree to the normal to the plane. Okay, so this line is 90 degree. This line is 90 degree. Any line you on the plane will be 90 degree to it. Okay, so we are basically finding our equation of the plane based on this fact that R minus a vector will be perpendicular to your end. For all our vector lying on the plane lying on the play. Okay. So R being any gendered position vector. This condition will always hold to that is to say R minus a vector dot with n should give you zero. Right, so please note down this itself acts like a vector equation of a plane. However, we can further generalize it like this. Okay. And many times what we what we do is we write this term as a. We write this term as a constant. Let's say I call it as a D. So the typical equation of a plane in vector form would look like our dot some vector plus a constant equal to zero. For example later on we'll see many such examples but I'm giving you one such case here. Let's say if I have a vector equation like this, then basically it will represent a plane. Okay, so this is an example of a equation of a plane in vector form. Okay. The one thing for sure is when you look at this vector, you can easily figure out what is the normal to the plane. So this vector basically carries the information of normal to the plane. Normal to the plane. Okay. And this constant is basically obtained from the negative of the dot product of the given position vector on the plane with this normal vector. We will come back to this little later on in this topic. Okay. So please note down. This is the vector form of the equation of a plane. Is it fine? Any questions? Any concerns? So if the normal and a point on the plane is given to you, this is how we end up getting the equation of the plane. Let's take a simple example based on this before we go to the Cartesian form. Let us say, yes, this is acting like the D for you. D is just another, I'll tell you why we call it as a D also. When we derive the Cartesian form, we'll come to know why we call that as a D, but D is just a constant. Okay. Now, let's say I give you a question that there is a plane. By the way, plane is also called pi. Okay. Many books will say there's a plane pi 1. There's a plane pi 2. Okay. So pi 1, pi 2 are just named given to them. So let's say this plane contains a point on it, which is 2 minus 1, 3. Okay. And the normal to the plane. Normal to the plane. Okay. Normal could be going down also going up also. It doesn't matter actually. Okay. And let's say the normal to the plane is minus i or you can say direction. Ratios of the normal is let's say minus 1, 1, 2. Okay. Sorry for this. Yeah. My question is, find the vector equation of the plane. Find the vector equation of the plane of the plane. Yes, if D will be zero origin will lie on the plane. Correct. So if I use this form, which I have over here, maybe I'll be using this as of now. So our minus a, a will be 2i minus j plus 3k. Okay. Dot with N, N will be a vector which is formed by the direction ratios, which is minus i plus j plus 2k. Okay. This is going to be a zero. If you open the brackets, you'll end up getting r dot minus i plus j plus 2k. Okay. Minus dot product of these two will give you if I'm not mistaken, it'll be minus two minus one plus six. Am I right? Correct. So this dot product with this will give you minus two. This will this will give you minus one. This will this will give you a six. So it'll become r dot minus i plus j plus 2k. And this is going to be a minus three equal to zero. So this becomes your answer to this question. Is it fine? Any questions? We will discuss that, Aditya, in some time. Give me one second. I'll show that if a dot N will be zero, then the pain was passed through origin actually, because D has something to do with the distance of the origin to that plane. Okay. I'll come back to that. Okay. Meanwhile, anybody has any question or any concerns with respect to this equation that we have found out. So vector form, everybody's fine, especially when you have been given a point on the plane and the normal to the plane, whether they give you the direction ratios or cosines, it doesn't actually matter. Okay. The process still remains the same. So take a vector connecting the generic point to that given point, its dot product with the normal vector should be zero, that itself forms the basis for the equation of a plane. Is it fine? Okay. Now, let us look into the Cartesian form. And I'll be deriving this Cartesian form from the vector form itself. So just now we saw that if there is a plane, fine, let's say, and I know a position vector on it, and I know the direction of the normal to it, or if I know the normal vector to it, the equation of the plane will be r minus a dot N equal to zero. Okay. Now what I'm going to do is I'm going to give it a Cartesian shape. I'm going to call this A as X1, Y1, Z1. And I'm going to call this vectors direction ratios as A, B, C. Okay. All of you please pay attention. Don't get confused. This A, B, C are direction ratios and this A is a position vector over there. So please don't get confused. There are two A's being used but both stand for different things. However, one is a scalar quantity and there is a vector. So I'm sure you all are smart enough to know the difference between vector and scalar. Now, if I use this fact that r is nothing but X, I, Y, J, Z, K, please note r is any generic point on the plane, isn't it? So I can say r is your position vector of any generic point. So X, I plus Y, J plus Z, K. So r minus A, r minus A, sorry, r minus A would be nothing but X minus X1, I cap, Y minus Y1, J cap and Z minus Z1, K cap. So r minus A dot N. Okay. Now what is N in this case? And in this case is your vector made up of the direction ratios. So A, I plus B, J plus C, K. This should be equal to zero. That means X minus X1, I cap, Y minus Y1, J cap, Z minus Z1, K cap, dot with A, I, B, J, C, K equal to zero. Correct? On expanding this, we will end up getting A times X minus X1, B times Y minus Y1 and C times Z minus Z1 equal to zero. Right? Right? Everybody's happy so far? Any questions? Any concerns here? Okay. When you expand this, you end up getting AX, BY, CZ. And apart from that, you end up getting negative AX1, negative BY1, negative CZ1. Okay. This is what we used to call as a D because A, B, C is already used and hence we use the alphabet D to represent that constant. So everybody note this down that this will actually represent the general form or the general Cartesian form of the equation of a plane, Cartesian form of equation of a plane. So any plane existing in this world, if you write the general form, the general Cartesian form, this will look like AX plus BY plus CZ plus D equal to zero. Okay. Now you would think as if this is an extension of your 2D line equation, right? So in your 2D line, if you remember in class 11, the general form was AX plus BY plus C equal to zero. In the same general form of your 2D line, if you just add an extra dimension of Z, that means you make it as AX plus BY plus CZ plus D equal to zero, then that same 2D line actually becomes the equation of a plane. In fact, a 2D line is actually a plane projected on one of the planes, right? So I've shown this to you earlier also. If you write 2X plus 3Y equal to 5, it is actually a plane equation but you actually see it on an XY plane and hence you see it as a line. Okay. So line is nothing but it's just an equation of a plane which has been projected on the XY plane or YZ plane or ZX plane depending upon which plane are you dealing with. Okay. And that's the main reason why many of the concepts of planes will have a close resemblance with your concepts of 2D lines. You will soon realize that when we are taking other topics. Okay, you'll see that line in 2D and plane in 3D, they have a lot of similarities in terms of formulae, not only in terms of their equations, but of course in terms of their formulae as well. Yes, Anusha. Okay, now coming to your question, Aditya. If your D is zero, that means if your D is zero, that means your plane has to pass through the origin, isn't it? Because let's say if the plane is passing through the origin, then you'll have zero plus zero plus zero plus D equal to zero, which ultimately means D is equal to zero. Okay. So if there is no constant term, that curve will definitely pass through the origin and that is applicable in general to any curve. If a curve doesn't contain any constant term, that means zero is the only constant term, that means that curve will definitely pass through origin. Okay. Now, a few things I would like to discuss over here. This stage that you see here, this is very important stage of your derivation. Okay. Why I call this important is because many a times they will give you a point or they will give you certain points are lying on a plane and they will ask you to get the equation of a plane. So many a times I have seen people starting with the fact that let the plane equation be this. As you can see here, there is three unknowns involved ABC only, unlike the last one where there will be four unknowns involved. So if you know a point on a plane, and you need to assume the equation of a plane to begin with, then this is the right point to actually start the problem. Okay. That is why see I'm talking from my experience. Okay. So I've seen people. In fact, I also do the same. I know a point on the plane and of course other things will also be given to me. Okay. Just one point is not good enough for me to get a unique plane. Then my launching point would be this equation. Okay. So this is a good interesting. I know I can say expression to start with, you know, in order to solve the question of course more things will be required for you to get your A, B, and C. Okay. So please note this down. Okay. So please understand that when you're comparing R dot N plus D equal to zero. Okay. This this term that you see is actually your R dot N. Right. That means this ABC, this ABC, which happens to be, let me write it like this first, the coefficients of X, Y, Z, the coefficients of X, Y, Z. In the equation of a plane of a plane are nothing but they are the direction ratios of the normal to the plane. Okay. So when you see a plane C equation, okay, the coefficient of X, Y, Z, they convey a meaning to you. And the meaning is it gives you the vector or it gives you the direction ratios of the normal to the plane. Okay. So you will ask me said, okay, ABC gives you the, that gives you an idea about the normal to the plane. What does D give you? Okay. What does D give you? Now, in order to show that I'll have to take up the very first form of the equation of a plane, which I call as the normal form. And there I will show you what is the significance of this D? What is this constant actually shows? Now, don't say that it shows A dot N. A dot N is just an operation. But what is the geometrical significance of D? That will be clear once I take the normal form. Okay. So if let's say somebody says, I give you a plane 2X plus 3Y minus Z plus 1 equal to 0. Can you give me or can you tell me a normal to this plane? Can you give me the normal vector to the plane? What you need to do is just take these coefficients of XYZ, make a vector out of it. That's it. This will become the normal to the plane. Okay. Or anything proportional to it. Okay. So these, these constant that you see they have a meaning and the meaning is nothing but it gives you the direction ratios. Or if you just attach IJK to them, it will give you a normal vector to the plane. Please understand this fact. Very, very important. By the way, whatever apps conveyed so far, if you have understood it, whatever is going to come your way, you will be able to get through very, very easily. Okay. The chapter is almost over here itself. So all we need to do is just extend this understanding to address other sub topics involved in the chapter. Okay. What is the significance of one? As I told you, I will tell you in some time. Give me one minute. I'll be discussing about that also. Is it fine? Any questions? Anybody? Okay. So before I go to the normal form, I will like to take one question based on this. Can I go to the next slide? Have you all submitted your projects and all in school and other schools? Have you started submitting your projects? Maths projects? I say you're completing your practicals. Okay. So you're going to school for that or you're doing it from home? Last minute. Many people do that. There is no math project for you all. Okay. I mean, ISC has that I know for sure. CSE should also have no physics also. So what did they did? They have done away with the projects or like because of an exception here or something like that. Or you have a chem project. A very simple question here, find the equation of a plane which passes through 231 and having 532 as the direction ratios. Of the normal to the plane. By the way, if I am the question center, I would never use this sound brackets. I use some different kind of a bracket here. Maybe the angular brackets. But depends upon them also. Yeah. Please do this and give me the answer quickly. Give me the vector equation also and the Cartesian equation also. Vector equation Cartesian equation. Oh, is my voice breaking? Maybe I'll try switching off my camera for some time. Maybe. Now is it proper Anusha? Still breaking. Is my voice breaking for everybody or is it only for? Okay. It's fine for Gayatri. It seems. Yes. Vector form. R minus a dot n equal to zero. You may write it like this also because it helps you to simplify faster. Because ultimately after this you will be doing the next step itself. So R is R. N is nothing but a vector made out of the normal. That's it. Minus a dot n. So dot product of this position vector with this normal vector. So that will give you 10 plus nine plus two. Okay. In other words, it become our dot five I plus CJ plus two K minus 21 equal to zero or you can write it as equal to 21. Also, so this is your answer. Okay. Now once you've got this in vector form converting it to Cartesian is a cakewalk. All you have to do is just attach an XYZ to these five three two. So XYZ and this number 21 will remain as it is. So this becomes your Cartesian form. Okay. So many people ask me, sir, can I do this also? Since I know a point on the plane to three one. Can I use the fact that a X minus X one. Can I use the Y minus Y one CZ minus one equal to zero. Can I use this? Yes, very much. Where ABC are your five three and two. So if you do something like this X minus. I think this was two, I believe, right? What was the point two, three, one. Yeah, two, three, one. Oh, sorry. Sorry. Two, three. And this was one. Okay. So when you expand it, you automatically get the same answer. Minus 10 minus nine minus two equal to zero, which is as good as. Okay. Is this fine? Any questions? Any concerns? All right. So can I move on to the normal form now? Next is the normal form. See normal form. The question center would ask you. Or the question center will provide you. Let's say this is a plane. Okay. I want the equation. And I only know two things about the plane. I know how far is this plane from the origin? That means I know this distance B. Okay. And I know what is the, let's say I call this as M. I know what is the direction cosines of. Oh, M. Okay. So here L M N. L M N. R DCs of. Oh, M. Okay. Now I want to know what is the equation of this plane? What is the equation of this plane? Okay. Now, since you have already been appraised of the. Requirements to get a plane that that means I need a point and a normal to the plane. I think you are well equipped. Now to get the equation of the plane on your own. So I would request you all to please. You know, work this out and tell me what is going to be the equation of such a plane. Whose distance from the origin is small p. And the direction for science of the normal drawn from the origin. This is a normal drawn from the origin onto the plane. So normal drawn from the origin onto the plane has the direction cosines L M N. How will you solve this question? Or how will you deal with this situation? So one thing is sure that you know the normal to the plane. Okay. So normal to the plane is fine. We know L I M J N K will be the normal to the plate. So normal requirement is over. But who will fulfill the point requirement? I don't know a point on the plane right now. So where is the point? I need to know position vector lying on the M is there. But what is M? I want to know that. How will I find M? Or what is the coordinate of M? Of course, you're right. M is our desired point. But what is the coordinate of M? That's the question. So who will tell me the coordinates of M? The root of that ugly is the expression for the coordinate of M. Right. That's absolutely right. See all of you please pay attention. Okay. The position vector of M. Okay. Is nothing but this vector itself. So L I M J N K. It gives you a unit vector. I'm sorry. It gives you a unit vector. If you multiply it with the magnitude of. O M that will give you the O M vector. So this anyway is a unit vector. Right. So you probably know that the vector form by the direction. Coscience is a unit vector. And this is the magnitude. So magnitude times the unit vector will give you the vector. That is, you know, the normal way of generating a vector. And the length of O M is P itself. Correct. So the position vector of M is nothing but. P L I cap. P M J cap. P N K cap. So if you want to use a Cartesian coordinate system, you can say this point is PL comma PM comma PM. Okay. Now it is up to you whether you want to use Cartesian form or vector form to get the equation. But I think the job is almost over. So R minus a dot any equal to zero. That is our equation of a plane. So R dot N minus a dot N equal to zero. So R N is L I M J N K. And minus a dot M. Now this is like your a vector. This is your, this fellow is your a. Okay. This dot product with this. So that will give you correct me if I'm wrong. P L square. PM square. P N square. Is it fine. Everybody's fine till this stage. Any questions, any concerns that you have these. Let me know. R N. This is your N. Okay. And minus a is this guy. This fellow is a dot N. Equal to zero. So let's simplify this little bit more. So I'll give you R dot L I M J N K. And here if you see, if you take a minus P common, you get L square M square N square. And what is L square M square N square. This guy is actually equal to a one. So you end up getting the equation of a plane to be this. Okay. So this is the vector form. If you write the same in Cartesian form, it will look like this. L X. M Y. And Z minus P equal to zero. So this is your Cartesian form. Getting my point. Okay. Now everybody please pay attention. This equation I am now going to compare with the general form of the equation of a plane. Okay. I'm going to now compare it with the general form of the equation of a plane. Okay. So now pay attention. Very, very important. ABCR direction ratios of the normal. L M N other direction cosines of the normal. Please understand. ABCR direction ratios of the normal to the plane. L M N R direction cosines of the normal to the pain. Okay. So what I'm going to do is I'm going to divide by under root of a square B square C square throughout. Okay. So when you divide the direction ratios by under root of a square B square was D square. What does it actually give you? This is actually become the direction cosines. They become the DCs. Correct. So Diaz will become the DCs. Correct. So can I say this will become LX. M Y N Z plus D by under root of a square B square C square equal to zero. Okay. Now please let me tell you that they may differ from each other in terms of sign. Okay. So but that is not my interest area. So this is now in telling you that this fellow P that you have and this fellow this. They are actually same thing in terms of magnitude. They can differ in sign. That depends upon C. As I told you if from direction cosines, sorry, action ratios you want cosines, you can get two different answers plus minus both. But right now my interest area is not resolving whether it is plus or minus. But what I want to discuss here is that the modulus of minus and the modulus of this they will match. Okay. And since you know modulus of P is a P because P represents a distance on the plane. And there's no point putting a modulus on under root of something because there's this that is anyways positive. So this is a very, very important observation that you should note down. Okay. So when I was talking about the general form of the equation of a plane ABC as I told you. They represent the direction ratios of the normal to the plane. Right. And that constant D that you had, if you take a modulus of that constant and divide by under root of the squares of the coefficient of XYZ. This will actually give you the distance of the plane from the origin. Distance of the plane from the origin. So today when we are doing the official formula for distance of a point from a plane, we will see that we will obtain this expression from that discussion as well. Okay. So this is something that we will take up once again when we are trying to talk about and when we are discussing about the distance of a point from a plane. Okay. So what is the summary of this entire event summary of the entire event is the normal form of a plane. So this equation is called a normal form of a plane is given as LX plus MI plus NZ equal to P. Correct. Where LMN are the direction cosines of the normal drop from the origin onto the plane and P represents the distance of the origin from the plane. Okay. Second thing that we learned here was in the equation of a plane in the equation of a plane. This and this represent the direction ratios of normal to the plane. And this number D, if you divide by under root of A square plus B square plus C square, it actually gives you the, I can say modulus of this, this actually gives you the distance of the origin from the plane. Okay. Yes. And it is very similar to what we learned in our 2D lines also. That is why I kept, I know I was saying, I was telling you earlier that you'll find a lot of resemblance. Okay. A lot of resemblance at different fronts between a straight line in 2D and a plane in 3D, 3D geometry. Is it okay? Any questions? Anything that you would like to ask here or should we start solving questions? Yes, that's what that A dot N actually divided by, I'll just tell you, so that A dot N magnitude divided by modulus of N actually gives you that P. Got it, Aditya? We will derive this. We will derive this as well. Okay. We will derive this as well in, when we are talking about distance of a point from a plane. Let's take questions. Okay. Let's take this question. Find the vector equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector. Okay. That means normal drawn or normal to the plane is 2i plus j plus 2k. So now the two information given to you is distance of that plane from the origin and of course normal is anyhow given. So the point is not given, so as to say. So as I told you, directly or indirectly, you should be knowing that information. So please do this question and give me a response on the chat box. You may use the formula just now derived. No need to start it from scratch. You can use the derived formula in the previous line. Okay, we have a, are you sure? I can only see one correct answer so far out of at least five or six of you have responded. So as of now I can see the right answer coming only from Aditya. Okay, guys. I think most of you have not understood this properly. What did I tell you? r dot l i m j n k equal to p is the vector form. Are you getting my point? Okay. So the vector given to you, this is not a unit vector. This is a vector which is made, this is any vector collinear. You can say this is a non-unit vector along the normal. Okay. So if you want to use this figure eight, if you want to use this figure eight, you should multiply, you should do, sorry, you should do a dot product of r with the unit vector along the normal. Are you getting my point? Correct. So when you do this, when you do this, you will end up getting, in fact, two answers will come from here. By the way, when I am finding this, it will become two by, if you divide it by the magnitude of this vector, magnitude will be three actually, right? So if you divide this by the magnitude of this vector, which is two square, one square, two square, you'll end up getting two by three i cap, one by three j cap, two by three k cap, right? So it's two by three i, one by three j, two by three k equal to eight. That is one possibility. In fact, you can get r dot two i plus j plus two k equal to 24, which is nothing but two x plus two y plus, sorry, two x plus y, not two y, my bad, two x plus y plus two z equal to 24. That is one possibility. This is one answer. You could also have a plane which is r dot negative two by three i, negative one by three j, negative two by three k equal to eight. That will lead to r dot, okay? That is nothing but minus two x, minus y, minus two z equal to 24. That is nothing but two x plus y plus two z equal to negative 24. That's another answer, okay? See why? Because the same normal that you have, there could be, let's say this is our origin. One plane could be here, one plane could be here, another plane could be in other direction and the same distance eight units. Are you getting my point? Same normal, isn't it? So this, this plane also could be at eight units and this plane could also be at eight units, correct? Anyways, I was expecting any one of the answers to be given, but most of you have misjudged the fact that you have to have a unit vector over here, unit vector normal to the plane. Hence the components should be the direction cosines, getting the point, okay? So people who made a mistake, now you are aware, what is the mistake? Okay. Yes Siddhant, you had a question, sorry, yes tell me. You can speak it out also, no problem. Or you got it, okay fine. Is this okay? All right, good. Let's take another one. See vector, three dimension geometry, it's a very simple topic but provided you should understand the basics. If basics is clear, you are going to get the topic anyhow. I think I went a little higher, I have to have some problems from here. I already did. Okay, let's take this simple question also. Find the unit vector perpendicular to the, to this plane. And let's say also add one more part to it and also find the distance of origin from the plane. Distance of origin from the plane. Very simple, should not take you more than 10 seconds. Answer should be there on the chat box already. Correct. No Anusha, again you are making the same mistake. Janta, you are still making the same mistake. Hariharan has got it right. See this is a vector perpendicular to the plane or normal to the plane. So you want a unit vector. So everybody has done this fine. Okay, no doubt. Divide by its magnitude. Okay, fine. That is nothing but 2i plus j plus 2k by 3. Okay. But remember, this is actually the d part of it or you can say mod d part of it. Till you divide mod d by under root of a square plus b square, you won't get p. Correct. So you are basically, you have an equation of this nature. 2x plus y plus 2z equal to 5 or let's say minus 5 equal to 0. So this fellow is your d. Right. So to find p from d, it is mod d by under root of a square plus b square plus c square. You have to do this activity. So that's mod of negative 5 by under root of 2 square plus 1 plus 2 square. So the answer is 5 by 3 units. Okay. Are you getting my point? Or if you want to directly use the formula, you need to bring it to this form. Okay. So in order to bring it to this form, you must make this as a unit vector. So for making it as a unit vector, so what will you do? You will divide by 3 throughout. So you divide by 3 throughout. So this 5 will also get divided by 3. So you are dividing left side in order to make a unit vector out of this. Then you have to divide the 5 also by 3. Then this can say you can actually take a modulus because the question center can give a negative value also. Take a modulus of whatever quantity that you get over here. That will be the distance of the origin from the plane. Is it clear? Any questions? Let me check if I have one more. No, I think, yeah. This we can take. Find the distance of this plane from the origin. Plane and simple. By the way, later on when you learn the formula of a distance of a point from a plane, maybe you will not use whatever we have learned right now. So, but as of now, whatever we have learned, let us utilize that to get the answer. Correct. Absolutely right. Okay. So as I already told you, this is your D part. Okay. A is your 2. B is minus 1. C is minus 2. D is minus 9. So this P will be mod of D by under root of A square plus B square plus C square. Right. That's nothing but 9 by under root of 2 square plus 1 square plus 2 square. That is 9 by 3. 3 units. Yeah. 1, 2, 2, 2, 2, 1, 2, 1. The modulus is going to be 3. Yes. And one more is that gives you 7. That is, I think, 3, 6, 2. 3, 6, 2, if there is the components are 3, 6, 2 of a vector, it's modulus is going to be 7. So some quick formulae that you can use there. All right. So we are not going to talk about. Sorry. We are not going to talk about the intercept form of a play. Intercept. See, I'm not going to teach you a new concept. Whatever we are going to learn is just based on the initial discussion that we have. I want to get somehow normal to the plane. And one point at least. That is, that is more than enough for me to get to the equation of any plane that I want. Okay. So all these forms that I'm introducing it to you, they directly or indirectly give you that to information. You just have to utilize that and use the fact that R minus a dot n equal to 0. That's the whole principle on based on which the plane's concept is based out of. So what is the intercept form of a plane? So let us say there is a plane. There is a plane which cuts your coordinate axes. I'm just doing a right-handed coordinate system here. So there is a plane which cuts your coordinate axes at, let's say the following points. It cuts the x-axis at a 0 0. That means it's x-intercept is a 0 0. And let's say it cuts the y-axis at 0 b 0. And let's say it cuts the z-axis at 0 0 c. Okay. So let's say this is the plane. I'm just doing. Please don't get this wrong. It is not a triangular plane. Plane is a infinitely extending you know, geometrical figure. But I'm just connecting the part where it is cutting the x, y, z. So this is your part of the plane. Okay. So this is a part of the plane. Okay. So what is the equation of the plane? So how would you find the equation of such a plane? Now I'm putting the ball in your quote. Tell me I know the x-intercept, y-intercept and z-intercept. Okay. So x is x-intercept is a, y-intercept is b. Of course, a, b can be negative also depending upon how it is cutting it. And z-intercept is c. What is the equation of this plane? How will you figure it out? Okay. There are so many ways to solve it. So many ways to solve the question. Good. Okay. Area of a triangle. How will that help you get the equation? It's actually not a triangle. It's a plane. I'm just, I'm just connecting the points where, which is cut by that plane. So don't think it's a, it's a triangular plane. No, no, no. Plane is, plane is infinitely extending in all directions. Right. So that triangle is just a part lying on the plane. Okay. All that is not required, can't you? You don't need ortho-centers, centroids, et cetera, to get anything. You don't need area of the triangle. Okay. Go for it for them. I mean, you're on the right track. Go for it. And do let me know the final result. Whatever you're getting. Of course, your answer will only contain a, b and c. Of course, variables x, y, c will be also there. So give me a response of whatever you have thought on the chat box. Whatever you're getting from your idea, just solve it and tell me what are the answer coming up. And then we'll discuss it out. Very good, Aditya. I think your answer is absolutely right. So Aditya has got it. Anybody else? Anybody else? Now what I'm going to do is, I'm going to take a generic situation over here. I'll come back to this question. This question is a very specific form of a generic situation. So let's take a generic situation. Let us say I have a plane. Okay. And this plane, this plane contains three position vectors. I'm just making a plane here. So let's say this plane houses three position, three points on it. Okay. Normally I don't want to use PQR because R is already used as a generic position vector. Maybe let's say I can use UVW. Let's say UVW, R your three position vectors on this plane. Okay. Right. Now. Giving me three points. Is it good enough information for me to get the equation of a unique plane containing them? Yes or no? I think knowing three points on the plane is good enough to get the unique plane which contains them. So how would I get the equation of a unique plane containing them? So how would I get the equation of that unique plane? So you would be thinking, sir, I already know three points on it. I need one, but I'm given three points. Okay. So there is no dearth of points. What is required is the normal. So how do you find the normal to a plane, which houses these three position vectors? So what do you do? You will make any two vectors connecting any two vectors. Need not be the vectors which I'm showing you. You know, make any two vectors using these three points. Okay. So this vector is V minus you. This vector is W minus you. Right. So can I say the normal vector would be the cross product of these two? Am I right? The normal would be the cross product of these two. So once I know the normal vector. And there are only three points given to me. Can I not use? Can I not use the initial vector equation, which I discussed with you are minus eight dot n equal to zero. Right. So let us utilize that. So first of all, when you expand it, this is what you're going to see. I'm just writing it down for you. So V cross W, you will see. You will see V minus V cross you, which is actually you cross V. Okay. And you will see W cross you from here. In short, you will see you cross V, you cross V, V cross W plus W cross you. Correct. This is your N. Right. Now R minus a, a could be any of the three vectors. Let me take a U for the timing. Dot N will be equal to zero. Correct. Which means R dot N minus U dot N will be equal to zero. Right. Now U dot N means you are taking a dot product with dot product with this whole N vector. Okay. So I'm just writing it down. U cross V, V cross W, W cross you. Okay. Now let me expand this also. So here also ideally I should write the same thing. R dot, R dot, U cross V, V cross W, W cross you. Okay. Now when you expand it, what is R dot U cross V? Can I say it is going to be box R U V? Okay. R dot V cross W is box R VW. And this will be R W U box or scalar triple product. Okay. And here you realize that you will end up getting U dot this as zero. And here it will get box U VW and box of U W U will, that will also be zero, isn't it? In other words, this becomes your equation of, please note this down. This becomes your vector equation of the plane passing through three points or passing through three position vectors, U VW. Okay. Now I'm actually going to use, I'm not actually going to use this, but this is just an alternative representation. I'm going to use our initial, you know condition over here, which I'm going to show you a little while, but first note this down. Please note this down everybody. So what I'm going to use to solve this problem is not this formula, but this is just an alternative representation of an equation. And I've seen many questions framed on this fact. So they will give these as your options and say which option represents the equation of a plane passing through three given points. Okay. But this is not a working formula to solve the equation of a plane. The reason being, who will sit and calculate these four, you know, STPs, right? It's a waste of time, right? Okay. So this is good to know, but this is not a working rule. Okay. What is working rule here? Let us come back to the figure. I hope you all copied this. Yeah. Please take your time. No worries. Yeah. The working rule here is if I take any generic position vector, let's say R. Okay. The working rule is these vectors, which I'm showing you right now on your diagram, R minus U, V minus U, W minus U, R coplanar. This is the working rule to get the equation. R coplanar. Sorry. I'm just sort of playing that. It's coplanar. Okay. So using this fact, you are going to get the equation of a plane passing through three points, which will also help us to get the equation of a plane in the intercept form, which I just gave you a little while ago. So all you need to do is this implies that the STP of R minus U, V minus U, W minus U is going to be zero. Right. So this itself is a working principle. This itself is a working principle. Okay. Of finding, in fact, you can see that this particular thing will come at this stage. If I use n as here, so if I substitute over here, I'll get R minus U dot V minus U cross W minus U. Okay. As a zero. This is actually a representation for box product of R minus U, V minus U, W minus U. Okay. So they're not different from each other. But yes, when I expanded it, it became useless for me. This is, this is all I need. This is all I need to get my equation of a plane. Okay. So in order to solve this question, I can use my determinant. So let us use this to solve our question. So R is nothing but x i, y j, z k. U as per our question, I'm not mistaken. U is AI. V is BJ. And W is CK as per our question. See here. This point is what? This point is as good as saying AI only, right? This point is what? As good as BJ. This point is what? CK. Okay. So think as if you are now been provided with UVW. And of course, R is this. We need to get the equation. Now everybody please pay attention. So I will make a determinant whose first row is R minus U. R minus U will be X minus A, Y, Z. Correct. Second row is V minus U. V minus U. Correct me if I'm wrong. It's minus A. Okay. B, 0. And W minus U. W minus U will be minus A, 0 and C. Okay. So put this to 0 and expand this. This determinant will give you the equation directly. Are you getting my point? What I'm trying to say? So R minus U is your first row. V minus U is the second row of the determinant. Correct. And W minus U. That is your third row. So let us expand it with respect to the first row. So X minus A. This will be BC minus 0. Minus Y minus AC. And Z. In fact, that will give you plus AB equal to 0. So if you simplify this, it will become XBC, YAC, ZAB equal to ABC. So see ABC, I'm sending it to the right side. Now divide by ABC throughout. So when you do that, you end up getting the equation to be something like this. And this is something which I believe is familiar to you. Please recall your intercept form of the equation of a line in your 2D plane. X by Y by B equal to 1. It's just that now we have an extra dimension added to it. Fine. And you can also write down the vector form from here also. R dot BC ICAP. AC J-CAP. AB K-CAP equal to ABC. Okay. You may also write it like this. R dot 1 by A ICAP. 1 by B J-CAP. 1 by C K-CAP equal to 1. Both are fine. Is it fine? Any questions? Any concerns? So through this form, I basically took up a small theory where I was, where I wanted you to tell how to find the equation of a plane passing through three given points on that plane. So intercept form was a very special case under that. Okay. So please note this down. And if you have any questions, do let me know. Next slide, I will write down the Cartesian form for the same. So for this, this is the vector form. I will write down the Cartesian form. Part in yellow. Which part? This part in yellow or this one? Yeah. See when you write a box product, STP, scalar triple product, right? So let me, let me take you back to your vector days. I think some vector concepts have to be revised. If you have a box of ABC. Okay. If you write, if you want to write it as a determinant, how do you write it as a determinant? The first vector components, you write, let's say your vector is A1, A2, J, A3, then second vector components, you write third vector components, you write this is how you calculate the STP. Yes or no. So this becomes the, the component of this becomes your first row, component of this vector becomes second row, component of this vector becomes the third row. So here also, once we have got the fact that the equation of our plane would be obtained by using the fact that R minus U, V minus U and W minus U are coplanar. So when three vectors are coplanar, their STP is zero. So I got this equation. So R minus U, that means if you subtract AI from here, what are the components X minus A, YJZK. So that is what I wrote over here. I'm basically taking this, this situation where our plane is cutting the coordinate axes. So my U is nothing but AI. This is my U thing like that. This guy is my, this guy is my V. This guy is my W. So I'm just putting that in the equation over here. So R minus U components are here, V minus U components are in the second row, W minus U components are in the third row. Expand the determinant. That automatically gives you the equation. I will generalize that also. In the next slide, I will generalize this also. So I'm going to give you a Cartesian form for this vector form. Oh, I didn't get that P, Q, R, Z, equal to one substitute three points and solve. I'll come to that method also. Okay, all of you please pay attention. When I wrote the, let me just draw a diagram first. So let's say this is a plane pipe, which contains U, V, W. Then we got the equation of this plane as box product of R minus U. By the way, please don't get me wrong here. You can actually, you know, take this in any form you want. For example, you always have to take R minus U, V minus U, W minus U. No, not necessarily. I could have taken R minus V and here I could have taken U minus V and here I could have taken W minus U also. Any three vectors formed by any three vectors formed by these four points can be taken. Okay, so in this example, I've actually taken this vector, this vector and this vector. You can take any four vectors, any three vectors formed by these three points. Okay. It's not necessary that I have to write it like this. Now, the very same thing that I have written over here, had I given this, had I given the coordinates of these points? Let's say X1, Y1, Z1, X2, Y2, Z2, and X3, Y3, Z3. Okay. Y3, Z3. Then the same expression that I have over here. So this would become X minus X1, Y minus Y1, Z minus Z1. This would become X2 minus X1, Y2 minus Y1, Z2 minus Z1 and this would become X3 minus X1, Y3 minus Y1, Z3 minus Z1 determinant equal to zero. So this itself will become the Cartesian form of the equation of the plane. Is it fine? Any questions? And again, you have full, you know, liberty to choose any format of this determinant. Okay. For example, you may have a determinant whose first row is X minus X3, Y minus Y3, Z minus Z3. And here you can take, let's say difference of X1 and X3, Y1 and Y3, whatever you are using in one column, maintain the same consistency in the other column also. Don't start jumping around. Okay. So essentially you have to take any three vectors and make them co-planar. Correct. So when you are making them co-planar, you are automatically getting the STP to be zero. That means automatically this determinant would become zero and automatically it gives you an equation. That is my equation of the plane. Is it fine? Any questions? Okay. So in light of this, if you want to use this formula to solve the previous question, so your X1, Y1, Z1 was, I'm just rewriting the whole, this thing in the intercept form, this was A00. Okay. Your X2, Y2, Z2 was zero, B0. And your X3, Y3, Z3 was zero, zero C. Correct. So if you use this, you will get a determinant X minus A, Y minus zero, Z minus zero. Sorry for that crooked line. Yeah. And this is X2 minus X1. X2 minus X1 will be nothing but minus A. Y2 minus Y1, B. Z2 minus Z1, zero. Again, X3 minus X1, minus A. Y3 minus Y1, zero. Z3 minus Z1, C. So put it to zero, expand it, you'll end up getting the same stuff. Okay. So same as previous slide. Is it fine? Any questions? Any questions, any concerns? Let's take some problems based on the same so that you have a hands-on experience on solving that question. Find the equation of a plane which passes through these three points. Also find the unit vector perpendicular to this plane. Now it is up to you whether you want to solve this problem coming from the scratch or you want to already use the derived result. It is up to you. Okay. So take your approach and do let me know the final result. Especially the equation of the plane. Unit vector perpendicular to this plane we can know from the equation itself. So please solve the first part at least and give me a response on the chat box. Okay, very good. A Hadithya answer is different from the other one. Okay. So two different answers I've got already. Okay, we'll check. Who is correct? Who is not? Anybody else? See, I'll show you various ways to solve this question. Okay. Aiyo. Hariharan's answer is completely different. There was some resemblance between Hadithya's answer and Weber's answer but Hariharan is totally different answer. Okay. This is what hurts me. Different people giving me different answers. Okay. So four options. Four answers I've already got. Okay, Ruchita. Fifth answer I've got now. Aiyo. Before I see 7th and 8th and 10th and 11th and 100th answers let me solve this question. See, let us use the determinant that we had discussed. Okay. Take any one of the points. Let's say I take the first one and do x minus 2, y minus 2, z plus 1. Okay. Now the second though you take a difference of 32 coordinates and write them down. For example, if I do, you know, difference of b minus a. So 1, 2, 3. Okay. And you can take a difference of let's say c and b. So 4 minus 4, 4. Okay. Put it to zero. Right. Let's expand it. You can do also one simple activity here since there is a 4, 4, 4, you can take a 4 common and you can make it vanish because anyways, there is a 0 on the right side. Okay. So make it simple as possible. Why to aggravate the simplification process. So this is going to be 2 plus 3, 5. Okay. So coefficient of x has to be 5. Okay. Those have got it well done. Next. Negative y minus 2. This is going to be 1 minus 3, which is negative 2 and z plus 1, z plus 1. Minus 1 minus 2, which is minus 3 equal to 0. Okay. Let's expand this. So you'll end up getting 5x plus 2y minus 3z first of all and constant terms will be minus 10 coming from the first minus 4 coming from the second and minus 3 coming from the third. So the answer to this question is 5x plus 2y minus 3z equal to 17. Let us check who all got this right. Kinshukh Kalia has got it right. Ritu H has got it right. Aniruddha, no, no, no, no. Aditya has got it right. Vabhav, constant is wrong Vabhav. Is it fine? Any questions? Okay. Now, this is the one way to solve the question. See, if you want to take a vector root also, that also you can do it. Okay. Let's say, let's solve this question completely from vector point of view. Right. So from vector point of view, if I want to solve this question, what I'm going to do is I'm, I know two position vectors on the, I know three position vectors. Let's say, you know, A, B and C. Okay. So these are your three position vectors, A, B and C. A, B and C. So I will get the normal, right? How do I get the normal? Take a cross product of any two vectors. So let's say, let's say I, let me, let me name them as, okay, it's fine. Okay. Take cross product of any two vectors, any two vectors. See, I'm writing B minus A, C minus A. That doesn't mean you will also do the same. Any two vectors take the dot, cross product. I'll give you the normal vector. Okay. So let us do this activity. So B minus A, C minus A cross product I want. So B minus A, B minus A will be one, two, three, okay. C minus A will be five, minus two, seven. Okay. So this is your end vector. So let's expand it. So that'll give me i cap, 14 plus six, 20 minus J cap, seven minus 15 minus eight and plus K cap. That is going to be minus two, minus 10, which is minus 12. Now, once you know n, use the formula R dot n minus A dot n equal to zero or R dot n minus B dot n equal to zero or R dot n minus C dot n equal to zero. That will all give you the same result. Please take my word for it. Okay. So R dot 20 i plus eight J minus 12 K minus A dot n. Now, see, A was two i, two J minus K. Okay. So let me write that down also. So A was two i, two J minus K. So what will be A dot n? A dot n will be 40, 16 and plus 12. Okay. That's nothing but 68. So minus 68 equal to zero. So automatically you drop a factor of, you drop a factor of four. So five i, two J minus three K minus 17 equal to zero. Correct. And that is as good as saying five X, two Y minus three Z minus 17 equal to zero. That's the same as the answer that we had got here. Correct. So this is coming from your basics of your, you know, vector form of the equation. The another approach which normally people use in school also. Third approach. Okay. So three, three approaches are there. So I'll give you the third approach. We need to maintain, means see what I meant to say was Aditya, let's say you're doing, you're doing three minus two. Then you should not do two minus four like that. I was trying to say getting my one take any three vectors. Okay. The fact that they are co-planar is all I need to use to solve this question. Okay. Now third method, which I'm going to discuss here. Everybody please pay attention to two minus one. Let me write down the points first of all, because I'll keep, I'll keep shuffling to the right and left of the screen to two minus one. What was the other one? 342. 342. And what was the third one? Seven something seven zero six seven zero six. Now this is a method which normally school in school, also people will use. They will say, let the plane be, let the plane be a X minus any one of the point they will choose. Let's say X minus two B X Y minus two CZ plus one equal to zero. Remember when I was discussing the equation of a plane in the cartesian form, I told you that this may be your starting point of a question. If you know any point on the plane, you can start the equation of a plane by this. Now this plane is also be supposed to be satisfied by 342. So C since 342 satisfies it. Can I say, let me call it as one. So can I say if I put a 342, I'll get a to be three C equal to zero. Since seven zero six also satisfies one, you'll end up getting five A minus two B plus seven C equal to zero. Okay. Now using this two equations two and three, you need to get the ratio in which ABC lie. You don't have to find exactly ABC value. In fact, you cannot find ABC value because two equations three unknowns, you cannot solve it. Okay. So what do you do in such cases? We follow the class 10th cross multiplication method, which you had all learned in your, you know, 10th days. So this is what we do see. I will write a minus B and C. Okay. And the cross multiplication technique that you have learned. So under a hide this 14 minus minus six, which is 20. Okay. Then hide this seven minus minus 15, seven minus 15, which is minus eight. Okay. And under C minus two minus 10, which is minus 12. So just drop this minus sign here, minus sign here. Okay. And take a simple ratio of you can divide by four. So five two minus three. Okay. So take these values and dump it in, put this value and dump it in your first equation. That is this equation. So take these values and put them over here. Okay. So you automatically get five X minus two. Plus two Y minus two. And minus three Z plus one. Okay. So this will end up giving you five X plus two Y minus three X minus two Z minus two. So, That is the same answer as we have gotten. So it is up to you which method you want to use? It is up to you, which method you want to use to solve this question. So three methods I have shown you. Was solving the same question. But if I were you, I would use the first, you know, method to solve it because if I'm a competitive level exam aspirant, I would definitely like to save my time. Okay, so let us answer the second part of the question, unit vector perpendicular to the plane. So the unit vector perpendicular to the plane will be nothing but five i to j minus 3k divided by its magnitude, okay? So I think this will come out to be 25 plus 429, 29 plus 9 root 38. In fact, plus minus both are possible because you can have a normal going in either of the two direction. Is it fine, any questions, any concerns? Any questions, any concerns? Any method that you would like to copy, let me know, I can drag the screen to that side. This is not required, you already know this is. Clear everybody? Okay. So the next thing that we are going to talk about is angle between planes. Okay, let's talk about angle between two planes. So if you see here, we have actually covered up certain types of situations that will be given to us. If somebody gives me a normal and a point, I know how to deal with that person or how to deal with that question. If somebody gives me the distance of the origin and the direction cosines or ratios of the normal from the origin to the plane or, it's actually not a normal from the origin to the plane, it could be like any normal to the plane, okay? So that itself is taken up in the normal form. If somebody provides me the x, y, z intercept made by that plane on the coordinate axis, I know how to deal with that also. If somebody gives me three points lying on the plane, then I know how to get the equation from that case as well, okay? So please understand, these are different, you can say varieties of questions, but they all are based on your basic understanding. Okay, so don't get perturbed at all, this situation I've never seen before. See, try to make sense from that situation, what all you can find out that will help you to get the equation. Okay, so if you have no other way out, just try to get a point and the normal, that is good enough for you to crack that question. All right, so we'll not talk about equation of, sorry, angle between two planes, angle between two planes. So angle between two planes is nothing, but it is the angle between the normals to the plane. Okay, so let us say if somebody gives you two planes, let's say one plane is like this, okay? And another plane is like this. Let's say this plane, blue plane is your pi one plane, and this plane is your pi two plane. And I want to know what is the angle between the planes? This angle, let's say theta, okay? So please understand here, the angle between the planes is nothing, but the angle between the normals. So let me just draw normal to pi two. Let me call this as N two. And let me again draw a normal to pi one. Let's say this is your N one. So this angle will also be theta, do you all agree? So angle between the two planes is nothing, but the angle between the normals to the two plane. So this basically makes your life simple. So let's say pi one plane equation is R dot N one plus D one equal to zero. And let's say pi two equation is R dot N two plus D two equal to zero. Then the angle theta can be found out by using N one dot N two by mod N one, mod N two. In case you want to know, in case you want to know, okay? The acute angle between them. If theta is acute, take a mod. Okay, if you want to make it as an acute angle, just take a mod, is it fine? So there's nothing different from whatever we have learned in our vectors. So the dot product is the, you can say the fallback point for us to get the angle between any geometrical figure whether it is two 3D lines or whether it is two planes. Okay, so if you have been given the same two equations in let's say Cartesian form, something like this. Then the cost of the angle is given by a mod A one, A two, B one, B two, C one, C two. If you want your theta to be acute, you have to take a mod by under root of A one square, B one square, C one square, A two square, B two square, C two square. Simple as that, right? So whether in vector or whether in Cartesian, you can safely find out or you can easily find out the angle between the two planes. So here, let us quickly note down the condition for parallelism, condition for parallelism. So if two planes are parallel, please note, N one will be lambda times N two. That means N one and N two will be proportional to each other. N one, N two is proportional to each other. So when two planes are parallel, both the normal vectors would be proportional to each other because they will be like collinear vectors when they are parallel, isn't it? And you will realize here that A one by A two, B one by B two and C one by C two, ratio will be the same, okay? That means the coefficients of their X, Y, Z terms would be proportional. And what is the condition for perpendicularity? Condition for perpendicularity would be their dot product of N one and N two would be zero. Okay? And in terms of Cartesian, A one, A two, B one, B two, C one, C two would be equal to zero. So these are the conditions. So I've given it in vector form as well as when it is given in Cartesian form, depending upon what situation you are dealing with, you can use it. Is it fine? Any questions, any questions, any concerns with angle between the two planes? Again, I'm repeating this, dear students. There is nothing new that you are learning here. Let me remind this again and again. Everything comes from your basic understanding of vectors. Let's take questions. Let's take questions. Find the vector equation of a line passing through, okay? So this is a mix of a line and a plane question. Find the vector equation of a line passing through this point and perpendicular to this plane. So this is a plane and this is a line which is perpendicular to this plane and it is passing through a point whose position vector is three i minus five j plus seven k. Or in other words, three minus five comma seven. This actually tests your basic concepts. Nothing related to whatever we have discussed so far. So I just picked and happened to get this question so I thought I would do it. Done. Find the vector equation of a line passing through this and perpendicular to this. So assign. Hello, am I audible? My test, my test, my test, yeah. My mic is working. All right, Aditya. See, guys, in order to get the equation of a line you need a point on the line. So this is a point on the line and you need the direction of the line, right? I need a vector in the direction of the line, correct? So the point is already provided to you. Three i minus five j plus seven k, okay? Put a lambda and you need a vector in the direction of the line. Now remember, the planes x, y, z coefficient themselves is the normal to the plane, right? So direction of the line is matching with the direction of the normal to the plane. So use three i minus four j plus five k itself over. That is the answer for the equation. Actually, I wanted to test you whether you know the fact that the normal to the plane is actually in the direction of the line and the normal to the plane is obtained from the coefficient of x, y, z. So that will behave as your b vector here. A is already given to you, got it? Any questions? Can I go to the next slide? Okay, so we'll take some in fact, there are a lot of questions pending to be taken. Okay, we'll start with this one. We'll come to the intercept form a little later on. Find the angle between these two planes. Now see the question setter has tried to confuse you by giving one equation in Cartesian and one equation in vector. But I'm sure you are smart enough to figure that how to manipulate with this. Yeah, find the angle between the two planes. Give me any of the two angles. That's fine. Can I go to Kinshukh? Oh, yes, yes, yes, yes, yes. Yeah, I'm so sorry. It has to be 2z. Can't be xx more than that, okay? This is z. Yes, simple. So as I was talking about n1 and n2, we can figure that out very easily. n1, you can take the coefficient of x, y, z that will act like your n1. And n2 is very explicitly mentioned here, six i plus three j plus two k. So in order to get the angle between them, you can use the fact that the angle between n1 and n2, between the two normals, that is going to be your angle between the planes. I.O. Yeah. So this is going to give you, this is going to give you 12 plus three minus four upon three. And as I already told you, this is seven. Familiar figures, save your time. So theta is cos inverse 11 by 21. Clear? Good enough. So as I told you, I have not taken a concept on intercept form. I'll take a question on that as well. I.O. They have given the answer also. Okay, anyways. A plane meets the coordinate axis in ABC such that the centroid of the triangle ABC is the point PQR. If I were the question center, I would have just framed. Then which of the following equation is the equation of the plane? However, they have already said show that it is this. So, okay. So just write it down on the chat box once you're done. I hope the question is clear to everybody. Question is the points where the plane is cutting the coordinate axis, you are joining them to make a triangle, right? That triangle centroid is PQR. So what should be the equation of the plane? That is what they're asking you. And they have also given the answer so that you can check. Okay, so please do this and let me know with the done on the chat box. Done, right? So let me just quickly draw that figure. So let's say these are your coordinate axes. Let's at this point, the X intercept is A. So this is A00, Y intercept is B. Now I'm showing it on the positive directions of these coordinate axes. That doesn't mean our intercepts will always be positive. Okay, so A, B, C could be negative quantities as well. Okay. So what is given to you that the G, the centroid is PQR, okay? So we all know that A plus zero plus zero by three is P. That means A is three P. Zero plus B plus zero by three is going to be Q. So B is three Q, okay? And zero plus zero plus three by three is equal to R. That means C is three R. And in the equation of the intercept form, where you had used X by A, Y by B, Z by C equal to one, just use three P, three Q, three R, three R, okay? So that boils down to the equation that we are supposed to prove. Easy question, all right. Next we'll talk about of the perpendicular foot or length. I mean, both the concept I'm going to talk into the same of the perpendicular from a point on the on a plane, on a plane, okay? We'll also talk about image, the image concept, okay? Now this concept that I'm going to talk about will also help you to get the distance of origin from a given plane, okay? So this is a more, you can say, generic concept. So let us listen to this. So let's say this is our point A, X1, Y1, Z1, okay? I'm taking it in the Cartesian form right now, but I could take vector form also that doesn't change the concept. So let us say this is a plane pi whose equation is AX plus BY plus CZ equal to zero. And from a point A, I'm dropping a perpendicular, okay? And I want to know the coordinates of M. And I also need to know this distance. Let's call this distance as, what should I call it? Suggest me a name to call the distance. D is already used. P, okay, so fine. I mean, I don't want you to get confused with the P of the normal form. Let's use, let's use T, okay? So let's say this distance is a T. Now, how would I get the point M first of all? And of course, we will also find out the distance T from here. Now, what are the first thoughts that run in your mind when you are given such a situation? What will you do? Focus on how would you get M? How would you get the foot of the perpendicular? Any thoughts that run in your mind right now when you think of this situation? You can share it with me. You can say, I'm going to do this, then I'm going to do this, and then I'm going to get this, and from there I will get M, something like that. Is some kind of a thought like that running in your mind? I think L would be a good word to you. Yeah, length L. Yes, anything that runs in your mind? See, if I have to solve this question, I will first focus on getting the equation of AM line, the line which contains AM, okay? So in order to get the equation of any line, I need two things. I need a point and I need the direction of that line and both the things that are available to me because I know A and I know the normal to the plane which itself acts like a direction of that line, correct? So can I say this is the equation of a line passing through A and perpendicular to the plane, isn't it? Yes or no? Now, I will say let the parameter of M be lambda, okay? So for a particular lambda value, that means I choose a particular value of lambda for which I will get my M. In other words, M coordinates can be taken as X1 plus A lambda, Y1 plus B lambda and Z1 plus C lambda, correct? Now, in order to get this lambda, I would use the fact that M satisfies pi, M satisfies pi, isn't it? Because M is our point on the plane, right? So if this satisfies that, that means A times X, B times Y, C times Z equal to, sorry, plus D equal to zero, okay? So from here, all of you please pay attention. So from here, you get lambda A square plus B square plus C square, okay? And the other terms would be, if I'm not mistaken, AX1, BY1, CZ1 plus D equal to zero, right? Any questions so far? Now please note A, B, C, AX1, BY1, CZ1 plus D, et cetera, they are all known to you. So lambda can easily be found out from here, correct? So find a lambda from here and put this in this equation. You are done with the foot of the perpendicular. Of course, I'm not going to write that down because it will look very ugly, but you know what to do next, okay? But I'll be finding the length of the perpendicular for you accurately so that formula is there in your formula list, okay? So for M, I'm not going to solve it to the last word because it is going to just lead to a ugly figure over here. So you just have to put this lambda value that you will get, let me write it down, lambda value will be minus this term by A square plus B square plus C square, okay? So just put it over here and you're done. Now in order to find the length of the perpendicular, all of you please pay attention. You already know that M coordinate is this, right? So L is nothing but the distance AM. So L will be nothing but, all of you please pay attention. This minus this square, this minus this square. So this means this minus this square, this minus this square and this minus this square under root, which will actually give you just under root of A lambda square, B lambda square, C lambda square. I'm not putting lambda value here right now because I don't want to make that expression ugly. So this is mod lambda under root of A square plus B square plus C square, correct? Now what is mod of lambda? Mod of lambda will be nothing but mod of, so from this expression I can say mod lambda will be mod of AX1, BY1, CZ1 plus D by A square plus B square plus C square, correct? That is your mod lambda. And you already have under root of A square plus B square plus C square. Please note that there is no point putting a mod around already positive quantity. So I've not put mod around the denominator. So this will partially cut this out and give you under root over here. So your distance between the given point X1, Y1, Z1 and the plane AX plus BY plus CZ plus D equal to zero is this. Now don't you see a resemblance here as well? You see, this is the same formula that we were using in the straight line, 2D lines in our class 11. It's just that now an extra dimension has got added to it. Okay, so that is why I said you'll find a lot of resemblance between 2D lines and 3D planes. Getting my point. And earlier I was talking about the distance of origin from the plane. So for origin, you need to put X1, Y1, Z1 as zero, correct? So that will just remain mod D by under root A square plus B square plus E square, right? Because for origin, for origin, X1, Y1, Z1 is zero, zero, zero. Okay. Is this fine? So please note this down a very important formula that is going to be needed while solving questions. For the image, you all know what to do for the image. For image, you can use the fact that, I'll draw the figure again. Let's say A dash is the image. So you can use the fact that M is the midpoint of, M is the midpoint of A, and M is the midpoint of A and A dash. So if you know M, you know A, you can easily find out A dash, okay? So this you can use for finding the image. Any questions? Any questions, any concerns anywhere? Anywhere you want me to take the screen, take the, and move the screen so that you want to copy it, do let me know. All right, let's take a question. Okay, the question just asks you for the distance of the point from this plane, but let me add one more part to it. Also find the foot of the perpendicular and image of that point. Let's say I call this point as P. Let's find the image of P in the plane. Okay, so foot of the perpendicular is called the projection of the point in the plane or on the plane, and image is basically fitting the plane like a mirror. Okay, so there are two different things. Don't get confused. So we'll do this problem and then we'll take a small break. Yes, ready? I hope ugly values for lambda are not coming out because that's the only thing I hate. Lambda value should be like very simple. One minus one is the best values. Okay, let's check. Minus one is coming out. Thank God. Let's check Arjunth. Let's keep our fingers crossed. Yeah, so the first thing is we need to get the equation of a line passing through two, one, zero and perpendicular to the plane. So as I already told you, that equation is very simple to find out. X minus two by two, Y minus one by one, and Z minus zero by two. Correct, everybody is fine. Equation of line through PM. See, if you have any doubts now in these basic things, please get it clarified. See, this chapter is very easy, but there are some students who missed that initial connect. So one that missed that initial connect, they'll keep struggling in the subsequent part of the chapter. So toward the end of the year, most of the people will come to me asking help only for 3D geometry. Because initially, what happens is like, as the year passes out, all of us get worn out. I'm sure you have all got worn out by this time, right? So much of exam happened and you're writing your project reports, all those lab records, everything is happening at the same time. So somewhere the attention span becomes reduced and you get de-focused. And you miss out on those initial, crucial part of the topic. So many of you, how did you write this equation? I don't know that. So if you are struggling with that aspect, please get it clarified right now itself. I can understand. I have full, I also pass through the same phase in my life, okay? So if you have missed out some crucial aspect of this 3D geometry chapter, it will hit you for all the other concepts which are coming later on. So get it clarified ASAP. Don't keep any doubt. All your chats are personal to me, right? You can write it down, sir. Please can you explain how you're getting this, okay? So don't keep anything with yourself because all these things are going to help you solve problems later on in 3D geometry. Now, let's say lambda for M, I consider the parameter to be lambda. So let lambda be the adharka number for M. So M coordinate, I can write it as two lambda plus two, lambda plus one and two lambda. Now this should satisfy pi that given equation of the plane. So two times two lambda plus two and lambda plus one and two times two lambda plus five should be zero. Right? So I think this will give you nine lambda, okay? And this will be four plus one, five, five plus five, 10. Aiyo Ramakrishna. Oh, Archit, you're saying you're getting minus one or something? I got minus 10 by nine. Who is wrong? Yes, no, minus 10 by nine, no? This is what I don't like, minus 10 by nine and ugly figures are coming. Anyways, so I need to also got it. Thanks, Anil. So put this lambda value here. So M will become minus 20 by nine plus two, which is minus two by nine, if I'm not mistaken. And this is minus one by nine and this is minus 20 by nine. Okay, so this becomes your foot of the perpendicular. Next thing is distance. Okay, let's say I want to find out the length of the perpendicular. For length, you can directly use your formula. You don't have to find the distance formula also, however, both of them will give you the same answer. So just use the formula mod of AX1 by one CZ1 plus D by under root of A square plus B square plus C square. Don't write D square, right? Many people write D square also. No need to write D square. So AX1 will be four. BY1 will be one. CZ1 will be zero, zero plus five, mod by under root of a familiar figure, two, one, two, that is three. So answer is 10 by three units. Those who solved it, are you getting 10 by three only? Confirm it. Yes, okay. Next, image. So let's say P dash is the image of P in the plane pipe. So let's say I call it as alpha beta gamma, use the midpoint formula. Alpha plus two by two is equal to minus two by nine. Beta plus one by two, beta plus one by two is equal to minus one by nine and gamma plus zero by two is equal to minus 20 by nine. Okay. So correct me if I'm wrong. This is going to be minus four by nine, minus two. Minus four by nine, minus two will be minus, minus, minus 22 by nine. This will be minus two by nine, minus one, minus 11 by nine. And this is as good as minus 40 by nine. So this image will be minus 22 by nine, minus 11 by nine, minus 40 by nine, right? So all the things from the same, you know, core principle, okay, from the basic idea. So please remember this basic idea. You will be able to solve all questions related to either the image or the foot of the perpendicular or the length of the. So for the length, you don't have to go this way. Okay. Don't do this entire exercise if they're just asking you for the length, for the distance. Okay, don't sit and find M. Getting M is not required for, you know, getting the length of the perpendicular. Use the formula directly. Don't waste your time if they're not asking you for the foot of the perpendicular. Just distance, I would not use this formula. I will not use this approach. I would just use the formula over it, this formula and I'm done. Okay. All right, so we'll take a break right now. Break. Six. Let's meet at six, oh, it's already 30. Okay, six, 15 to six, 30. On the other side of the break, we'll talk about parallel planes, distance between parallel planes. We'll also talk about the concept of family of planes, line of intersection and the equation of the bisector of a pair of planes. See you on the other side of the break. So the next concept that we're going to talk about is distance between parallel planes. Distance between parallel planes. So let us say we have been provided with two planes. Let us say we have been provided with two planes which are parallel to each other. Now, as we have already seen in the condition of parallelism, that if two planes are parallel, their normals are proportional to each other. That means you can actually make their normals equal also. That means coefficient of x, y, z can be made equal also. So for example, let's say this is one plane and this is another plane and I purposely kept their coefficient of x, y, z as equal. Just the same way as it happens in case of a line in 2D. So in 2D, when we have two lines which are parallel, we can actually make the coefficient of x and y as equal also by multiplying it with the proportionality constant, one of the equations, okay? So how do I find the distance between the two planes? How do I find the distance between the two planes? So for this, the approach is almost the same as what we had used in case of a line also. What do we do? We choose an orbit point. Let's say I call this point to be p, okay? And I want to find out what is the distance of p from the other plane? Let's say this is pi one plane, this is pi two plane. So what is the distance of p from the pi two plane? So that distance itself will give you the distance between the parallel planes. So what I'm going to do in order to find p, I'm going to put two of the coordinates here, let's say x, y as zero, okay? So z will automatically become minus d1 by c, correct? So now think this is your x1, y1, z1. And you want to find out the distance of this point from the other plane, okay? So the distance, as we have already seen in the previous slide, that will be A x1, B y1, C z1, plus d2, because there's a d2 here, upon under root of a square plus b square plus c square, correct? So let us put x1, y1 as zero, zero. So that will give us zero plus zero and z1 will become minus d1 by c. Minus d1 by c will give you something like this, okay? So c, c gets canceled. So it becomes mod of d2 minus d1, or you can also write d1 minus d2. Within the mod, it hardly makes any difference whether you write d1 minus d2 or d2 minus d1. So this becomes your distance between the, distance between the parallel planes. Please note this down. And again, this formula is resembling the formula which you had learned for your 2D lines. In 2D lines, we used to take the difference of their constants and divide by the square of the coefficient of x and y, isn't it? So here we are again doing the similar activity. Is it fine? Any questions? Just to get a point on the plane, you could take any of the two as zero, but don't put all of them as zero because then d1 will become zero, which might not be true. So any two of them you can make zero. You can make x and z as zero, zero and get your y. Or you can put your y, z, zero, zero, get your x. It's not going to affect your answer. I just needed a point on that plane, getting my point. So for that, I basically took any two coordinates, x, y, zero, zero each and got my z coordinate. See, it is up to you, whichever point you want to take because the plane, I'm assuming that the plane is going and cutting, in this case, it is cutting the z axis. Unless and until the plane is parallel to your z axis. In that case, there will be no z term appearing. Are you getting my point? So when I'm taking my x and y as a zero, zero, I'm assuming that the plane is going and cutting the z axis somewhere. So if the plane were not cutting the z axis, then there would have been no z term actually. That means c would have been zero. In that case, this would not be valid. So this is more generic case to be more precise, not a specific case. So let's take a small question, a very small, just to practice this formula. Find the distance between, find the distance between the planes given by, pi one is two x minus y plus two z plus three equal to zero and pi two is minus four x plus y minus four z minus one equal to zero. Oh, sorry, I missed two. Done. Once done, you can give me a response on the chat box. Two different answers I'm getting. Why is that confusion? I should get a uniform answer, right? Okay. Now, when you're using this formula, which I derived just now for you, this formula, please note this formula works when you have made the coefficient of x, y, z same in both the equations. But to confuse you, I had kept, I have given you different coefficients for x, y, z in the second equation. However, they're proportional. Okay. So what are you supposed to do? You are going to, you are supposed to make both of them as same first of all. So for that, you either divide the second one by minus two or you multiply the top one with minus two. It is up to you. So let's say I divide this by, I divide this by minus two. So I'll end up getting two x minus y plus two z plus half equal to zero. Correct? Now I have made the coefficient of x, y, z the same in both the equations. Now I can go for the formula. This is my D1, let's say this is my D2. So as per the formula, the distance will be mod D1 minus D2 by under root of a square plus b square plus c square, which is three minus half mod by under root of two square plus one square plus two square, which is actually a three, right? So this gives you five by six units. Is it fine? Any questions? Any questions, any concerns? So with this, we are now going to move towards the concept of family of planes. This concept is also written by the name of chief of planes also. Okay, the word chief comes from, the chief is basically a word which is used for the arrangement of pages in a book, okay? So the pages in a book are like, they're like chief, okay? It's a chief of pages. So it appears as if there are a lot of planes passing through a single line. So the family of planes concept is almost like how you can say how the pages in your book are basically attached to the rib of the book, correct? So let's say this is one plane, okay? And it meets this plane, let us say, this line or on this line, okay? By the way, this line is called the line of intersection, line of intersection. In shortcut, I will call it as LOI, okay? So the meeting point of two planes, if at all they are meeting, that is called the line of intersection of the planes, okay? So let's say pi one and pi two are the two planes, they are meeting on this LOI. Through this LOI, you can pass, you can pass so many different planes, right? Let's say I'm making one plane passing through it. So let's say this is one plane. So this is one plane passing through it. I could make another plane passing through it. In fact, so many planes can be made to pass through it. It's just making the diagram a little bit more ugly. So I'm just removing it, okay? So they will all form, these are, they will all form a family, okay? So family of planes is nothing but it's a concept which is very similar to what we did in the straight line. In straight lines, if you remember, through the intersection of two lines, you can pass several lines and those all lines will actually form a family, okay? In the same way, through the intersection of, through the line of intersection of two planes, okay? Maybe pi one and pi two planes would be given to you. So through the intersection of those two planes, you can pass several planes and those planes will actually form a family member. They will all form a family, right? So here we are going to learn two concepts. The one concept that we are going to learn is how do we find the line of intersection? Line of intersection of two planes. That is one concept we are going to talk about. And second thing is how do we deal with the concept of family of planes? Okay. Now all of you please pay attention. Let us go back to our straight lines chapter. In straight lines chapter, we had learned that the equation of a line in 3D is given by this. Correct. X minus X1 by Y minus Y1 by B, Z minus Z1 by C, right? This equation is actually called the symmetrical form of the equation of a line, okay? This equation is actually known to be the symmetrical form of the equation of a line. Why? Why the same symmetrical form? Because these expressions look very symmetric to each other. Okay. But if you mention the equation of a line by mentioning the two planes whose intersection generates that line. So in our example, these two planes together, they generate that given line, isn't it? So when you write the equation of a line by mentioning the two planes which generate that line or whose intersection generates that line, then that form is called asymmetrical form. Are you getting my point? So many a times the question setter will give you something like this. You will say there is a line and you will mention two planes after that. And many people will get confused. Are you there saying line and they're giving two plane equation? How is that possible? It is very much possible. So they can give you a line by mentioning two plane equation together. So it is up to you to understand that, oh, when these two planes meet, that line is generated. Okay. So sometimes there are questions which ask you to convert asymmetrical to symmetrical form. So there are some questions which will be based on finding symmetrical form from asymmetrical form. So we will do that as our first exercise, how to get the symmetrical form of the line of intersection. Then we'll come to the concept of family of planes. Okay. So in order to explain this process, I will take an example. Okay, let's take an example. Let's take a simple question. Let us say the question says, find the symmetrical form for the asymmetrical form and that asymmetrical form is this. Now, as you can see, I have given a combined equation of two planes and I'm calling that actually as a line. So this is actually making a line, let's say. This is a line's equation. So two equations of a plane combine and they generate a line. That means the intersection generates a line. Okay. So when you mention the two planes whose intersection generates a line, then we call that equation to be an asymmetrical form. Now from here, how do we get a symmetrical form? So if you recall, for a symmetrical form of a line, we need two things. We need a point, okay? We need a point on the line and we need the direction of the line. Direction, I mean direction to which the line is collinear too. Okay? So these two things are required. So now how do you get a point? Everybody please pay attention. To get a point, we put any one of the variables in these two equations as zero. So somebody was asking, I think Viva was asking me, why did you put zero? So in order to get a point, you can actually assume that that line is cutting one of the planes. Okay? So let's say I'm assuming that it is cutting the XY plane. So for XY plane, let's put Z as zero. Now, if it is not cutting XY plane and I'm putting Z as zero, then what will happen? It will lead to an inconsistent equation. Okay? Let us see what happens when I do the same activity over here. If I put Z as zero in both the equations, it actually gives me two simple equations, which I don't think so is inconsistent. Yeah. We can actually solve for X and Y here. Correct? So I would request people to solve for X and Y and tell me what points do you get? What is the point that you're getting? You can multiply this top with a two. Okay? So that will give you two X plus two Y equal to 12. Okay? Subtracted. So Y becomes minus 17. Correct? Yeah. So Y becomes minus 17. So X becomes 23. So you know a point on the line. And this point can be found out by different people in a different, different way. Don't expect that Aditya's answer will match with Shrita's answer. Shrita's answer will match with... No. It is up to them. And Y to put Z as zero, I can put Z as five also. It is up to me. Okay? So don't expect that there would be a consistent answer coming from everybody. Everybody will have a different set of points. Okay? All I need is one of the points. That's it. I don't care who is finding which value of the point, but I need a point which is going to satisfy both the equations simultaneously. Period. Whatever is that point? Okay? Now, many people ask me this question. Said, why don't we put X and Y, Y and Z both as zero, zero? Right? So don't be too greedy. If you're doing that, it'll become X equal to six and here it'll give you X as minus five by two. So there'll be an inconsistency arising. So that scenario might never happen. So when you're putting X and Y both as zero, zero, that means you are claiming that the line of intersection will definitely cut the Z axis, which might not be the case. Anyways. So having got the point, our first requirement is fulfilled. Okay? Next, how would I get direction? Can anybody give me a direction towards getting a direction? Can anybody guide me how to get the direction of the line? From the two planes. Okay? So let's say these are the two planes whose intersection is creating that line. Let's say this is one plane and this is another plane. I am going, yeah. I want to know the direction of this line. How will I know the direction of this line? This is what I'm looking for. Oh, is it breaking? One second. Not as better. My test, yeah. Yes. Can you tell me how to get the direction of this line? Given that I know these two planes. Very good, Aniruddha. Very good. So Aniruddha is suggesting that, yes. This line direction would be perpendicular to the normal to both the planes. See everybody, please pay attention. Let's say this is, let me close my windows. Yeah, sorry. So please note that this line is not only going to be perpendicular to this fellow, it is also going to be perpendicular to this fellow. As you can see here, correct? So if let's say n1 vector and n2 vector are the normals to pi1 and pi2 respectively, then the direction will be along n1 cross n2. Getting my point. So what is n1 here and what is n2 here? Let us figure it out. So look at the equation. Tell me n1. n1 will be i plus j plus k. That is your coefficient of x, y, z. As we all know, in a planes equation, the coefficients of x, y, z, that is the direction ratio of the normal. So you can make a vector out of it. Similarly, n2 will be 2i plus 3j plus 4k, correct? So n1 cross n2 will be ijk111234. Let's calculate this quickly. So that's going to give me i times four minus three, which is one and minus j times, I think that'll give you a two and k times one. So i minus 2j plus k, okay? Now what I'm going to do is, I know the point and I know the direction ratios of the required line and what are they? One minus two, one. So once I know the point and I know the direction ratios, what are we waiting for? This becomes your symmetrical form of the equation of a line. Is this fine? Any questions? Any concerns? Okay. So a question related to finding the symmetrical form of the equation of a line can be asked to you as well. Any questions here in the approach of finding the symmetrical form of the equation of a line? Do let me know. Fine, no questions? Okay. Now let us talk about finally the concept that we are interested in is the family of planes. Okay, family of planes. So if you have been given that there are two planes which are intersecting on a line of intersection. Let's say this is one plane and this is another plane, here. Okay. And they're passing through a line of intersection. Okay. Then all planes, that pass through the LOI equation of family of planes that pass through the line of intersection of pi one and pi two is given by pi one plus lambda pi two equal to zero. Okay. So as to say that if let's say pi one is a one x, b one y, c one z plus d one equal to zero and pi two is a two x, b two y, c two z plus d two equal to zero. Then this equation will become something like this. a one x, b one y, c one z plus d one lambda times. Very similar to whatever we had done in case of a line. Okay. Fine. Where this lambda could be any real number. So for, for different, different family members, lambda will be different, different, right? As I already told you in our line chapter also for pi one itself, lambda will be zero, right? Because pi one itself is a part of that family. For pi two, lambda will be infinitely large. Okay. So whatever lambda value you choose, you will accordingly get a family member for the same. So what type of questions can come under this topic? You may be given that there is a plane passing through intersection of so and so plane. And that plane is satisfying certain condition. Maybe they will give you a point on it or they will give you the distance of that plane from origin or they may give you that the plane is perpendicular or parallel to so and so planes. So many things can be given to you and they will ask you for that equation. So the word here to identify is the word intersection of two planes. The moment you read this phrase that there is a plane passing through intersection of two given planes, automatically the family of planes idea should appear in your mind. Okay. So let's take a problem to identify what is the nature of the question that we will be getting under this. This, yeah, it will be perpendicular. The LOI will be perpendicular to any plane because see that LOI is on the plane. No. Let's say I take any plane. This LOI is like this. So perpendicular to this plane will be perpendicular to LOI. Any family member if you take passing through that plane till that LOI is on that plane, any line, the normal to that will be perpendicular to LOI. This is something which I started my discussion of this plane with. On a plane, if you make any line, every line on the plane would be perpendicular to the normal to the plane. See, if there's a normal to the plane, no. Any line that you make like this or like this or like this, they will all be perpendicular to this normal. Okay. Maybe let me just show you in the diagram wise. Okay. They will all be perpendicular to this normal. Let's take a question. Yeah. Find the equation of a plane containing the line of intersection of these two planes and passing through 111. Everybody try this out. Plane and simple question. Plane and simple question of a plane. Done. Okay. So we can start, we can assume that the required plane that we are looking for is this. Again, please let me tell you that you can put lambda with the other term also. It is not a, you can say compulsion that if you have to always put lambda with the second equation. Okay. And depending upon that, your lambda value may be reciprocals of, let's say if somebody is putting a lambda along with the first plane and somebody is putting a lambda with the second plane, their lambda values will be reciprocals of each other. So again, don't try to compare each other lambda values also. Now, in order to find this lambda, use the fact that this plane must be satisfied by 111. Correct. That means one plus one plus one minus six plus lambda two plus three plus four plus five should be equal to zero. Okay. That means lambda is three upon how much has come out to be 912. Correct or? Sorry. 14. Yes or no? 14 only? All right. So put this lambda back over here and your job is over. So x plus y plus z minus six plus three by 14. Two x plus three y plus four z plus five equal to zero. Now, please understand here that the simplest way to simplify this is multiply this entire expression with the denominator of lambda. So that automatically makes the process slightly easier to simplify. So that'll give you a 20x. This is going to give you a 23y, correct? And we are going to get 14 plus 12, which is 26z. And minus 64 plus 15. Okay. Minus 64 plus 15 is minus 49. Is it fine? Any questions? Any concerns? Oh, sorry. Minus 84 plus 16. Sorry. It's minus 69. Is it fine? Any problem with this? Everybody has got this? Okay. Let's try one more question. Oh, I think I have, okay. All right. So let's take another question. This question I'll be writing down on the board here. Okay. So there is a plane AX plus BY equal to zero. Okay. And this plane is rotated through an angle of alpha. This plane is rotated by an angle of, by an angle of alpha. About its LOI, about its line of intersection with the plane, with the plane Z equal to zero. Okay. Find the equation of the plane in the new position. Find the equation of the plane in the new position. In the new position. I hope the question is well understood by all of you. Okay. So let's say this is a blue plane, the first one. And this is the second plane Z equal to zero. This is the line of intersection, LOI. Now this plane is rotated by an angle of alpha. Says that it comes now to this position, the green position. I'm not drawing a very big figure because it's going to dash again, so there are values. So this rotation is alpha. Now it could be alpha in either direction also. It could be down also. It could be up also. So you should get two answers for the equation of the plane in the new position. Okay. We'll end up getting two planes. Let's do it. Done. Okay. Let's discuss it. Aditya has given one response. See, here we have to understand that the plane in the new position and the plane in the old position and of course the other plane. They are all family members, isn't it? Okay. So can I say, let the equation of the plane in the new position be? Let the equation of the plane in the new position, in the new position be? AX plus BY plus lambda Z equal to zero. Can I say this? Correct. Now please understand that this plane, this plane makes alpha with this plane. Okay. That means the angle between them is alpha. Correct. So can I do this? Can I use my angle between the two planes formula? Right? So what is the angle between two planes formula? So cos alpha, let me write a mod, mod cos alpha is A1, A2, which is A square, B1, B2, which is B square. And of course there is nothing to multiply herewith divided by under root of A square, B square, lambda square under root of A square, this is equal to cos of alpha. So in fact, you can, since this is positive, it doesn't actually need a mod, but you can just cancel this and put a under root over it. Okay. Now without my lambda, I cannot complete my answer. So for that, I need to do some small trick over it. I'm going to reciprocate it and square it. And this is as good as saying one plus lambda square by A square plus B square. Send this one to the other side. So secant square alpha minus one is lambda square, A square plus B square. That means your same tan square alpha is equal to lambda square, A square plus B square, which means lambda value is plus minus under root of A square plus B square tan alpha. So just put this value, just put this value. Let me just make a long arrow here. Let's put this value over here. Okay. So when you do that, you end up getting your two planes, which is A x plus B y plus under root A square plus B square tan alpha z equal to zero. And the other one will be A x plus B y minus under root A square plus B square tan alpha z equal to zero. So these two will become your answers. Is it fine, any questions, any concerns? Any questions, any concerns, do let me know. All right, so we'll now move on to the next concept. Angle between a line and a plane. Angle between a line and plane. First of all, let me define this expression. What is the meaning of angle between a line and a plane? So let's say this is a plane. Okay, and let's say this is a line. Okay, let me make a line like this. So what is the angle between this line L and this plane pipe? So please note, angle between a line and a plane is defined as the angle between the line and its projection on the plane. What are the meaning of projection? Projection means from every point if you start dropping perpendicular, okay. Whatever line it traces, that will be called as a projection. Okay, so this line that you see, this line that you see, this is called the projection of L on the pipe plane. Projection of L the pipe plane. So the angle between them is what we call as the angle between a line and a plane. Now tell me, how would you find this out? How would you find this angle alpha? Any suggestions from your side? Let's say I have given you this line equation. Okay, let's say this line equation is X minus X1 by A1. Oh, sorry. Y minus Y1 by B1 and Z minus Z1 by C1. And let's say this plane equation is also given to me. This plane equation is A2X B2Y C2Z plus D equal to zero. Yeah. So how would you find the equation between the line, angle between this line and the plane? Now please, yes, Kinshok, absolutely right. So please remember that this alpha means this is 90 degree minus alpha. That means this angle between the normal to the plane and the line is complimentary of the required angle. Isn't it? So can I say cost of 90 degree minus alpha would be nothing but A1, A2, B1, B2, C1, C2 by under root of A1 square, B1 square, C1 square under root of A2 square, B2 square, C2 square. Okay, that means sine of alpha is going to be this. So alpha is sine inverse of A1, A2, B1, B2, C1, C2 by under root of A1 square, B1 square, C1 square, A2 square, B2 square, C2 square. Clear, any questions? Okay, so when it comes to an angle between line and a plane, please use sine inverse, not cos inverse because of the fact that the normal to the plane will actually make the complimentary of the angle that the line actually makes with its projection on the plane, right? So because of this fact, sine alpha comes here. Because of that fact, sine comes over here. Is it fine? Clear? Note this down. We'll take a small question based on the same. Yeah, yeah, I think just now I got a notification that my internet is unstable but it keeps on becoming stable and stable. Time and again. I've kept my camera off. I hope now it is better. Let's take a question. Okay, let me just give you a question from my side. I don't think so. I have taken a question on the concept. Let's take a simple question. Find the angle between, find the angle between the line x minus two by one, y plus one by two, z minus three by zero, and the plane, and the plane. Two x plus y plus two z equal to three. Now, to be, yeah, so, see, sine of alpha should always be positive, right? Because the angle between, see, actually, what are you finding? You're actually finding the angle between two lines only here, right? One is the actual line and the other is the projection, right? Between any two lines, the angle will always be between zero to 180 degree, correct? So please understand here that, please understand here that, that when you're using this formula, this will automatically have to be kept as positive, right? Because sine of any angle which lies between zero to 180 degree will actually be, will actually be positive only, right? So here, here, what we have basically getting is, we are always going to get an answer which is, you know, you can either say it in, you can either say it is in the first quadrant or in the second quadrant. So obtuse, acute difference you will not be able to make. Other than the fact that the original, see, the normal to the plane, many of you might take the normal to the plane like this also, right? In that case, this angle will become obtuse in nature. Are you getting my point? But if you're using sine of something, you will always should get a positive answer because whether acute or obtuse, it is within the first and the second quadrant only. So normally when we define the angle between them, we always mention the acute angle. So as to now say, okay? So we'll always mention the acute angle in this case. However, for the same value of sine alpha, alpha could also be obtuse. I'm not denying it. Let's say you get a half. So you state 30 degrees only. You will not state 150 degrees. Even 150 degrees could have been possible, isn't it? Because the person who is looking at it could be looking at from this point of view or he could be looking from this point of view. It depends upon the person, right? So, but we always state the acute angle unless until the option is guiding you to mark something which is not there. Let's say 30 degree was not there and 150 degree is there. So maybe 150 degree is what I'll be going for it. But as you can see here, that if you're talking about angle between the line and the plane, whatever is the position of the line, it is always going to make an acute angle only. So acute angle is what we mentioned. If the line was like this also, let's say, then you would have mentioned this angle, getting my point. So the answer is always acute in this case. Yeah, coming back to the question. Done? I mean, just leave your answer in terms of sine inverse something. Don't need to calculate it. Yeah, correct kinship. So the answer in this case, sine alpha would be nothing but one into two, two into one, zero into two, my under root of one square, two square, zero square, under root of two square, one square, two square. So that's going to be a four by three root five. So sine inverse of four by three root five. Of course, this is not a well-known figure. So you'll not be able to write down the exact values. Good enough. So moving on to the last topic of this particular chapter. In fact, I would like to do more and more questions. So maybe when we start with the bridge course, sorry, bridge course, nothing, crash course, I will take up some specific problem solving sessions on 3D geometry. So just one and a half hours will just solve problems. Okay. I will let you know when I'm going to schedule that, not in the immediate future, maybe in 2022, we will do that. Okay. All right. So bisectors of pair of planes. Now, let us say these are the two planes. I'm just showing you the cross-section area, cross-section of the two planes. Okay. So let's say these are the two planes. Okay. And you want to know the equation of the bisectors of these two planes. Again, as I told you, I'm showing you the cross-section. Okay. Plane exactly I'm not showing. Okay. So you're looking at a plane from, you can say from the edge of the plane. Okay. So this and okay, let me draw one in some different color. So the blue one is bisecting or one of the angles. Okay. And the green one is bisecting another angle. So this is plane pi one. This is plane pi two. And these are your bisector planes. So if you've been provided the equation of the planes. Okay. And you are asked to find out the equation of the bisector planes. So what do we normally do in this case is, we use the fact that this plane B one or B two, if you take any point H comma K comma, let's say L on it, its distance from the two planes, pi one and pi two would be equal. Okay. The distance and the distance will be equal. Okay. So in order to basically find the equation of B one and B two, you are using the locus concept that the distance of any point on B one and B two from the two given planes, pi one and pi two are equal. Okay. So even if you take any point here, then this distance and this distance will be equal. Correct. So using that fact and using the distance of a point from a line, you can say mod of A one H B one K C one L plus D one by under root of A one square B one square C one square is equal to mod A two H B two K C two L plus D two by under root of A two square B two square C two square, right? So what do we do in this case? We remove the mod and we generalize it. So generalized by putting H as X, K as Y and L as Z. So and remove the mod. If you remove the mod, it will become something like A one X B one Y C one Z plus D one by under root of A one square B one square C one square equal to plus minus A two X B two Y C two Z plus D two by under root of A two square B two square C two square. So plus minus basically signifies that there are two possible planes. So two bisectors can be obtained from this equation. Please note this down. Again, this is very similar to what you had done in your bisector of lines in 2D, no different. Just add one more dimension Z is sitting. Any questions, any concerns? Now under this bisector, why couldn't we do this for vectors? As in you want to know the equation of a vector which is bisecting. So how do you use the concept of, I mean, what concept will you use for vectors to get the same? What is the analogous concept? What is the concept which will be helping you in vectors to achieve the same? What is the principle behind you getting the equation? There's no concept like distance of a position vector on a vector from another vector. Yeah, all right. So with respect to the bisector equations, there are three types of questions that are normally asked. We'll talk about them one by one. Maybe I'll not be able to complete all of them here. So one type of question that they ask is, they will ask you for a bisector bisecting or bisector that bisects the angle containing the origin. Bisects the angle containing origin, okay? Second type of question that is asked is bisector that bisects the angle containing certain point, alpha, beta, gamma, okay? So you can say that the first case is a specialized version of the second case. And the third type of question they can ask you is the acute or obtuse angle bisector, okay? So these are the three types of questions that is normally asked with respect to, you know, bisector of a pair of planes, okay? Now, in the next few minutes, I may take around five, 10 minutes extra. So please cooperate. I'll be quickly giving you the working rule for all the three, okay? Just the working rule. The derivation for those working rule is exactly the same way as what we had done for the similar concept when we had taken 2D lines. So please, you know, go to that video of 2D lines. If you want, I can send it across to you once again, okay? So the working principle, which I'm going to give you in the next few minutes, they are derived from your concept of the same scenarios when applied to 2D lines, okay? So let's talk about the first scenario. How do you get the equation of a bisector that bisects the angle containing the origin? So let me write it down separately, bisector of the angle containing origin. So what are the meaning of this bisector of the angle containing origin? See, when there are two planes, okay? Let's say one plane is this plane, another plane is this plane. I'm just making some orbit plane. In one of the angles, there would be origin line. Let's say origin is line here, okay? So this is the angle where the origin falls. So if I want to find out the equation of a bisector, which bisects, okay? The equation of the bisector, which bisects this angle, then this is my desired bisector, okay? So we'll say this bisector bisects the angle which contains the origin. Are you getting my point? What are the meaning of the statement bisector of the angle containing the origin, okay? So here the working rule is you write both the equations of the plane keeping your D1 and D2 as positive, okay? So please ensure you have made D1 and D2 both as positive, okay? That means they should not be negative, right? How many people say, sir, what if one of them is zero? If one of them is zero, then the origin is actually lying on one of the planes, okay? So they will be either both positive, both negative or one positive, one negative. So you have to first ensure that before you use the formula which I'm going to give you in a few second time, you have to make D1 and D2 positive both. Let's say D1 is negative, then you have to multiply with a minus one throughout, okay? And if the other one is already positive, don't do anything to it. So once you have converted D1, D2 as positive, then please note down the bisector of the angle that contains the origin, that contains the origin would be this one. I'm writing it down now. A1x, B1y, C1z plus D1 by under root equal to A2x, B2y, C2z plus D2 by under root of this. Okay, plain and simple. That means use a positive sign, but again, I'm repeating this. You should have made D1, D2 positive. This is very, very important. Without doing this, don't do anything. Don't use any of the formula, okay? So this formula to be utilized only after you have made D1, D2 positive. That means with the change the equation, you need to work on this formula, okay? I'll give you a simple example maybe. Let's say I have this plane, x plus y plus z minus three equal to zero. And the other plane, let me take some figures which are like easy for us. And the other plane is something like this plus five equal to zero, okay? I want to find out the bisector of these two planes, pi one and pi two, which contains the origin. Now I can see here that this is positive, no problem, but this is not positive. So what I will do, I will rewrite it like this first of all, okay? An old equation, I will erase, let me not erase like this. Else you will lose track of what was the question. So please erase this from your mind, right? So this is your A1, B1, C1, D1, A2, B2, C2, D2, okay? So then use this formula which I've given you. Oh, sorry. I think this will come out to be three, so I did not worry about it. This will come out to be, okay? So simplify this, that will be your answer, okay? So your answer will be minus three x, minus four y, minus three z equal to two. Clear what I mean by saying that you have to use this formula only after making D1, D2 positive? Any questions, any concerns here? Okay. Next, could you scroll down? Yes, why not Shraddha? The second type of question that is asked is bisector of an angle containing an origin. Bisector of an angle containing a certain point, sorry, origin I've already taken, containing certain point alpha, beta, gamma, okay? Now here, what do we do? What do we do here? Everybody please pay attention. Let's say pi1 and pi2 are given to you. So here we don't have to worry about making D1, D2 positive. All we do is we check that when we put alpha, beta, gamma in the left-hand side of both the equations, okay? So let's say you're putting alpha, beta, gamma in both the sides of these two equations. What values come out, okay? If these two values both are of same sign, okay? Then the bisector equation will be, so if they are of same sign, let me write it like this to be more precise. Let's say this is K and this is L, okay? If K and L are of same sign, okay? Then the bisector equation will be this, okay? And if K and L are of opposite signs, then your bisector equation will be this. Sorry for that overwriting there. Yeah, please note this down. So same sign put a plus in between, opposite sign put a minus here. Simple as that. Now, if one of them is zero, that means that point alpha, beta, gamma is lying on one of the planes. So there is no question of the angle containing that point to be bisected. So they will either be both positive or both negative or of opposite signs, okay? So if they are both positive or both negative, use the first form, use the first equation as the bisector equation. And if one is positive and other is negative, use the second equation as your bisector equation, okay? Coming to the last part, I will not take much of your time. How do you get acute and the obtuse angle bisector? So here, the working rule is very much similar as what we had got in our 2D lines. Ginshukh, that is why I told you know, there is a video which I'll be sending it across to all of you. Watch that video, okay? The answer to the derivation is, the answer to your question is hidden in the derivation which I have given in that video, okay? Now all of you please pay attention here. If you want to know which is the acute angle bisector and which is the obtuse angle bisector, the approach is, as I'm going to mention over here, please make your D1, D2 again positive, okay? So make your D1, D2 again as positive, okay? Now calculate a value K which is given by A1, A2, A1, A2, B1, B2, C1, C2, okay? Calculate this value, call it K. Now if this value K, if this is positive, okay? Then the obtuse angle bisector, then the obtuse angle bisector would be A1X, B1Y, C1Z plus D1 by A1 square, B1 square, C1 square with a positive sign and the acute angle bisector would be the one with the negative sign. By the way, I'm not writing everything down, I'll just put a ditto here and put a minus sign, again a ditto, okay? So I hope you'll be able to understand. I don't want to write this lengthy term over it. And if K is negative, if K is negative, then your obtuse angle bisector would be A1X, B1Y, C1Z plus D1 by under root of, under root of A1 square, B1 square, C1 square with a negative sign, A2X, B2Y, C2Z plus D2 by under root of this. And the acute angle bisector in that case will be the same thing, but with a positive sign here. Okay, so the way I remember this is, whatever is the sign of K, with the same sign, you'll always get the obtuse angle bisector. So if this is positive with the positive sign, you'll get the obtuse angle bisector and the other one will become acute. If it is negative, then with the negative sign, you'll get the obtuse angle bisector and the other one will automatically become the acute. In the second case, there is no requirement for D1, D2 to be positive. That requirement is only in the first and the third case, a right web of, okay? So this is very, very important. If you don't do this, your entire formula will go for a toss, okay? Now many people would be having this doubt, why didn't I take K as zero? Because if K is zero, the two planes will become perpendicular because this expression is the expression that we saw in condition of perpendicularity. So if the two planes become perpendicular, there is nothing like acute and obtuse angle, right? So we will not be talking about the acute and obtuse angle bisector in those cases. So we'll stop here because I think I've already taken 15 minutes of your extra time. One small concept is left, which is basically the concept of a position of two points with respect to a plane. We will take that in one of the classes in between our crash course, okay? And we'll take a lot of problems also in that session, okay? So we'll stop here. We have almost completed 99% of the chapter. 1% that concept is left. We'll take that up little later on. Thank you. Bye-bye. Take care. Good night. Stay safe.