 this question. Let's not talk about the next question here. Okay, a typical question. Find the equation of a plane passing through A, B and C. Also find a unit vector perpendicular to this plane. A very stereotypical question. Okay, you must know how to solve these because much more difficult questions will be framed using these simpler concepts. You can again type the answer privately to me or send me a WhatsApp. Sir, I sent it to you. Yeah, yeah. Can you send me something like an equation? It doesn't look like an equation. Yes, Ardhara is absolutely correct. Aditya, do send me a Cartesian equation. Sir, is it 1 by root of 38 of 5i cap plus 2j plus 3k. Equal to something, right? Or plus something equal to 0. Equation means something has to be equal to, right? Sir, that is n vector like that. You have mentioned the equation of the unit vector, right? Yes, sir. The unit vector is correct. I want the equation first. And that too also there is a small mistake. Aditya, in the sign of k that you have written. Okay. Just check. Yes, sir. So there are two parts to the question. One is the equation of the plane and other is the unit vector perpendicular to this. So if you get one, I think the second one will be easier to find out. Yes, I'll send the equation. Sir, I had sent it to you. Yes, yes. That's correct. Yeah, absolutely correct. Aditya, also correct. Just write 5x and 2y. Okay, now let's discuss this very simple question. Let's do this in Cartesian form. Now remember when you have been given three points on the plane, let's say three non-colonial points first of all, because when you have a collinear point, they can be infinitely many planes passing through it. But when you have three non-colonial points, you can only have a unique plane passing through it. So let's say x1 has coordinate, a has coordinate x1, y1, z1, b has coordinates x2, y2, z2 and c has coordinates x3, y3, z3. I had already discussed with you that we use the concept of coplanarity, correct? Coplanarity whenever the word comes, scalar triple product must appear in our mind. And when you talk about scalar triple product, we talk about coplanarity of three vectors. So I'll take another point, let's say r, x, y, z, okay? And I claim to say that this vector, this vector, and this vector. So basically I'm saying ar vector, ac vector, ab vector, they are coplanar, correct? That means the i, j, k components of these vectors written in determinant form should be zero. So what is the i, j, k component of ar vector? x minus x1, y minus y1, z minus z1. What is the x, y, z component of ac vector? x3 minus x1. Again, you can write ab first, but it doesn't really matter. You can write in any one of the rows, any one of them because ultimately you're equating it to zero. So let me write ab first just to maintain an order of one, two, three. So y2 minus y1, z2 minus z1, zx3 minus x1, y3 minus y1, and z3 minus z1. Now let me tell you it doesn't really matter which three vectors you are taking and writing their stp as zero. You could have taken bc vector also, you could have taken ab vector also, you could have taken ra vector also. But r must be there, else your equation will not be there. So take any three vectors formed by these four non-colonial points, in fact four coplanar points to be more precise. So when you use this in our given problem, so let's take our x1, y1, z1 to be our point c because it looks simpler of all. x minus 7, y minus 0, z minus 6. And take the difference of any two points you want, doesn't really matter. So let's say I take the difference of ab, so I get a 1, I get a 2, I get a 3, then I take the difference of bc, let's say. So I get a 4, I get a minus 4 and I get a 4 again. Equated to zero. Just drop the factor of 4 throughout just to save our time. So I'll make it as 1 minus 1 and 1. 1 minus 1 and 1. And let me expand it, so it's x minus 7 times 2 plus 3 which is 5, minus y times 1 minus 3 which is minus 2, and z minus 6 times minus 3 equal to 0. So if you expand it, you get 5x plus 2y minus 3z and constants will be minus 35, minus 35 plus 18 which is minus 17 equal to 0, that becomes the answer of the question. If you want to write it in the vector form, you can write it as r dot 5i plus 2j minus 3k is equal to 17 or minus 17 equal to 0. So this is the vector form. Now they're asking a unit vector along the normal, just take this vector and divide by the magnitude of this vector which is under root 25 plus 4 plus 9, that's 5i plus 2j minus 3k over root 38. Make sense? Okay. Now a very interesting thing that we can also discuss when we are talking about equation of a plane passing through three points. So let's say I have these three points a, b and c. These are my three non-colonial points. Now when I say these three points form a plane, that means they are coplanar, correct? So can I say any, let's say I take any generic point r, can I say r minus a vector is coplanar to is coplanar to, is coplanar to, let's say b minus a and c minus a vector, correct? Now you could have written that also in another form without stp, that is you could have written r minus a vector as x times b minus a plus y times c minus a. Now remember our first discussion on coplanarity of points was based on expressing a vector, let's say a, b, c, r, 3 vectors which are coplanar, then one can be expressed as a scalar combination or the linear combination of the other two. So this vector is coplanar to this and this. So it could be expressed as a linear combination of other two. So x and y here are some scalar quantities. Do you remember this? All of you remember this? Yes or no? Yes. Yeah, so there's another way that you can actually represent the equation of a plane which is having a point a, b, c on it is a plus xb minus a plus yc minus a, okay? When you group up your a together, you get a 1 minus x minus y plus xb plus yc, okay? Is that fine? Correct? Now you could also choose to do away with one of the constants. You can choose your one of the constants as a 1 and make this as a plus some lambda b plus some mu c. Are you getting it? Okay. So this is another form of the equation of a plane but this is very less used. I mean you will not see this in at least your NCRT books, okay? In some international books, yes, they talk about it but you will not find them in your NCRT book. Are you getting my point? So you could also write down the equation of a plane passing through three points in this form. This is called by the way a parametric form because lambda and mu are parameters over here because as lambda and mu change, then only you get different, different points on the plane. Are you getting it? So if I give you an equation which is in this form, will you be able to write a Cartesian form for it? Let's take a question on this. So there is an equation of a plane given like this. Following plane it should be not planes in Cartesian form. So basically they have given you an equation in this form. r is equal to a plus lambda b plus mu c. So it's just another way of saying that r i minus j i plus j plus k and i minus 2j plus 3k, these are coplanar points. So these are coplanar points. It's just another way of saying it. That means they all lie on a plane. Please reply privately to me once you're done. No Shreya. Oh, you may be correct also. By the way, let me write it like this. r minus this. Send it to you sir. Send it to me? Yes sir. Yes. So r minus a vector is coplanar to b and c. That's what I'm trying to say. So r minus a, when you write it, you can write it as, you know, i plus yj plus zk minus i minus i plus j minus i. Just so you open the bracket, it'll become plus j. So it becomes x minus 1i, y minus 1j and zk. So this will come in your first row. So x minus 1, y minus 1, z will come in the first row. The other two vectors, b and c vectors will come in the second row. 1, 1, 1, 1 minus 2, 3. Okay. Getting the point. So when you expand this, it becomes x minus 1, 3 plus 2, which is 5, minus y minus 1, which is 3 minus 1, which is 2, and z times minus 3 equal to 0. When you expand it, you get 5x minus 2y minus 3z. And you get a minus 5 and you get a plus 2 and a minus 3 equal to 0. Is that fine? Everything is okay so far. Sorry, there is no. So wanted to y plus 1. Y plus 1. Y plus 1. Ah, yes, yes, yes, yes. This is y plus 1. I'm so sorry. Yeah, y plus 1, plus 1. So it will become a minus 2 here. This will not be there. This is minus 2. So if you expand it, it becomes this. Okay. Now, whenever you are in confusion, there's another way to deal with it. You can just write it as xi plus yj plus zk and compare both the sides. See, another way of doing it, but probably a slightly lengthier way to do it. So I've been given that r is equal to i minus j plus lambda times i plus j plus k plus mu times i minus 2j plus 3k. Okay. So what I'll do is first I'll write my r as xi plus yj plus zk. Okay. And I'll compare and I'll compare my i coefficients, j coefficients and k coefficients on both the sides. So x is equal to 1 plus lambda plus mu. Y is equal to minus 1 plus lambda minus 2 mu and z is equal to lambda plus 3 mu. Now, from here, I have to eliminate my lambda and mu. So what I'll do is I'll take any two of the equations. Let me take the first two, not first two, first and the last. Okay. So let me subtract it first. So when I subtract x minus z is equal to 1 minus 2 lambda, sorry, 2 mu. Correct. So mu is nothing but 1 minus x minus z by 2. Okay. Put it in the last one. So lambda is going to be z minus 3 mu. Mu is 1 minus x minus z by 2. Okay. So once you know these two values, put it in the second equation. So y is equal to minus 1 plus lambda. Lambda is going to be z minus 3 by 2 1 minus x minus z minus 2 mu. So minus 2 mu will be just minus 1 minus x minus z. Okay. And just simplify this further. So if you open the bracket, it becomes y is equal to minus 1 plus z minus 3 by 2 plus 3 by 2 x plus 3 by 2 z minus 1 plus x plus z. Okay. So minus 1 minus 1 will become so y is equal to minus 2 minus 2. Just correct me if I missed out on anything. This will become z will become 5 z by 2 plus z. Sir minus 3 by 2 and it's 3 by 2 x. So they won't cancel. Oh, my mistake. My mistake. So I'm sorry. I didn't see that x. I'm sorry. Yeah. So minus 2 and minus minus 2 and plus 3 by 2 I have right now minus minus everywhere. So it's minus 7 by 2. So this is going to be minus 7 by 2. So let me just remove the constants from everywhere and z term would be 5 z by 2 plus plus 2. So 7 z by 2 7 z by 2. So Z Z gone. And one more that is there. Okay. And you have 3 by 2 x and y is on the other side. So multiply with the 2 and bring the y on the other side so it becomes 3 x minus 2 y plus 3 z. Why am I getting 7 z 2 z 5 z just check if the values are correct x plus lambda plus mu. That's correct. So why is equal to minus 1 plus lambda minus 2 mu. That's also correct. And z is k lambda plus 3 mu. That's also correct. And if you subtract it becomes 1 minus 2 mu. So there's a small mistake here. It's a plus here. So this will also become a plus. So we understood the method. Understood the method. So just just write out this method also you should get the very same answer. Okay. So I'm not wasting time doing it. Okay. Is this clear? So all you need to understand is when you have been given such a scenario. It's as good as saying R minus a vector is coplanar to is coplanar to vector B and C. Okay. So this is another way of expressing R minus a BC STP is equal to zero. So this is equivalent to saying R is equal to a plus lambda B plus mu C. So don't be surprised when you see such a equation given to you. Don't misunderstand this as a equation of a line because it resembles very much like a equation of a line. If just this mu C has not had not been there, it would have resembled a equation of a line. Is the idea clear how this works? And this is not there in your NCRT. Okay. Now let us take few mix and match of these concepts that we have learned through problems.