 So let me begin with this question a very simple question, which is based on the normal form So I hope you can read the question Find the unit vector perpendicular to the plane r dot 2i plus j plus 2k equal to 5 Okay, just a quick recap of what we did in the last class regarding the normal form So basically normal form is where your Distance from the origin is stated. So let's say this distance of the plane pi from the origin over stated as p and Not only that, you are provided with the direction cosines of this normal. Let's say l mn, right? So what are the equation for such a plane equation for such a plane becomes r dot li plus mj plus nk is equal to p Okay, this is the vector form for it. If you write the same in Cartesian form It becomes simply replace your r with xi plus yj plus zk That means you do this activity Okay, and when you just do the dot product, you end up getting lx plus my plus nz is equal to So this is the Cartesian form of the equation. Okay based on this form itself is the question Find the unit vector perpendicular to the plane this Okay, so now try to compare this and the given equation Sure says done anybody else. Yeah, simple Just compare these two equations. So this we know that it's a unit vector. It's a unit vector, which is perpendicular to the plane All right. So all you need to do is make this a unit vector. So the unit vector Which is perpendicular to the plane is just this vector divided by the magnitude of the vector That is 2 square plus 1 square plus 2 square, which is nothing but 2i plus j plus 2k over 3 Okay Now there's one more thing that is associated with it that if you would make this as a unit vector By dividing by 3 throughout then the constant on the right hand side signifies the distance from the origin So if I write the equation like this r dot 2i plus j plus 2k Over 3 then the right hand side becomes 5 by 3 right because you're dividing by 3 on both the sides And this number starts behaving as the distance of the plane pi from the origin Then from the origin Okay So last class I had also told you that The learning here is that whenever you have been provided the equation of a plane in the form of a x plus b y plus c z plus d equal to zero These numbers a b c and d are not ordinary numbers they signify something or the other about the plane the numbers a b c actually signify the Direction ratios of the normal to the plane I'm sure you have noted this down last class also Okay, and mod d by under root of a square plus b square plus c square it represents the Distance of the plane from the origin distance of the plane from the order So please keep these two things into mind So whenever you see the equation of a plane you should know that the coefficients of x y z are nothing But they are direction ratios of the normal and the constant divided by under root of The square of the coefficients of x y z added up will give you the distance of the plane from the origin Is that here everyone So we'll take an x question on this now the same concept Just formulated as a question Find the distance of the plane to x minus y minus 2 z minus 9 equal to 0 from the origin I Think you should not take more than two seconds to answer this Hello, so what are the distance tell me for effort So we have just discussed the distance is mod d by under root of a square plus b square plus e square We'll also discuss this concept when I'm doing the distance of a point from the plane But since we took up that normal form, I thought I will take a problem on this So mod of minus 9 by under root of 2 square minus 1 square minus 2 square that's again 9 by 3 that is 3 units done straight away Just by appreciating the normal form you can solve this question. Is that fine? No questions. Let's have another one then, okay Find the vector equation of a line Okay, so I'm asking your equation of a line this time Passing through 3i minus 5j plus 7k and perpendicular to the plane 3x minus 4y plus 5z equal to 8 So there is a scenario like this that there is a plane And there is a line which is perpendicular to this plane Okay, so this is your right angle Okay, so let me call this plane as pi plane. This has a line Yeah, we have to find the equation of the line Just type done when you're done. Okay, so so that I know who all are done So done Done everyone Yes, sir Just for more problems. These are not the difficult ones. They're just a warm up problem Okay, so I'm sure the direction ratios of the line will be same as the direction ratios of the perpendicular to the plane And you know that 3i 3 minus 4 and 5 represent the direction ratios So you can write the equation of a line directly as x In fact, they're asking you the vector equation. So let me write vector equation r is equal to 3i minus 5j plus 7k plus lambda times The coefficient of these xyz so form a vector by using these coefficients because those coefficients represent the normal direction ratios Okay, so this is the answer simple as that if you have to write the same thing in Cartesian You will write it as x minus 3 by 3 y plus 5 by minus 4 and z minus 7 by 5 Is that fine? Let's take another one then so here goes the next question from your ncrt book itself Find the foot of the perpendicular drop from the origin to a plane Sorry, the foot of the perpendicular dawn from the origin to a plane is already given 1 comma 2 comma minus 3 Find the equation of the plane And there's one more thing if o is the origin and the coordinates of p is 1 2 minus 3 Then find the equation of a plane passing through p and perpendicular to op. So let's do both the questions even though it is or So let me just draw the situation for you. So this is your plane Let's say this is your origin And the foot of the perpendicular drop from the origin here. Let me call it as point uh m That's 1 2 minus 3 It's the first question But First question is super easy Since you know the points of uh the foot of the perpendicular and you know origin is 000 You can always know the direction of the normal Correct So the direction ratios of the line om would be nothing but the coordinates of m itself which is 1 2 minus 3 correct and Once you know the point you can use r minus a dot n is equal to zero correct So r dot n is equal to a dot n So a itself is also this point and n is also the same point. So a vector itself is your 1 2 minus 3 by the way, I can represent vector like this also fine. So no need to put ijk every time But when you're using it in school, please confirm from your school teachers that you can write a vector like a column matrix Okay So r dot n r dot n n is your i plus 2 j minus 3 k is equal to a dot n So a dot n would be 1 plus 2 square And 1 square 2 square and minus 3 square correct So r dot i plus 2 j minus 3 k is equal to 9 plus 4 13 plus 14 Okay Write it in a Cartesian form. It will become x plus 2 y minus 3 z is equal to 14 Now if you directly want to write in Cartesian form Recall that you can use this formula A x minus x 1 b y minus y 1 c z minus z 1 equal to zero Okay, directly you can write in Cartesian form where a b c are direction ratios of the normal x 1 y 1 z 1 other given point or is a given point on the plane So I can write it as 1 x minus 1 plus 2 x minus 2 I'm sorry y minus 2 And minus 3 z plus 3 equal to zero Correct. So if you expanded it becomes x plus 2 y minus 3 z minus 1 minus 4 minus 9 equal to zero which is nothing but x plus 2 y minus 3 z is equal to 14 So multiple ways to reach to the same answer So first part is done by everyone. Is there anybody who is feeling trouble in finding this? Guys, let me tell you 3d geometry is very easy if you understand things well But if you don't it becomes equally difficult I have been getting complaints from a lot of students that I don't understand 3d geometry I'm not able to imagine it She's very simple. Just try to understand the basic logic behind these formula Our second part done Yes, no trying So aren't they like the same thing same question right? Okay, so no need to do it Or is the origin coordinate of p is find the equation of a plane passing through p and perpendicular to op so same thing So no need to do it. Let's move on to the next question Find the equation of a plane says that the image of the point 1 2 3 in it is minus 1 0 1 Image means not the projection image means the The one you have when you have a plane mirror as your plane itself. So let's say this is your plane And the point here is 1 2 3 when I say image. I mean this point I mean this point. So let's say this is your object. This is your image Many people miss construe Image to be the projection of the point on the plane So the midpoint of onm. Let's say m that is the projection of onm. That is not the image So read the question very carefully whether they're asking the projection or whether they're asking you the image Done anyone? Yes, sir two minutes ending it Two minutes. Okay fine Yes, sir. I've sent it Oh, you could have just typed it on the chat box also, no problems. Okay, so let me see Oh sure. That's not correct At least it doesn't match with the one given in the in the answer for it Just check once more Yes, sir So I sent an answer Okay Likith is correct No No, Samyukta Samyukta Paras Paras you send me the equation of a line it seems I need the equation of a plane Likith is correct so far Yes, now you're correct Shreya Yeah, I realized my mistake Okay, see here, uh I was expecting everyone to Answer this So two things you need to get the equation of a plane one is the direction of the normal and other is a point on the plane Correct Now how would I get direction of the normal to the plane? So direction of the normal to the plane is nothing but the direction ratios of the line oi correct So direction ratios of the normal To the plane is nothing but that's the direction ratios of the line oi correct How do you find direction ratios of any line when you know two points on it? Simple you take the ratio or you take the difference of their coordinates Correct. So direction ratios will you just subtract it to That is one minus minus one that's two then two minus zero again two then three minus one which is again two Okay, so these are the direction ratios of the normal Now we I also need a point on the plane I can get the point by using the fact that m is the midpoint of oni So what are the m coordinate one minus one by two which is zero Then two plus zero by two which is one and three plus one by two which is two Correct Now all I need Is the equation of the plane when you know a point and the direction ratios of the normal and that is a x minus x one This equation is very important. Let me tell you guys Very very handy very very useful equation When you know the direction ratios of the normal and you know a point on the plane, okay So just put the values now a is going to be two x minus x one is x minus zero b is again two y minus one c is again two z minus two Okay, you may drop the factor of two from everywhere because there's a zero on the right hand side So you'll end up getting x plus y minus one z minus two equal to zero Which is nothing but x plus y plus z is equal to three Makes sense Happy Okay, so don't give me the equation of line. I wanted the equation of a plane over here Any questions so far, please ask me Is that fine? Okay, great