 Can I have the next one everyone are you fine with this one? Yeah Yes, I've copied Try this one out. It's a good question. This is X minus one by two Y plus one by minus one Question is find the equation of the projection of this line on this plane Hope you know the meaning of projection now projection doesn't mean the image It means that what is the shadow of this line on the plane is the question to everyone See I'll draw it out for you So let's say this is a plane Let's say this is my line Okay, when I say projection of this line on this plane what I mean is The image of this line on the plane Image means a shadow not a shadow. Okay, so this shadows equation is what we call as a projection So we have to get the equation of the projection This problem actually Can be solved once we are able to find out the image of a point and projection of a point on a plane So let me take that opportunity first to discuss that with you and then we'll come back to this question Okay, so let me just take a Scenario where I have been given the equation of a plane Let's say AX plus BY plus CZ plus D equal to zero and there is a point X1 Y1 Z1 Okay, and I want to find out what is the foot of the perpendicular or you can say what is the projection of this point Let's say I call it as P and M. So what is the projection of this point? That is M on the plane Okay So I'll categorize this as question of image slash projection of a point projection of a point on a plane and I will also talk about the perpendicular distance No, how do I deal with such situations? Anybody who can tell me how do I get the point P and how do I get the length PM? So let me just ask you a few questions. I want M. I want PM Okay, and I also want the image of P. Let's say Q Q which is actually the image of P treating this plane as a mirror Any idea how to deal with this any suggestion Sorry, sorry, sorry, I didn't hear you. I was gonna say M lies on the normal vector of the plane and also satisfies the Equation of the plane but vector is not a fixed thing right for us. Normal can be here also, no Yeah, but then it also should normal can be here also normal can be here also normal can be here What is the meaning of a point lying on a vector should be on the line PM one day Yeah, you should say you should lie lie on the PM line I don't just lie on a normal vector that would be a wrong, you know terminology to be used So what do I do then if it lies on PM? Do I know the equation of PM line? Yes, I do know X minus X one X minus X one by a Y minus Y one by B Z minus Z one by C Correct see we know we know the direction of the normal to the plane, right? Right, so the normal's direction ratio to the direction ratios of the normal Would become the direction ratios of the line Yes, or no, and you know a point on the line X one Y one that one so you can write down Correct. Now. Let's say M corresponds to some value lambda So every point as you know on this line has a different different lambda so this point is a different lambda this point Is a different lambda this point is a different lambda this point is a different lambda this point is a different So even M has some different lambda. It's a value lambda here So can I express the coordinates of M in terms of lambda like this? Correct. So let's say this is the coordinates of M Now, how do I get lambda I get lambda by using this fact that this will satisfy The plane equation pi am I right? Yes, M lies on the plane. So it will satisfy the plane equation also So let me put this in place of X Y and Z. So a a lambda X one B B lambda Y one CC lambda Z one Plus D equal to zero. Okay. So if you find lambda from here, I'm just you know making things Look shorter. This is the value of lambda Am I right? Am I correct now place this lambda In these points you will get the coordinates of M Are you getting it? Yes, sir Okay, I'm not going to put that because you know how to do it. Well, yeah, we'll try to you know use that in the problem Which I've already cited before this Concept How do I get PM? PM is nothing but It's the distance between X1 Y1 Z1 and M point Correct. So it's the distance between P that is X1 Y1 Z1 and this M point Okay, what I will do is I will not substitute lambda first. I will simplify it and then put the lambda at the end So the distance formula is given by under root of Okay X minus X1. So what is X minus X1? Can I say it's a lambda? Y minus Y1, can I say it's B lambda and Z minus Z1. Can I say it's C lambda? Whole square under root clear So I'm finding the distance between this point and X1 Y1 Z1 So I'm taking the difference, right? So this is your X and this is X1 so X minus X1 whole square Y minus Y1 whole square Z minus Z1 whole square So if you see this, it'll give you mod lambda under root of a square plus B square plus C square Correct. Now put it back over here. Put this lambda back over here. So when you do that You get the perpendicular distance as mod AX1 BY1 CZ1. I'm just removing the negative sign because it has been modded by under root of sorry by a square plus B square plus C square Into under root of a square plus B square plus C square Okay, just cancel this and there would be an under root over here So the PM distance is going to be mod AX1 BY1 CZ1 plus D by under root of a square plus B square plus C square This is formula doesn't this formula resemble the distance of a point from a line in 2d? Correct. Just try to recall in 2d. How does the formula look like? AX1, BY1 CZ1 Yeah, the same formula. See again another resemblance with the line in 2d All you have is just an extra term CZ1 like that. Okay So if you remember this, if you remember your 2d very well, you don't have to mug up this formula separately Getting my point Yes, sir. Now, how do I find q? Very simple. You have already found out m You can find the value of q by using the fact that m is the midpoint of PNq getting my point By the way, I'm going to give you a simple formula to get m and q also very fast Okay, you don't have to follow this approach every time you can bypass this approach I'll give you a formula to get m coordinate and q coordinate By the way, this process itself is clear If you don't over rely on formula because you may forget the formula also But this approach you will never forget Yeah, it's clear Yeah, so I'll come back again to the problem. I know there's a problem pending on us So but before that I'll just give you a formula for formula for finding the projection of a point and image of a point on a given thing Okay so let's say there's a plane here and There's a point p x1 y1 z1 The projection of this point on this plane Let me write down the equation of a plane ax plus by plus cz plus d equal to 0 This projection is x2 y2 z2 Okay, and the image here. Let's say q is x3 y3 z3 Okay, the formula for projection is this So let's say x2 y2 z2 other coordinates. So you write x2 minus x1 by a y2 minus y1 by b z2 minus z1 by c is equal to negative a x1 b y1 c z1 plus d by a Square plus b square plus c square. This is a formula from here You can get your x2 y2 z2 because other things are already known x1 y1 z1 will be given to you a b cd etc. Would be known to you. Please do not use this in school Okay, we'll take a sample question also on this, but let me finish off first coordinates of image will be Following this formula x3 minus x1 by a y3 minus y1 by b Z3 minus z1 by c is equal to the only difference is we have a minus 2 We have just two coming up in between here And don't put a under root and all in the a square plus b square plus c square because we have a habit of putting an under root You know that habit has come because of finding the magnitude because of finding the distance formula So here be careful. Don't put an under root So please make a note of it not to be used in school. Please note that you will get a zero if you use this in school So if a formula is not an ncrt, you cannot use it Done noted Done, okay, so let's take a question on this Let me add one more thing find the length of the perpendicular foot of the perpendicular and also the image of this point in the plane Let me write it again for you. This is 2 x plus 4 y Minus z equal to 2 in case you're not able to read this this plane is this Now you can use the formula So please note the image also has to be found out So this is 2 x plus 4 y minus z equal to 2 and there's a point 7 14 5 P M Q so you have to give M number 1 number 2 P M Number 3 Q Okay, so let me show you how the formula can be used So let's say I want to find the coordinates of M So I'll call it as that's the x 2 y 2 z 2 and this coordinate. I'll call as x 3 y 3 z 3 So use the formula here x 2 minus 7 by a is 2 Okay, why 2 minus 14 by B, which is 4 and Z 2 minus 5 by minus 1. Okay, this is equal to minus a X 1 that is you have to put this point in this plane so 2 into 7 Plus 4 into 14 minus 5 minus 2 by a square B square C square a square B square C square Getting my point. Okay. So let's simplify this so this will give me X 2 minus 7 by 2 y 2 minus 14 by 4 Z 2 minus 5 by minus 1. This is going to be 14 plus 56 56 minus 7 So it's a 71 70 minus 3 which is 63 and This is going to be 21. Oh Perfectly divisible. So it's minus 3. Okay, so x 2 is 2 into minus 3 plus 7, which is nothing but 1 Y 2 is 4 into minus 3 plus 14, which is 2 and Z 2 is minus 1 into minus 3 plus 5, which is 8 So here you go. The coordinates of M is 1 to 8 Clear how it works So this formula is useful in competitive exams. You can save a lot of time Now once you've won once you've got M things are easy now PM distance can be easily found out by using our distance formula 7 minus 1 whole square is 6 square 14 minus 2 whole square is 12 square and 5 minus whole square is 3 square Okay Clear. So how much does it come out to be? 17 and the root of 36 plus 144 plus 9 It's a 1 7 root of 189, right? right root of 189 is like You can write it as 3 root 21 also correct If you use the formula directly, you can get the same thing. What was the formula formula was perpendicular distance is mod a x 1 b y 1 c z 1 Plus d by under root of a square plus b square plus c square correct So put this point 7 4 and 15 over here So you'll end up getting this same term over here. So I'm not repeating it. So it ended up getting Mod 63 by under root of 21. So mod 63 is 63 root 21 correct So 63 is written as 3 into 21 and root 21 you can cancel off a root here Again, the answer is same 3 root 21. So same answer we got at both the places So whether you use the formula or whether you use the direct principle you get the same result clear Okay, let's find out q coordinate as well. So for q coordinate, I will use the formula x 3 minus x 1 by a y 3 minus y 1 by b Z 3 minus z 1 by c is equal to minus 2 a x 1 b y 1 c z 1 plus d by Don't put an under root over here. Okay So x 3 minus the point was 7 I guess a was to This was Y 3 minus 14 the point the B was for here z 3 minus 5 minus 1 and by the way, this is it was already known 2 into 63 by 21. So that's minus 6 correct Well, I'm sorry Okay, so from here I can get x 3 as minus 12 plus 5 which is minus 5 Y 3 as minus 24 plus 14, which is minus 10 and Z 3 as 6 plus 5 11. So this will become your image. Hope I have done Yeah, that's correct. So 5 minus 5 minus 10 11 Minus 5 minus 10 Is that clear? Okay, now since I've started talking about the Length, let me also tell you something very important That is the distance of a distance between Between parallel planes Okay, so let us say I have two planes which are parallel to each other Pi 1 plane and pi 2 plane Remember when the planes are parallel, they only differ in their constant term just like a line does Okay, so let's say it is a x plus B y plus C z plus D 1 equal to 0 Other will be a x plus B y plus C z plus D 2 equal to 0 The only difference would be in their constant terms Okay, in vector form if you write it'll be r dot n plus D 1 equal to 0 and the other will be r dot n plus D 2 equal to 0 so their Perpendiculars normals will either be the same or proportional So you can always make them same by dividing or multiplying with the proportionality constant Now if I want the perpendicular distance between these two planes, let me call it as B. How would I find it? Very simple. I will choose a point on this plane Okay, so on the first plane I will choose a point. How will I choose a point there also? I will you know Take a shortcut. I'll put x and y both as 0 so z will automatically become minus D 1 by C Okay, and I would think that as if I'm trying to find the distance between this point Let me call this point as a from this plane pi 2 Okay, so let's find out the distance between a point and pi 2 plane So let me use the formula a x 1 B y 1 C z 1 please note here d 2 will be there by Under root a square plus B square plus C square So your x 1 was 0 y 1 was 0 z 1 was minus D 1 by C Okay, so this point you have to use so that would become Minus D 1 plus D 2 by under root of a square plus B square now again This also resembles the formula between two parallel lines C 1 minus you see one minus C 2 by under root a square plus B square right, okay? Is that clear? Okay, in case of vector. How would the same equation become it will become more D 1 minus D 2 by modern The same end that you have in both the lines. So this will become your distance So, please remember in vector form also clear CLR CLR