 Then let's go back to the problem which we had left off. I think it was on board number 22 Yeah, this guy now give it a try So here the idea is if you want the equation of a projection we first need to know what is the point of intersection Okay, that we can find out easily And we have been also yeah, and we have been also indirectly given that the point one Minus one and three lies on this so you find the image of this here So once your P and Q are known you can always write down the equation of the line connecting P and Q. That's the whole agenda Got it. So see what what what does my requirement here? My requirement is the equation of this line of projection So I'll first find out P Then I'll find out Q Then I'll find out the line connecting P and Q Okay, so how will I find P? How will I find Q that I'll leave up to you? So please do it and tell me the answer. I'm waiting So I got the answer in Cartesian form That's fine. What's the answer? So I got x minus 9 by 2 by 2 is equal to y minus 11 by 4 by minus 1 is equal to z minus 10 by 4 Okay Anybody else? Okay, so let me take this up quickly So for finding P we'll assume that P P corresponds to some lambda parameter So let's say the parameter corresponding to P is lambda. So let me take it as lambda so I can write down the Coordinates of P as 2 lambda plus 1 comma minus lambda minus 1 comma 4 lambda plus 3 Okay, this must satisfy pi this must lie on this plane pi So I can say 2 lambda plus 1 plus 2 times minus lambda minus 1 Plus 4 lambda plus 3 is equal to 9 cool So let me solve for lambda. I think 2 lambda 2 lambda will get cancelled. So it'll be 4 lambda and You have a 3 plus 1 which is 4 4 minus 2 is 2 is equal to 9. So lambda is 7 by 9 7 by 4, sorry Correct 7 by 4 7 by 4. So the point P here would be 7 by 2 plus 1 which is 9 by 2 Okay, minus lambda minus 1 will be minus 11 by 4 and 4 lambda plus 1 will be 10 Yeah How would I find Q Q can be found out by using the formula which I just now told you but not to be used in school So let's say Q is x 2 y 2 z 2 So x 2 minus 1 by 1 Y 2 plus 1 by 2 and Z 2 minus 3 by 1 is negative Put this point over here. You get 1 minus 2 plus 3 minus 9 by a square b square c square a square b square c square So we have x 2 minus 1 by 1 is equal to y 2 plus 1 by 2 is equal to z 2 minus 3 by 1 And this will give you minus 11 plus 4 so 7 by 6 correct So x 2 is equal to 11 by sorry 13 by 6, right? Y 2 is equal to 7 by 3 minus 1 4 by 3 and z 2 will be 25 by 6 am I correct? Yeah, so this is your point Q So once you know those two points P and Q and you have to write down the equation of a line passing through them That's very simple. You can say x minus 9 by 2 y plus 11 by 4 z minus 10 and divided by the difference of these two points So 27 by 6 minus this will be 27 by 6 That's 9 by 2 by the way minus 13 by 6 that's going to be 14 by 6 14 by 6 is 7 by 3 okay Then subtract these two minus 11 by 4 minus 4 by 3 so minus 33 minus 16 minus 49 by 12 correct And finally 10 minus 25 by 6. That's 35 by 6 Is that fine? I don't think so you got the same answer Aditya So my direction ratios are not correct. Okay, so one thing you can multiply this with 12 12 12 in the denominator also Okay, and you can put your answer as this x minus 9 by 2 by 28 y plus 11 by 4 by minus 49 and Z minus 10 by 70. Okay, and the good thing is you can drop the 7 factor So x minus 9 by 2 by 4 is Y plus 11 by 4 by minus 7 is equal to z minus 10 by 10 Okay, and you can further simplify by multiplying by 4 in the numerator. Anyways Now there's another way to deal with this problem Can I get the equation of a plane which contains this line? Can I get this plane equation which contains this line guys? So can I get a plane which is Containing this line and perpendicular to this plane see what I'm trying to do is see my Intentions my intention is to get a plane So this was the line right whose projection we needed So if I make a plane which is containing this line Correct and which is perpendicular to it So can I say the projection the projection line will be nothing but a line of Intersection of this plane and this plane pi 1 and pi 2 plane Can I do it like this also? Let me know if you agree with what I'm saying see consider a plane which is perpendicular to pi 1 and Containing the given line to you Can I say the combined equation of pi 1 and pi 2 will give you the asymmetric form of the line of Intersection, which is nothing but the projection line Guys, yes. No, maybe Yes, sir. Yes. No, sir. Okay, so Can you get a plane which is perpendicular to this and containing this line? Yes or no Once you get that plane just write down this plane and the other plane that you get as an asymmetric form That will give you the equation of the projection of the line from there. You can convert it to symmetric form That's your call. Is that fine? Will you do that as homework and send it to me? So here my question was get a plane which contains x minus 1 by 2 y plus 1 by minus 1 and z minus 3 by 4 and Perpendicular to x plus 2 y plus z equal to 9. Okay, once you get this plane then this plane and The other plane that you get would be the asymmetric form will form the asymmetric form for the equation of the projection line clear Yes, sir Next concept that will take up is the concept of angle between a line in a plane So let's say there's a plane here and there's a line like this, which is passing through the plane So the angle between the line in the plane is basically defined as the angle between the line and its projection and its projection on The given plane, okay. By the way finding projection is difficult. That's why we also say that it's Supply it's complimentary of the angle between the normal to the plane and the line This will become your 90 minus theta. That's an easy way to look at the same scenario Okay, so you can say that it is also the Complementary of the angle between the line and the normal to the plane getting this so Let's say this line equation is r equal to a1 plus lambda b1 Okay, and this plane equation is r dot n Plus d equal to zero So can I say the angle could be found out by using the formula dot protect that is cos of 90 minus theta Okay, and if you find if you want the acute angle, it will be mod of n dot b1 by mod n mod b1 Okay Please note It will now give you sign of the angle. Okay, so don't forget here sign will come When you are finding the angle between a line and a plane your theta would be sign inverse of this However, you're using the dot product itself, but here because your theta Here is a complimentary of the required theta. That's why a sign will come. Is that clear? CLR, okay Can we have a quick problem on this? Let's take this is plus here Then what is the answer? Sign inverse 4 by Sign inverse 4 by 3 root 2 Okay, good That's correct So your sign of theta would be Sign of theta would be this is what we need these two other things which we are concerned with this and this So dot product of it will you what? 2 plus 1 plus 1 By and a root of this is going to be root 6 and this is going to be root 3 Okay, so 4 by this is root 2 into root 3 root 3 which is nothing but 4 by 3 root So theta is sign inverse of 4 by 3 root Fine Yes, sir. Let's take more questions. Yeah, this is based on family of planes Read the question carefully There's a plane a x plus b y equal to zero and it is rotated through an angle of alpha about its line of intersection with the plane That equal to zero Show that the equation of the plane in the new positions is given by so they lead to answers So that's why there's a plus and minus in the answer. How do I do this? So can we find the point where it intersects the girl? No need no need you can use family of planes for this see it the situation is there was This plane a x plus b y equal to zero plane and there is z equal to zero plane Okay Now through the line of intersection, which is this line Okay, this plane pi 1 is Given a twist of alpha. It can be in both directions. It can be up. Also. It can be down also So let's say I reach a new position here. Okay. I want the equation of this guy Now you can all see that the yellow planes and the white plane they all form a family member, isn't it? So can can I say let the required plane will have the equation let the required plane have this equation A x plus b y plus lambda z equal to zero Yes or no Now in order to find the value of lambda, I can use the fact that the angle between these two planes here is alpha correct And you already know how to find the angle between two planes when you know the Know their equations, isn't it? a1 a2 plus b1 b2 plus c1 c2 by under root a1 square b1 square c1 square under root a2 square plus b2 square plus c2 square So I can say cos alpha Okay, now this equation was a x plus b y equal to zero Or you can say zero z equal to zero So a1 a2 will be a square b1 b2 will be b square by under root of a square plus b square under root of a square plus b square plus lambda square Okay Now this you can cancel off and write a root over here square both the sides So when you square both the sides you get cos square alpha is equal to a square b square by a square plus b square plus lambda square Reciprocate it So a square plus b square Plus lambda square by a square plus b square is equal to secant square correct So you can write this as 1 plus lambda square by a square plus b square is secant square Send this one to the other side So lambda square a square plus b square is equal to secant square alpha minus 1 which is tan square alpha So from here it implies lambda could be written as a lambda square could be written as a square plus b square tan square alpha correct So lambda could be written as plus minus under root a square plus b square tan alpha correct just put this in place of Just put this in place of lambda over here So when you put the lambda value you end up getting a x plus b y Okay, plus minus under root a square plus b square tan alpha equal to z This is what was given to us Hence proved Is the idea clear? This question has come so many times. It has come so many times in comparative exams The same question clear everyone