 Now, guys, we are going to talk about the last concept for the day. That is the equation of bisector of angles between two planes. Everyone is here? Everyone is attended now? This is the last concept? Yes, sir. Yes, sir. Now, between two planes, let's say I'm drawing these two thin planes. The angle that is created, so there are two types of angle. One is the acute angle and another is the obtuse angle. So through these two planes, passing through the line of intersection of these two planes, I can make a bisector plane like this, which bisects the angle. So this is getting bisected. Or I could also have a plane that is a green plane like this. This plane bisects the external angle. So these two are your bisector planes. Let me call it as B1 and B2 planes. So now, if the two equations of pi1 and pi2, that is the planes whose angles are getting bisected, are provided to us, how do I get the equation of the bisector planes? So let's say I give you the equation of the two planes. How do I get the equation of the bisector planes? Any idea? Equate the angles made by... But don't you think there's a difficult way to get it? There's an easier way to do it. It's very long. Yeah, very long. See, very simple. I'll just draw the cross-section view. So let's say these are the two planes. Okay. And you're looking for the equation of these bisector planes, correct? Can I say bisector planes are locus of such points which are equidistant from these two planes? So this and this distance should be equal. Similarly, this and this distance should be equal. Yes or no? So we already know the equation. We already know the formula for distance of any point. Let's say I take any generic point, hkl. Okay. So can I say the distance of this point from pi1 and pi2 planes must both be equal? Okay, now we got it. Oh, you got it? So you can say a1hb1kc1l plus d1 mod y under root of a1 square b1 square c1 square will be equal to mod a2hb2kc2l plus d2 by under root of a2 square b2 square c2 square, correct? Now drop the mod from both the sides and generalize it by replacing your h with x, k with y and l with z. So you'll end up getting the equation as a1xb1yc1z plus d1 by under root of a1 square b1 square c1 square is equal to plus minus a2xb2yc2z plus d2 by under root of a2 square b2 square c2 square. Is that clear? Yeah. Now plus minus means there are two equations possible, thereby giving you both the equations of bisector b1 and b2. But the problem is how will I know which is b1 and how will you know which is b2? That's something which we'll discuss in some time. Okay. Meanwhile, just a quick question on this so that you are comfortable with this concept. Please note this down because I'm switching to the next board. Let me know once done. Yes, sir. Then we'll do the first part of this question first. Find the equation of the bisector of the angles of these two planes and specify which plane bisects the acute and which bisects the obtuse will take that later on. Sit down, sir. Done? Yes, sir. Yes, sir. So I'm directly writing the result. This by under root of this will be 3 as I told you, this figure is quite popular plus minus and this will be by 7, right? Now cross multiply, take a plus sign first. So take a plus sign first. You'll end up getting 14x minus 7y plus 14z plus 21 is equal to 9x minus 6y plus 18z plus 24. Bring it to one side. It becomes 5x minus y minus 4z is equal to 3. So this is one of the bisectors. Now take a negative sign. You'll end up getting 14x minus 7y plus 14z plus 21 minus 9x plus 6y minus 18z minus 24. On simplification, it gives you 23x minus 13y plus 32z plus 45 equal to 0. This is another bisector. Cool? Now how would I know which bisector is the acute angle bisector, which is the obtuse angle bisector? Now for that, you can use multiple things. You take one of the bisectors and you take one of the given planes. So let's say you pick up one of the original planes given to you, pi1, and you pick one of the bisectors, b1. Now find out the tan of the angle between them. You can always find the cos of the angle. For any two planes, you can always find the cos of the angle between the planes. Convert that cos to tan. So let's say you got this angle alpha and tan alpha mod happens to be greater than 1. If it is greater than 1, that means the bisector you had chosen was the obtuse angle bisector. And if tan alpha is less than 1, that means the bisector that you had chosen was the acute angle bisector. Is that clear? Now how is this one important over here? See, if it was an acute obtuse angle bisector, alpha would have been greater than 45 degrees. So tan alpha would have been greater than 1. That's how. If that angle was acute angle, if that angle were acute angle, then alpha would have been less than 45 degrees. So tan alpha would have been less than 1. Are you getting my point? Yes, sir. Now, many a times they ask you a very targeted question. They will say find the equation of the obtuse angle bisector of these two planes. So many people think that finding both the equations is just a waste of time. So can I have a direct algorithm or a formula by which we can jump to the equation that we need directly rather than finding both the equations and then figuring it out which is what? Okay. So I have an algorithm. Okay. I have a method by which you can directly jump to find the acute or obtuse angle bisector directly. Now listen to this formula very, very carefully because many people make a mistake in understanding it. Step number one, write down both the equations of pi 1 and pi 2 in such a way that the constant terms have been made both positive. So please note you should have made your d1 d2 both as positive. Let's say if one doesn't have a positive d1 or d2, what do you do to that? Multiply with a minus 1. Understood? Okay. Step number two, find a1 a2 plus b1 b2 plus c1 c2 value. Okay. Let's say this value is k. If k is positive, if k is positive, then a1 x, b1 y, c1 z plus d1 by under root of a1 square b1 square c1 square with a positive sign in between, with a positive sign in between would represent a obtuse angle bisector. Understood? And if k is, yeah, and the one with a negative sign will automatically become, I'm not writing the whole thing, will become the acute angle bisector. If k is negative, then negative sign in between will become the obtuse angle bisector. The same thing with a positive sign in between will become an acute angle bisector. So how do I remember this? I have a mnemonic for this. Positive means positive will be obtuse. Negative means negative will be obtuse. So whatever the sign of k, the same will be the sign to be used for obtuse angle bisector. Okay. So k positive means positive sign to be used for obtuse angle bisector. k negative means negative sign to be used for obtuse angle bisector. That's how I remember it understood so one quick question on this so for the last one negative one will be of accurate is it for the last question for the last question I give you yes the second part okay we'll take that question itself fine guys one more important thing that I wanted to tell you many people make a mistake there let me go back to the board number 31 see here once you have made your D1 D2 as positive okay you should use your new equation everywhere means everywhere means in finding the final answer so here your new equations will be used not the old ones are you getting my point it's better to discard the old equation the moment you have made your D1 D2 positive just cancel out the previous equations so you have to work with your new equations everywhere remember is that clear yes okay don't use your start using your old equations at this step that would make it wrong everything will go wrong okay fine now going back to the previous problem let's say I want to find the acute angle bisector so first I will do D1 D2 positive oh thankfully D1 D2 is positive already here so there's no need to make any change to the equation so these are your new equations also okay so here D1 and D2 are okay then find A1 A2 A1 A2 B1 B2 C1 C2 that's positive okay so positive means with a positive sign acute angle bisector will come sorry obtuse angle bisector will come and with a negative sign acute angle bisector will come so the acute angle bisector if you want so you can directly write with a negative sign guys okay so if you simplify it I am sure you know the answer what is that 20 3x minus 13 y plus 32 z plus 45 okay so this one here at the acute angle bisector okay and the other one which was 5x minus y minus 4z is equal to 3 this one will be your obtuse angle bisector okay so one last concept which is remaining let me take that up quickly position of two points with respect to a plane hardly will take me two three minutes of two points with respect to a plane so if you have been given a plane let's say ax plus by plus cz plus d equal to zero how do I know whether the two points a and b okay whether they lie on the same side or whether a and b lie on the opposite sides of the plane okay so let's say a is given by x1 y1 z1 b is given by x2 y2 z2 okay how do you know it's lying on the same side or on the opposite sides okay now the method for this is basically very simple I'll just discuss the result with you put this point in the plane so ax1 b y1 cz1 plus d evaluate this number and evaluate the value of this also so both these points you put in the plane okay if these two are of the same signs okay then both are on the same side and if these two are of opposite signs if these two numbers are of opposite signs then they would lie on the opposite sides if the idea clear now how does this come actually very simple if the points are on the same side then what will happen is the plane will sorry if the points are on the same side what will happen the planes the plane will externally divide the join of these two points isn't it so this point c will externally divide the join of a and b so there's an external division happening here okay whereas if they're on the opposite side the plane will internally divide the join of a and b are you getting it so here if I assume this ratio to be lambda is to one okay and here is to be lambda is to one the lambda here will come out to be negative lambda here will come out to be positive okay so let me express the lambda first so position of the point c could be written as x1 plus lambda x2 by lambda plus one y1 plus lambda y2 by lambda plus one and z1 plus lambda z2 by lambda plus one isn't it now lambda would be negative positive that is something which I'll take care later on so as of now what I'm I'm I'm taking I'm taking this point c coordinate to be x1 this is your let's say oh sorry this was your x2 y2 z2 right okay and this was your x1 y1 z1 yeah so I've taken it to be lambda is to one isn't it okay so the coordinates of c would be one into x1 lambda into x2 by lambda plus one likewise but this should satisfy this should satisfy ax plus by plus cz plus d equal to zero so when I put this point it'll become a x1 plus lambda x2 by lambda plus one b y1 plus lambda y2 by lambda plus one and c z1 plus lambda z2 by lambda plus one plus d equal to zero multiply throughout with lambda plus one so it becomes ax1 plus lambda x2 y1 plus lambda y2 cz1 plus lambda z2 plus d lambda plus d equal to zero take all the lambda terms together so it becomes ax2 by 2 cz2 plus d is equal to negative ax1 by 1 cz1 plus d so lambda is negative ax1 by 1 cz1 plus d by ax2 by 2 plus cz2 plus d now for internal angle bisector sorry for external angle bisector lambda should be negative if lambda is negative that means this quantity itself should be positive correct let me call this quantity as beta beta should be positive okay if beta is positive means both the numerator and denominator must be of the same sign that's what I wrote over here that's what I wrote over here getting my point similarly if lambda were to be negative sorry lambda were to be positive that means this quantity beta has to be negative and if beta has to be negative it means these two quantities must be of opposite sign correct these two must be of opposite sign then correct that's what I wrote over here is the idea clear fine then I'll stop here thank you so much we have officially completed the syllabus