 Let's take another question. Find the point of intersection of a line passing through 001 and intersecting the lines and intersecting the lines, these two are the lines with the XY plane. So let me explain this question. It sounds very peculiar, but the question says there's a line, okay, which passes to this point, okay, and it intersects these two lines. So what is the point of intersection of such a line with the XY plane? That's what they're asking you. Is anybody who is not able to understand the question, please ask. What happened, Aditya? So I'm not able to understand, see, I'll make you understand this question first. So let's say these are the two lines, L1, L2, okay, so I'll call this line as, let me use a black color, this is your L1, this is your L2, okay. So there's a line which passes through 001. Let me make this line with yellow color, like this, fine. So Aditya, what is trying to say is that this line, yellow line intersects these red lines, L1, L2, okay. Now where will this line go and intersect the XY plane? Yes sir, now I'll try it. Should I discuss or are you guys trying? You want some time? Do you want more time or should I start solving it? Sir, from my side you can start solving. Okay, are there, anybody who is trying? Okay, so let's discuss this, fine. See here, first of all, let's find out, let's find out a point on this. So for L1, L2, we need the equation, okay, so let's focus on L1 first. So x plus 2y plus z equal to 1 and minus x plus y minus 2z is equal to 2. To get a point, let me put z as 0 here, okay. So it gives you x plus 2y is equal to 1 and minus x plus y is equal to 2. So y is equal to 1, okay. So z0 and y is 1 means x has to be minus 1, correct. So here one of the points are known to me on this line which is minus 1, 1, 0, okay. Let me also find a point on L2. So in L2 again I can put x plus y is equal to 2 and put z as 0, so x is equal to 2. So x is 2, y is 0, okay. That's correct. Ardhara, you are absolutely correct. So Ardhara has given me the right answer. Okay, never mind. Let's continue. So when x is 2, y is 0 and z is also 0. So basically you have a point again, 2, 0, 0 over here, fine. Now let me also find out the direction offline, direction ratios of L1, okay. So for direction ratios of L1, direction ratios of L1, let me use the vector product ijk, 1, 2, 1, minus 1, minus 1, 1, minus 2, okay. So it's i times minus 4, minus 1 is minus 5, minus j, minus 2 plus 1 which is minus 1, plus k, 1 plus 2 which is 3. So that's minus 5i plus j plus 3k, okay. So the direction ratios of this line are minus 5, 1, and 3. Let's find the direction ratios of line L2, that is ijk, okay. And the direction ratios for this I'll have to use 1, 1, 0 and 1, 0, 1. So i1 minus j, again a 1 and k minus 1. So direction ratios of this line are 1, minus 1, minus 1, fine. Now let the line that we desired have a direction ratio of abc, okay. Now use the condition of intersection. Guys, how many of you still remember the condition of intersection of two lines? So the condition of intersection of two lines is x2 minus x1, y2 minus y1, z2 minus z1 by a1, b1, c1, a2, b2, c2. Hope you understand what are the meaning of x1, xy1, z1, x2, y2, z2 and a1, b1, c1. Basically these are the equation, these are derived from the two lines which are intersecting. So condition for intersection is when two lines intersect, okay. This must be 0, why did I write conditions, condition, okay. This you have to remember because this is going to be asked in slightly higher version of the equation to you in school as well. So now let's say I use the fact that my L3 line intersects L1 line. So what is the condition of intersection for this? If L3 intersects L1, can I say x2 minus x1? So let's take the difference of the points first and you can take it in any order you want. So difference is going to be 1 minus 1 and 1 and then write down the direction ratios of L1 that is minus 5, 1 and 3 and then write down the direction ratios of the L3 line. This should be 0, am I right? So this is 0, expand it, what I will get, c minus 3b plus 1 time minus 5c minus 3a plus 1 minus 5b minus a equal to 0, getting it, okay. So let me simplify this, what does it give you? How many a terms are there, minus 3a minus a which is minus 4a, how many b terms are there, minus a to b, how many c terms are there, minus 4c, okay. Drop a factor of minus 4, drop a factor of minus 4, so it becomes a plus 2b plus c equal to 0. So this is my first equation. Similarly let's use the condition of intersection of L3 with L2. So if L3 intersects L2, what will I write, okay. So let's look at here, so first take the difference of the points, difference of the point will be 2, 0 minus 1, 2, 0 minus 1, 2, 0 minus 1 and a, b, c will come in the action ratio point and 1 minus 1 minus 1, 1 minus 1 minus 1, let's expand it. So it becomes 2 times minus c minus of minus b, so it's b minus c and minus 1 time b plus a. So you end up getting b minus a plus b minus 2c equal to 0, this is your second equation. So first equation, second equation. So roughly I've understood what I'm trying to do, everyone. Then only I'll proceed further. So what I did first, I got the equation of both the lines, correct. And I then use the fact that my required line L3 intersects these two lines. So from there I got these two conditions of intersection. Getting my point? Now I have to get my ratio of a, b and c. How will I do that, c, everyone? There are two ways to do it. Many people, they write a and b, many people write this in terms of c, okay. While I prefer some other method, what I do, I use cross multiplication method. I'm sure you would have done this in class 10th while you're solving system of simultaneous equation. So what I do? So you told in the class also? Yeah, I told in the class also. So a, then hide this minus 4 minus 1, which is minus 5. Under minus b, you have minus 2 plus 1, which is minus 1. And in the c, you have minus 1 plus 2, which is 3, okay. So your a by minus 5 is equal to b by 1 is equal to c by 3, okay. So my desired line will become x minus 0 by minus 5, y minus 0 by 1 and z minus, sorry, z minus 1 by 3, okay. Now you want to know where does this line intersects the xy plane. Remember in xy plane, z is 0. Yes or no? Yes or no? So z is 0 on the xy plane. So put z as 0. So when you put z as 0 over here, x automatically comes out to be 5 by 3, right? And y automatically comes out to be 1 minus 1 by 3, correct? So the point of intersection of this plane with the xy plane is going to be, I'm writing the answer here at the corner, 5 by 3 comma minus 1 by 3 comma 0. Are you getting my point? Everyone is that clear? Okay. Now I'll make this concept simplified once I talk about family of lines. Once I talk about family of lines, I'll make the same problem, a very simplified one. You can solve this within half the time that you have taken to solve this, okay. But it's always important that you understand the underlying concept. So so many concepts got tested over here. The concept of interline of intersection, first of all, the concept of condition of intersection of two lines, all of those concepts got tested in this problem. So I'll come back to the same problem with a simpler solution, okay. But let me take before it the concept of family of planes. Any question, anybody? All right. So now we'll discuss about family of planes. Now it's very similar to the concept of family of lines that you had learned in your class 11. Okay. So how do you form a family? Basically, there are two planes which will be given to you, okay. So I call them as the parent plane, because these two parents, the family will be generated, okay. Now through the line of intersection, let me make a line of intersection over here. So this line of intersection, there can be so many planes passing. For example, they can be a plane like this, let me choose a color here, yeah. So they can be a plane like this passing, okay. They can be a plane like this passing. So there can be so many infinitely planes passing through the line of intersection of these two planes. So all these planes will constitute a family. Are you getting my point? So these two planes, let me call as pi 1, pi 2 planes. Those are the parent planes and through the line of intersection here, there are so many planes which can pass and they all constitute a family. So how do you write the equation of any general family member? So that's the equation of pi 1 is given to us as a 1 x plus b 1 y plus c 1 z plus d 1 equal to 0. An equation of pi 2 plane is given to us as a 2 x b 2 y plus c 2 z plus d 2 equal to 0, okay. Then the equation of any family member could be written as lambda 1 plus lambda pi 1 plus lambda pi 2 is equal to 0. That is a 1 x b 1 y c 1 z plus d 1 plus lambda times a 2 x plus b 2 y plus c 2 z plus d 2 equal to 0. Are you getting it? Now here lambda is what? Lambda is some real number. And as you vary your lambda, you get different different family members. So just like you had done in case of a line, whenever you differ your lambda, you get different different family members belonging to that family. So lambda is like an identity card for a family member. So for a different lambda, you get a different family member. For example, the plane pi 1 itself, it has a lambda value 0 because when you put lambda as 0, all you are left with is the equation of the plane pi 1. For the plane pi 2, lambda is infinity. So for different lambdas, you get different different family member. How do I get a lambda for a given question? For that, an additional condition would be mentioned to you in the question itself. For example, they will give you a question like this. Let me just take an example question. Let's say they give you a question that find the equation of a plane and give you an example. Find the equation of a plane which passes through the intersection of these two planes and also passes through or passing through 1, 1, 1. Now the moment you read this phrase that there is a plane which is passing through intersection of two planes, that itself should remind you of the family of planes concept. So now let's see how do I solve this. So first I will say, let the equation of the plane that I am looking for or let's say the family member which I am looking for has this equation. Getting my point. Now to find lambda, I would use the fact that this plane must pass through. This plane must be satisfied by the point 1, 1, 1. So put x, y, z as 1, 1, 1. So you'll get 1 plus 1 plus 1 minus 6 plus lambda. 2 plus 3 plus 4 plus 5 equal to 0. So the lambda value that will come out from here will be 3 over 14. Okay. Put this value of lambda back here, okay? And when you do that, you end up getting x plus y plus z minus 6 plus 3 by 14 times 2x plus 3y plus 4z plus 5 equal to 0. Now in order to simplify this, I have a suggestion. Multiply this entire stuff with 14, okay? Just because there's a 14 in the denominator. Just to simplify whatever you have got. You may have your own ways of simplifying it. So you can follow that. So when you multiply it throughout with 14, it becomes this. And when you finally expand it, 14 plus 6, it becomes 20x. And 14 plus 9, 23y. And 14 plus 12, 26z. And minus 84 plus 15. Minus 84 plus 15 is minus 69, I guess. So this becomes your desired equation of the plane. Is that clear to all of you? Yes, sir. How it works? Now, how does the same thing look if I have been given the equation of the planes in vector form? Now in vector form, you don't have to convert it to Cartesian. You can deal directly in vector form. So let's say the same two equations I gave you in vector form. r dot n1 plus d1 equal to 0. And this is r dot n2 plus d2 equal to 0, okay? So when you're writing it in vector form, you can write it as r dot n1 plus d1 plus lambda times r dot n2 plus d2 equal to 0, same method. But you can further simplify this by taking r common from these two. So you can have r dot n1 plus lambda n2 plus d1 plus lambda d2, okay? And once they have been given, once they give you a condition to find lambda, you can use it over this equation. There's no need to convert to Cartesian. Please don't convert unnecessarily vector to Cartesian and show your weakness in dealing with vectors, okay? That gives a very wrong indication to the examiner that since the child is converting vector to Cartesian, he is not very confident about vectors. So whatever you have been given the condition to find lambda, you can use it over this itself. Is that fine? Any questions so far?