IIT JEE perpendicular planes (part 1)

We define the equation of the plane containing the straight line-- They give us the equation right over here-- And this plane has to be perpendicular to the plane containing the straight lines. This thing, right over here, x/3 is equal to y/4, is equal to z/2. And also it has the line x/4 is equal to y/2, is equal to z/3. Now just to visualize this properly, let me draw the planes. The first plane in question that we need to find the equation of-- Let me draw it. So we need to find the equation of this plane. And I'm going to try to do this whole problem using vector notation, because it's easier in my mind to manipulate vectors in a way that you can find perpendicular vectors and equations for planes. There's other ways that you could do this problem. So we've defined the equation of this plane right here. We need to find the equation. Now we know this plane contains this line, and this line, if we look at right over here, it contains the point 0, 0, 0. If x and y and z are equal to 0, then all sides of these equations are 0. So it contains the point 0, 0, 0. Or you could even view it as it contains the position vector 0, 0, 0. And it also contains the point 2, 3, 4. Because if x is 2, y is 3, and z is 4, then all of these are going to be equal to 1. So let me draw that right over here. So it also contains the point, or the position vector, we could say 2, 3, 4. So if we wanted to just draw this line, to point specify the line, this line would look like that. But if we wanted to think about a vector that this plane contains, it would be the vector that goes from 0, 0, 0 to 2, 3, 4. So it would be this vector right here, which really is the position vector 2, 3, 4. And I'm going to write an ijk notation just to make it clear that this is a vector. So this vector that lies in the plane in question is the vector 2i-- I'll write it over here-- 2i plus 3j plus 4k. I just wrote in unit vector notation. I'll say a, I'll call this vector a. Vector a is equal to that. And I'll write a over here. Now we need to figure out the equation of this line. And the best way to do it, or at least the way my brain is telling me to do it, if I can find another vector in this plane, then I can find a normal vector to this plane. And then using a normal vector and some arbitrary other vector on this plane, I can then figure out the equation of the line. Because the normal vector dotted that arbitrary other vector has to be equal to 0. Now if none of that makes any sense, maybe as we go through the problem it will make more sense. So what I'm going to do is use these other lines, or really the other vectors, that are implicitly defined by these lines to figure out another vector in this plane over here. So let me draw what I'm talking about, that might be easier. So let me draw this other plane. And I'll draw it in white. And this other plane also contains 0, 0, 0. Or actually, both of these lines over here contain 0, 0, 0. So let me draw it over here. So this other plane, I think what I just said, if I do it step by step, it will make a lot more sense. So the other plane, that's when it comes vertically above this plane, and it goes vertically below the plane as well. And we see the plane that contains these two lines is perpendicular to this. They're perpendicular to each other, and we need to find the equation of the yellow plane. Now this blue line over here contains the point 0, 0, 0 and the point 3, 4, 2. Same logic that we've got to get to this point. If this is 3, this is 4, this is 2, they all equal 1. The equation holds. So let me just draw some arbitrary place. So 3, 4, 2. And I want to be clear, I haven't even defined the coordinate axis. We don't know what angle we're looking at this from. This is just for visualization purposes. So this line right over here, I could draw it, would look something like this. It would look something like that. It sits on that white plane. We can do the same thing for this magenta line. The magenta line also contains 0, 0, 0, but it contains the point 4, 2, and 3. So let me just draw it over here. So that's the magenta line. And you wouldn't even see it if this was transparent. You would see the line, it would just keep going down like that. Now, I'm going to attempt to explain how I'm going to tackle this problem again. I'm going to use vectors defined, that are implicitly defined, by this blue line and this magenta line. I'm going to take the cross product of those two vectors to find a vector that is normal to this entire white plane. And that vector has to sit in this perpendicular plane. And then I can use that vector and vector a to figure out a normal vector for this plane. And then I'll be able to figure out the equation of the plane. Let me just do it, instead of keep trying to-- So what's a vector? What are two vectors that sit in this white plane? Well, the point 0,0, 0, or the position vector is 0, 0, 0. And the position vector 4,2,3 is in this plane. So the vector, I'll call that b, the vector b is equal to 4i plus 2j plus 3k is also in this white plane. And you really, I'm just subtracting this position vector from that position vector to get this position vector, which clearly sits in the plane. This vector starts at the origin goes to 4, 2, 3. This plane contains the origin is clearly sitting in that plane. And by that same logic I could figure out a vector. It's actually the position vector itself sits on this plane. So this is the vector I'll call that c is equal to 3i plus 4j plus 2k. The fact that everything is at the origin kind of simplifies it. The position vectors themselves sit on the plane. If this wasn't centered at the origin we wouldn't be able to do that. We would have to subtract between these two vectors, or between two other points to actually get vectors on the plane. But with that out of the way, I could take the cross product of b and c, and that will give us some vector that is perpendicular to both b and c. Which means that it's perpendicular, or it's normal to this entire white plane, which means it must sit in this yellow plane. So I need to find, I want to find, a normal vector that's in the yellow plane, just like that. Let's call that vector d. So the vector d is going to be equal to this. And so the cross product of b and c, this is how I remember, is the determinant of you take ijk. I'm going to explain this in a lot more depth in the linear algebra playlist. But it's determinate of that, The next row is vector b. So it's 4, 2 and 3. And then the next row is the vector c. So 3, 4, and 2. And so this is going to be equal to-- let's think about it. So for i you cross out that column, that row, it's 2 times 2 minus 3 times 4. So that's 4 minus 12. So it's negative 8. Negative 8 times i minus-- remember you have to swap signs. You go plus, minus, plus when you're taking the determinant-- so you ignore the j, so it's 4 times 2. 4 times 2 is 8, minus 3 times 3, which is negative 1. So this would be a negative 1 here, but I already have a negative out there. So it's going to be plus 1j and then finally, the k. You cross out that row, that column, 4 times 4 is 16. 16 minus 6 is 10. So plus 10. Let me make sure I got that right. So the first one, I always have to make sure. So i got 4 minus 12 is negative 8. Then j, cross that out, 4 times 2 is 8, minus 9 is negative 1. But the j has a negative, so that's right. And then k, 4 times 4 is 16. 16 minus 6 is 10. So that looks right. So plus 10k. So this vector right here, this is the vector d. This is the vector d. I just figured it out. Now, if we take the cross product of d and vector a, the vector that was-- you can kind of view it as lying along this line over here-- then we will get a vector that is normal to this plane. We could call that vector n for normal. That is normal to the yellow plane. And it would actually sit in the white plane. So let's figure that. We'll see in a few steps, probably in the next video, why that is actually useful. So let's find what d cross-- so if I take the cross product-- d cross a is going to be equal to our normal vector. Which is equal to the same thing, ijk. We're going to take d, which is negative 8, 1, 10. And take the cross of that and the vector a, which is 2, 3, 4, and we will get-- and actually I'll continue this in the next video. Just so that I don't make these videos too long.