IIT JEE perpendicular plane (part 2)

Where we had left off, we had used the information from these two lines to figure out two vectors, which are really just almost, you could just say, two position vectors lying in this white plane. And we took their cross product to find a vector that's perpendicular, that's normal to this entire white plane. And that was vector d there. And vector d will have to sit, especially to visualize it, if you start the vector-- you can always position a vector anywhere you want-- but if you start it at 0, it will clearly sit in that plane. But since you could always position it anywhere, you could say, OK, let's start it at 0. So it will sit in this yellow plane that we need to find the equation for. And we know vector d sits in it. We already know that vector a sits in it. We got that early on in the first video on this problem. And so we could take the cross product of d and a to find a normal vector to it. And then using that normal vector, and then thinking about what an arbitrary point, or what an arbitrary vector on this plane might look like, we can then figure out the equation for this yellow plane. And we already touched on it a little bit in the linear algebra playlist, if any of this stuff makes you confused. And just a reminder, we take the cross product of two vectors in three dimensional space, it will give you a third vector that is perpendicular to both. That's the whole tool that we're using over here. So let's just take the cross product of d and a. So this is going to be equal to-- so for the i, we have 1 times 4, which is 4, minus 3 times 10. So 4 minus 30 is negative 26. So we have negative 26i. And then we have negative or minus for j. We have to swap signs for the j. And so we have negative 8 times 4 is negative 32. Negative 32 minus 2 times 10, so minus 20, is negative 52. So we put a negative 52 here. Or we could just say positive 52. Because it's minus negative 52, which is the same thing as plus 52j. And finally, for the k-- cross out that row, that column-- negative 8 times 3 is negative 24, minus 2 times 1. So minus 2, it's negative 26. So minus 26k. So this normal vector right here or an a vector that is normal to the yellow plane, the plane that we have to find the equation of, is this over here. And just because it kind of pops out at me that all of these are divisible by 26, I'm going to define another normal vector. I don't know, let's call it-- we could call it anything. Let me call it-- I don't know, let's call it p. Or let's call it-- I don't know, let's call it normal 1. I mean, just to show it's a different vector. I'm just going to divide-- although essentially it's pointing in the same direction. It has the same direction, just different magnitude-- I'm just going to divide all of these by negative 26. So if I divide all of these by negative 26, I'm just scaling the vector. But it's still going to go in the same direction. Because that's what I care about. I care about just finding any vector that is normal to this yellow plane right here. I want to find any vector. I already found one. And now I'm just going to scale it down so that it's a simpler vector. So if I divide everything by negative 26, I get i plus-- or I should say, minus 2j plus k. So that might look something like this right over there. So this is our n1 vector that is normal to this. Now, just using this, we can now figure out the equation of this plane right here. So let's just think about it a little bit. Let's just say I have a point, or I could even say a position vector, x, y, z. And I know x, y, z sits on this plane. So x, y, z. It sits on this plane. And we know that the point 0, 0, 0 is on the plane. So we also know that the position vector xi plus yj plus zk have to sit on this yellow plane in question. I'm assuming that x, y, z is a point on this plane. I could have put little 0's here just to make sure that-- you know, this could be a particular point on this plane. I'm saying this is just any point that I'm defining to be on this plane. Now, this position vector will then be on this plane. And since this vector is normal to this position vector right here, or another way to say it, it's perpendicular, or it's orthogonal. If I were to take the dot product, if I were to take that and then dot it with this vector right over here, and dot it with that vector right over there-- so I would take the dot product of that and this-- yj plus zk. If I take the dot product of these two things, it has to be equal to 0. If I take the dot product of two vectors that are orthogonal to each other-- and by definition, these two vectors are. I'm saying that this guy is in the plane. This vector, we already figured out, is perpendicular to the plane. Which means it's perpendicular, it's orthogonal to everything in the plane. So if we take their dot product, it will be equal to 0. So what does that give us? So if we take their dot product, we get x times 1. So if we take-- let me just go term by term. So if we take x times-- the coefficient there is 1. So we have x plus y times negative 2. So I should really call that-- well, I could just call that negative 2y. And then finally, I have z times 1. There's a 1 in front of this k. It's implicitly there. So z times 1. So plus z. That's the dot product. And that has to be equal to 0. Or another way to write it, just to simplify it a little bit, is that x minus 2y plus z is equal to 0. And that is the equation of our plane in question. Or I should actually write it in yellow, because it's the yellow plane. x minus 2y plus z is equal to 0. And I know this might have been a little daunting or a little confusing, and if you actually had to do this in exam, you wouldn't have to do all of the explanation. But just to visualize what we did, we used these two lines-- you can kind of view the vectors we used as kind of the slopes of these lines in three dimensions. Don't worry about it if that confuses you. But we used the vectors in-- both of these specify a different vector. You could keep scaling them into whatever. But we just took their cross product to find a vector that is normal to this entire white plane. And it would have to sit in the yellow plane. Now, if it sits on the yellow plane, we could take the cross product of that with the first vector that we found that was in the yellow plane. A vector that sat on that line, or you could almost view it as the slope of that line in three dimensions. And if you take the cross product of that-- it's not exactly the slope. I don't want you all to nit-pick me. So no, it's not exactly the slope. But a vector in three dimensions is kind of specifying a direction, which is analogous to a slope in two dimensions. But if we take the cross product of these two vectors now, we'll get a normal vector. That's what we started this video doing. I scaled back this normal vector just to simplify the math. And I said, OK, if I have this normal vector, and any arbitrary vector on this plane, when I take their dot product it'll be 0. And so what that allowed us to do is figure out what are the constraints on that x, y and z. Any x, y and z that's on the plane, when I take the dot product of that position vector with the normal vector, it has to be equal to 0. So that gave us the actual constraints for the plane. Anyway, hopefully you found that fun.