 Hi, I'm Zor. Welcome to Unizor Education. We continue this course of advanced mathematics for teenagers presented on Unizor.com. I do recommend you to watch this lecture from this website because it has comments, notes, whatever, which can be actually used as a textbook. Today we will talk about perpendicular planes. One plane is perpendicular to another plane. Well, first of all, we do have to define the perpendicularity between the planes and this definition is very simple. Whenever we are talking about planes intersecting each other, they go something like this. This is for instance one plane and something like this would be another plane and they intersect something like this, I guess. I hope it's the right representation, generally speaking. Okay, I think it would be better if I would do it this way, like this. Okay, these are two planes intersecting each other and they form one, two, three, four, four dihedral angles. The dihedral angle basically is the angle formed by two half planes, right? So, what is the perpendicularity of the planes? Well, for each dihedral angle we know what the corresponding linear angle is, right? That's if you have this intersection line, you draw perpendicular to one and perpendicular to another. Two perpendiculars are correspondingly in two half planes, perpendicular to line of intersection and they form an angle called linear angle. So, if this linear angle is the right angle then this dihedral angle is the right dihedral angle and the half planes are considered to be perpendicular and obviously if you have two intersecting planes and you have four different dihedral angles, if one of them is the right angle then all others will be right as well. So, that's the definition. It's very simple. It basically completely reduces the definition of the right dihedral angle to the corresponding right linear angle. All right, so that's all about the definition. What else? Okay, that's it and basically right now what probably is the most important part of this lecture I will present four different theorems about perpendicular planes and how it's related to perpendicularity between the line and the plane. Okay, so, theorem number one. You have a plane, you have a plane, let's call it sigma and point A on a plane and a perpendicular line, line P. Line P is perpendicular to plane sigma and the intersection between P and sigma is point A. All right, now, the theorem states that any plane which contains this perpendicular P to the plane sigma would in turn itself be perpendicular to sigma. So, any plane whenever we will put any plane which goes through this perpendicular to the plane, this plane would be perpendicular to the plane in this sense of the corresponding linear angles. So, that's the theorem. I apologize for the noise, it's really construction out there. All right, so let's draw a plane which is perpendicular, which is going through the plane P and let's see if we can prove that the plane is perpendicular. Let's see this is our plane which goes through this perpendicular P. Now, this plane has intersection with sigma, let's call this tau and what we will do right now is let's say intersection is line X and I will also build a perpendicular to X, call it N, in the plane sigma. So, X is supposed to be perpendicular to P and N is supposed to be perpendicular to P because P is perpendicular to an entire plane sigma and both X and N are inside the sigma, right? Now, what's happening is the line X from A to across the intersection between tau and sigma is basically the H between this plane or half plane if you wish and this half plane. I mean we can always consider only this half plane of sigma. So, X is the edge between these two half planes so the linear angle which I can build would define the dihedral angle between plane tau and sigma. But what is the linear angle? Well, basically it's the angle which is formed by P and N. Why? Because P is perpendicular to the line of intersection because P is perpendicular to the whole plane and N is perpendicular to the line of intersection because that's how we constructed it. Within the plane sigma we constructed N perpendicular to X. So, it looks like both P and N are perpendicular to the line of intersection to the edge of the dihedral angle. But then, but they in turn form the right angle because P is perpendicular to N since P is perpendicular to an entire sigma. So, our linear angle is the right angle that's why the dihedral angle formed by sigma and tau is the right angle. That's it. So, if you have a perpendicular plane then every plane which contains this perpendicular would also be perpendicular to the original plane. That's the first theorem. Theorem number two. Okay, so we have two planes perpendicular to each other. Okay, I will try to repeat the same. Okay, this is one plane and this is another. I will use half planes. It's easier. So, they are perpendicular to each other. Now, the theorem states that within plane sigma, so sigma is perpendicular to tau, that within plane sigma there is a line which is perpendicular to an entire plane tau. So, again, if you have two planes perpendicular to each other then within one of these planes exists the line which is perpendicular to an entire plane tau. Okay, how can we do it? Let's say this line of intersection. This is the edge. G is intersection between these two planes. Okay, now let's pick any point A on this intersection and let's build a plane, a new plane, which is going through this plane, this point A perpendicular to the line G. Now, we know that we can always build a plane perpendicular to the line at any point. So, this is the one, the point. And the plane would intersect sigma and tau along some lines. So, let's consider this is the line of intersection and this is the line of intersection. So, this is the plane which we call raw. So, the plane raw goes through A and it's perpendicular to line D. Now, since it's perpendicular to line D, line D would be perpendicular to any line within the plane raw which goes through this point. Again, if D is perpendicular to an entire plane raw, it would be perpendicular to any line within the plane raw which goes through point A. So, if this is one line and this is another line, lines of intersection of this plane raw with sigma and tau correspondingly, then this angle is the right angle and this angle is the right angle. Therefore, angle formed by these two lines is a linear angle of the dihedral angle formed by sigma and tau because this linear angle is formed by one line within sigma which is perpendicular to the edge and another line within plane tau which is perpendicular to the edge. So, we have a linear angle and since we know that the sigma is perpendicular to tau, this particular angle, let's call it BC or something, angle BAC is right angle. As follows from the perpendicularity between sigma and tau because it's a linear angle, right? So, what do we have now? We have this line AB which is perpendicular to AC and also we have AB perpendicular to D because D is perpendicular to an entire plane raw. So, it looks like this particular line AB is perpendicular to two lines within tau, the D line and AC line. Therefore, AB is perpendicular to tau because it's perpendicular to two lines at the point of intersection. That's it. Okay. Theorem number three. We have two planes again perpendicular to each other, sigma and tau. Again, D is their intersection line. So, it's at the edge of the dihedral angle and the dihedral angle is 90 degrees. It's perpendicular by definition. Now, let's take any point A on sigma which is not part of tau and drop a perpendicular to an entire plane tau. Now, the theorem states that this perpendicular is completely within plane sigma and the intersection between this perpendicular point B with the tau is actually on the edge. So, basically, if you imagine, for instance, the surface of the earth as tau and sigma as the building, side of the building, so whenever you drop a perpendicular from the top of the building, from the roof, down along the wall of the building, this perpendicular goes along the wall. It doesn't really deviate from the wall. It goes straight down. So, that's what the theorem states. Okay. Now, what we will do is the following. I will draw a perpendicular to the D within sigma. So, let's say A B prime is perpendicular to D and A D prime belongs to sigma. So, from A, not only I drop the perpendicular to the tau, as I was saying, but also within sigma I drop the perpendicular to D. So, it's points B and, okay, let's just assume that this is B and this is B prime. So, what I'm going to do right now is to prove that B and B prime are exactly the same points. Or, if you wish, the line A B prime is not only perpendicular to the D within plane sigma, but it's also perpendicular to an entire plane tau, which means that it corresponds, it coincides with A B. All right. Now, how can I do it? Well, let's wipe out for a second our perpendicular to the plane and just leave the perpendicular to the H within sigma. And I guess the best way right now from this point have a perpendicular to D within plane tau. So, A B prime with D is perpendicular within sigma and B prime C is perpendicular to D within plane tau. Now, what does it mean? These are two perpendicular from B prime, one along the sigma perpendicular to the H, another within tau perpendicular to the H, which means that A B prime C is linear angle of the dihedral angle between sigma and tau. And since sigma is perpendicular to tau, by condition, the angle A B prime C is also perpendicular 90 degrees. Now, what's interesting is that A B prime is perpendicular to B prime C and A B prime is perpendicular to the D, because that's how we basically drop this perpendicular. So A B prime is perpendicular to two lines within plane tau, which means it's perpendicular to the tau. So, which means that this perpendicular, which I drop from A to the plane tau, should coincide with B prime, because there is only one and only one perpendicular, which can be dropped from a point to a plane. So A B prime is the perpendicular to the entire plane tau, which we are looking for. So that's the end of the proof. And the last theorem is the following. So we have two perpendicular planes, again, sigma and tau. Now, their intersection is line P. So now, let's assume that we have a plane which is perpendicular to both sigma and tau. Then the theorem states that it should be perpendicular to the line of intersection between them. So if there is a plane, let's try to draw it. This is a plane, sigma, which is perpendicular to both, plane gamma, which is perpendicular to sigma and perpendicular to tau. Then it's supposed to be perpendicular to the line of intersection between sigma and tau, the edge between them, line P. So again, if sigma is perpendicular to tau, if gamma is perpendicular to sigma and gamma is perpendicular to tau, if P is intersection between sigma and tau, then gamma is perpendicular to P. Now, how can I prove that line is perpendicular to the plane? Well, obviously, I have to find two lines which are perpendicular to. So here is how we will do it. Let's pick any point A on the line P on the intersection outside of gamma. And let's drop a perpendicular from A to gamma, AB is perpendicular to gamma. Now, the previous theorem says that if there are two planes perpendicular to each other, in this case sigma and gamma, and from one point which belongs to one of these two planes, A belongs to gamma, right? Since it belongs to line P. I drop a perpendicular to gamma, then it should completely be inside the plane sigma. So, what I know is that AB must be completely inside sigma. So, this perpendicular to the plane gamma from A to B, well, basically I'm trying to prove that this is point B. So, what I do know is that AB, which is perpendicular to the gamma, A belongs to sigma, sigma and gamma are perpendicular, so AB should belong to sigma. The same thing as with ground and the wall, right? The wall is my sigma and the ground is my gamma in this particular case, right? This is the building, sigma is the building, gamma is the ground. I drop a perpendicular to do within building. Now, absolutely similarly, A belongs, since it belongs to the HP, it belongs to tau as well, right? And since it belongs to the tau and tau is also perpendicular to the gamma. Now, tau is playing the role of a building and gamma is playing the ground. So, this AB is supposed to belong to tau. So, this perpendicular from A to the plane gamma is supposed to belong to both sigma and tau, which means it belongs to the intersection between sigma and gamma, which means that AB is basically along the p-line. So, that's my fourth and final theory, which I wanted to present today. These are properties of perpendicular planes. Now, as usually, let me just mention very, I would say, common thought, which I'm trying to express throughout the course. Does it have any practical purpose? Well, I mean, yes, obviously you can say that our architecture is kind of using some tools which help it to build vertical lines, etc. But it's really very, very rarely when we are using it right now and only for those people who are involved in these professions. Otherwise, these are pure abstract constructions, planes, perpendicularity, lines, etc. And the only purpose which it serves in my personal view is it just develops your imagination. You have to really imagine this picture in three-dimensional space and it really helps to develop your mind. That's why all these theorems basically are presented. You need logic and you need this space vision. So, I hope these theorems would kind of exemplify this particular approach. So, thanks very much. That's it for today. Good luck.