 Hi, I'm Zor. Welcome to Unizor education. Let's talk about lines which are perpendicular to planes. This is just part of the solid geometry course presented on Unizor.com. I suggest you to watch this lecture from this website because the site contains notes for the lecture, which basically are like a textbook for you. Also, registered students have the ability to take the course in like a regular educational environment style, with test exams, enrollment, etc. The course is free by the way, so don't hesitate to register and take exams wherever it's available. It's not for every lecture I have an exam, not for every topic actually, but many of them do have it, I do suggest you. So, let's go to lines which are perpendicular to planes. Well, first of all, we will talk about certain lines which are perpendicular to each other. And whenever I'm talking about perpendicular to each other lines in three-dimensional space, I would like to point that at this particular moment we're talking about intersecting lines. So, these lines, although they are in three-dimensional space, they do intersect. And since they intersect, there is one and only one plane which contains both of them. And within that plane, we can measure the angles between the lines in the regular two-dimensional sense. So, when I'm saying that two lines are perpendicular to each other, it means that the lines do have the common point, intersect, and within the plane which contains both of them, they are perpendicular to each other in a two-dimensional sense. All right, now. Now we are ready to define basically what a line perpendicular to a plane is. Let's consider you have a plane. And at certain point you have a line which intersects at this point of the plane. Now, here's the definition. If for any line within the plane which also passes through the same point, this line is perpendicular to our line we're talking about, line A on this drawing. Then the line A is perpendicular to entire plane, gamma. That's the definition. So, again, if A is perpendicular to any line which lies in the plane, gamma, and passes through the intersection between A and gamma, then A is called to be perpendicular to the entire plane. So, this line, this line, this line, no matter which line we draw in the plane, gamma, which is passing through this point, this line would be perpendicular. So, this is a perpendicular, and this is a perpendicular, and this is a perpendicular. All these angles are 90 degrees. They're just moving. Now, again, as I told from the very beginning, the line is perpendicular to line in a two-dimensional sense. And we can actually apply it because, as I was saying, all these lines are passing through point A. And this point A belongs to the line A, right? So, which means they have the common point, which means there is always a plane, this plane, or this plane, or this plane, which contains both, our main line A and one of these lines within the plane gamma. And within that plane, line A and whatever other line is within the plane gamma, these lines are perpendicular. So, let's draw this plane like this. That would be for this particular line, or we can draw a plane like this, for this particular line, or plane like this. So, any of these planes contain the main line and one of these lines in the gamma. And in each of these planes, this line is perpendicular to the A. So, that's the definition. Now, just to expand a little bit this definition, if line is not perpendicular to a plane, which means it's crossing the plane, it's not, it does not belong to a plane, but it's crossing at, you know, a certain non perpendicular way. Let's put it this way. Now, from the logical standpoint, how can I formulate it? Well, the way to formulate it is following. There is one particular, at least one particular line inside plane gamma, also passing through this point of intersection, which is not perpendicular to our main line, okay? If there is at least one line which is not perpendicular, it means that this line is not perpendicular to the plane. Now, is it possible that some of the lines can be perpendicular, but not all of them? Yes, the answer is yes. If you imagine, for instance, let's say this board is my plane gamma. Now, if I have a line which is positioned like this, for instance, I mean, obviously it's not perpendicular to a plane. However, there is a line here, and if you will consider the plane, which includes this line and this line in that plane, these lines are perpendicular. Now, these lines, for instance, would not be perpendicular. This one and my pen. But this one and this will. So, not every line is perpendicular to this one. And in this case, this line, which is not perpendicular, has its own name. Usually they're called slant or oblique lines. All right. So, that's as far as the definition are concerned. And by the way, this point of intersection of the perpendicular line to a plane, this point A, is called the base of the perpendicular. Well, that's kind of just terminology. Now, what's interesting about any kind of a definition which you can present in mathematics? Well, you can define anything. Question is whether this definition makes sense or not. Now, how can I make sure that this definition of the perpendicular line makes sense? Well, I have to basically show that these lines perpendicular to planes do exist. And it would be even better if there is some kind of a construction which leads to a uniqueness of the perpendicular line if, let's say, provided some kind of a point. So, I will spend some time right now to explain two concepts about the perpendicular line in the plane. The concept of existence and the concept of uniqueness. Both are very important because if something which we have defined does not really exist, then, well, what's the sense of the definition, right? So, I would like to show that this definition does make sense, that the perpendicular line and plane do exist and in some way, they are unique, provided the point of intersection. Okay, here is one. Actually, there are two different aspects. If I have a line and a point on this line, I would like to be able to construct a plane which is perpendicular to this line and this point. If I will be able to do it, if I will be able to construct that plane, then existence is proven. And then if I will also add to this the uniqueness of this plane, that would be even better, right? Okay, so let's just do this. How can I construct a plane which is perpendicular to this line at this point? Well, let's just recall that the plane can easily be defined by two lines, intersecting lines, which it contains, right? So, if I, instead of the plane, if I will be able to build, let's say, two different lines on this plane, that would be enough, basically, right? Okay, so how can I build, for instance, this line? Well, very simply, just draw any plane through this A. How can I draw any plane? Well, pick a point outside of the line A and draw a plane based on the point and a line, which we know is always possible. So, we have a plane. Now, inside of this plane, at this point A, I can put a perpendicular to line A. That's it. So, we have one line which is perpendicular. Now, similarly, I can just draw a completely different plane and in that different plane, now, how can it be different? Well, I pick a point which is not part of this plane, let's say. So, that's it. And draw the plane through this point and the line. Now, in that next plane, I will draw another perpendicular to line A. So, now I have two perpendiculars. And in theory, I have two lines already which are defining the plane. Well, okay. So, now I have a plane with two lines in it, perpendicular to line A. Question is, is it a perpendicular plane and line? Remember, we need any line which is passing through this point A to be perpendicular, right? And for this reason, I'm going to prove a theory. If there is a plane and a line, and this line is perpendicular to two different lines on this plane which are passing through this point of intersection, then any other line also will be perpendicular. And that's an interesting theory which I'm going to prove right now. So, let's draw a nicer picture through this theorem. And that would help us to discuss this existence and uniqueness of the perpendicular. All right. So, so far, I assume that there is a plane, there is a line, and this line is perpendicular to two different lines on this plane. Let's say line B and line C. Now, this is point B. This is point C. All right. Now, let's choose point A. Now, this point will be, let's say, P and point A prime. Now, I'm choosing A and A prime in such a way that A P is equal to P A prime on the same distance from the intersection point with the plane. So, I know that A and B lines are perpendicular, A and C lines perpendicular. And now I'm going to prove that any line D would also be perpendicular to A. Here is how I am going to do it. Let's put this point a little closer. Okay. This will be my point C. And what I'm going to do is I will connect B and C. And intersection with line D would be point D. Okay. By the way, lines are actually lines. In this case, I draw half lines. And this is the purpose, actually. If I have three different lines, B, C and D, I can always find two half lines B and C. So, one particular line, half line D would be in between. That's very important because I need this intersection point. D. All right. So, if D is somewhere here, for instance, I would choose this half line of C and this line of B. So, again, D would be in between B and C. All right. All right. So, now we will build triangles. One triangle, another triangle, and another triangle. And same thing here. Okay. Now. So, what do I have as given? I have that A, P is perpendicular. A, P is perpendicular to P, B, and P, C. I have it as a given, which means that the angle A, P, C is 90 degree and A prime P, C also equals 90 degree. So, let's consider this plane, which is defined by three points A, C, and A prime. It's a triangle. And within that plane, P, C is basically an altitude of this triangle. It's perpendicular to A, A prime. Not only that, since A, P, and P, A prime are equal, and these angles are straight, right angles, this one and this one, right? So, basically triangles A, C, P, and A prime, C, P are congruent, which means that A, C is equal to A prime, C. Now, similarly, let's consider triangle A, B, A prime. Now, B, P, also an altitude, and again, these are equal, which means A, B, and A prime, B, also are equal. Finally, let's consider triangles, which are flat from this side. A, B, C, and A prime, B, C. What do we know about these triangles? Well, all three sides are equal. A, B is equal to A prime, B, as you see. A, C is equal to A prime, C, and B, C is common. So, which means that triangles A, B, C, and triangle A prime, B, C are congruent, which means that angles A, B, C, this angle, angle A, B, C. And angle A prime, B, C, also are congruent. Now, the very last thing, consider triangle A, B, D, A, B, D. D is in the middle. And triangle A prime, B, D. A, B equals to A prime, B. B, D is common for these triangles, this one and this one. And the angle between them is equal. So, triangles are also congruent. And from the congruency, I have A, D should be congruent to A prime. Now, A, D and A prime, D are equal in lengths. P, D is common between these two triangles, A, D, P, and A prime, D, P. And A prime, P is equal to A prime, P. A, P is equal to A prime, P. So, these triangles A, P, D and A prime, P, D, they are congruent to each other. So, within the plane defined by A, D and A prime, this triangle A, D, P and A prime, D, P are congruent, which means this angle must be the same. And since these two are supplementary angles, which means every one of them is 90 degrees. So, there is a perpendicularity between line A and line D. So, if I have only two lines perpendicular to my line, which intersects the plane, then any other line in the plane also will be perpendicular. Now, let's go back to my existence and uniqueness theorems. So, this theorem is proven and we are going to use it to construct a plane parallel to line at a certain point. Okay, this is the point. So, according to the theorem which I have just proven, it's completely sufficient for me to build only two lines in that plane, which are perpendicular to my line A. Now, I don't have this plane gamma, but I can obviously construct two lines, which are perpendicular to one particular line at a given point, right? So, I construct these two lines first, choose any plane and construct one line, choose any other plane, construct another line. And now, using these two lines, I basically define the plane which it belongs to, which they both belong to, okay? And now, within that plane which I have just constructed based on these two lines, I can say what? Well, I have a plane and I have two lines which are perpendicular to A, which means every other line will be perpendicular to A, which means that the whole plane is perpendicular to two lines. Okay, now, how about uniqueness? So, if I have a line and I have a point and now I know how to build a particular plane which is perpendicular to this line at this point, why, how I did it? I just choose two lines which are perpendicular to A and draw the plane. Now, what if I choose different lines instead of these two, some other two lines? Will I have exactly the same plane or not? Okay, that's not such an easy thing. Let's just assume, okay, let's remove this. So far, we know about existence but we don't know about uniqueness, right? So, let's consider that we have two planes and both are, I don't know even how to construct, how to draw them, something like this, two planes. Both are perpendicular to my line A, let's say gamma and delta. And both are perpendicular at this particular point. Okay, so what should I do? These planes are intersecting somewhere, right? If the plane, if two planes have one point in common, they intersect along some line. All right, that's okay. Now, let's choose and all other lines and all points are completely separate. So, these planes coincide along this line and they don't coincide in any other point. There is no other point which belongs to both. So, all common points are on this line. So, let's choose some other point here, for instance, on one of the planes and draw a plane through A and this point. Now, it will intersect one plane along one line and another plane along another line, right? So, this plane which I have, it will intersect both gamma and delta, but intersections will be different, right? Okay, let's call this plane alpha, all right? So, alpha intersects with gamma along this line and with delta along that line and these lines are different. At the same time, both of these lines are supposed to be perpendicular to A. Why? Well, because one of them belongs to delta, for instance, and I know that all lines which pass through this point of intersection are perpendicular. Another line belongs to plane gamma, which is also perpendicular to this and which means it's also supposed to be perpendicular. So, what happens is the following. Within the plane alpha, where both these lines belong to, I have two different perpendicular to one line A, which is impossible. From the plane geometry, we know that through the point, we can draw only one perpendicular line, not two different lines. So, that's the contradiction, which means we cannot have two separate planes, which are perpendicular and intersect at the same point to the plane A. So, that's the uniqueness of this construction. So, for any line and point given on this line, there is and only one, there is one and only one plane, which is perpendicular to our line and intersects at this particular point. Now, a slightly different variation of the same thing. So, here I have a line and a point, and I have proven existence and uniqueness of the plane. Now, let's go from that point. Let's say we have a plane and a point on it, and I would like to draw a line, which is perpendicular at one particular point. Is it possible? Here is how we can do it. Actually, very simply. Let's draw any two lines here within the plane gamma B and C. Now, let's consider this point P and line B. I know how to build one and only plane, which is perpendicular to B. It's probably something like this. So, this plane, let's call it beta, is perpendicular to my plane B, my line B, and this is the line where it's crossing actually the gamma. Okay, what I do know is the following, that if there is a perpendicular to a plane gamma, it's supposed to be perpendicular to B, right? And I know that if I have this plane, all the perpendicular to B lines are within this plane, because the whole plane is perpendicular to B, right? Now, I'll do exactly the same with C. So, I have another line of intersection. So, there is another plane, something like this. Let's call it gamma, we have already gamma. So, let's call it delta for some sake. So, delta is perpendicular to C. Now, these two planes, beta and gamma, have some intersection. It's a line, two planes intersect along the line. So, this line belongs to this and belongs to this. Now, since the whole plane beta is perpendicular to B, then this line will be perpendicular to B. Now, since the whole plane gamma is, sorry, delta is perpendicular to C, then this line, since it belongs to both, beta and delta, will be perpendicular to C. So, it looks like this line is perpendicular to both B and C. And that's a sufficient condition for the line, this line, called A, to be perpendicular to entire plane. So, that's the existence of the line which is perpendicular to a plane at one particular point. Okay. So, what's the freedom of choice? Well, we just took different, two different lines. What if we choose different lines? We will have a different perpendicular. Well, the uniqueness is also exist here. Let's just talk about uniqueness of the perpendicular in this case. So, assume again that you have a plane and we have a point. And let's say we have two different perpendiculars to a plane. Well, if you have two different perpendiculars, we can draw a plane through them, right? Two intersecting lines. So, this line, this plane would intersect our plane gamma. So, A and B both are perpendicular to gamma. And the plane, which contains A and B, something like this, is intersecting gamma at some point. So, what happens in this particular plane? Well, it's alpha. In the plane alpha, we have a line and two perpendiculars to the same line in the plane. And according to the plane geometry, that is impossible. Well, so, basically, we have just proven that there is always a perpendicular to a plane at one particular point and it's unique. Now, the existence, we have proven using the construction, which is an active procedure. Sometimes there are some theorems where existence is proven without actual construction. That's more difficult. But if you have an actual construction, that's a very dangerous thing. That's a very direct proof of existence. If you can construct it, it means it exists. And then, uniqueness is also very important. Okay, that's basically it for today. I wanted to introduce the concept of a line which is perpendicular to a plane. And we have defined this particular construction, these elements. We have been able to construct a plane perpendicular to line and a line perpendicular to plane. At chosen point. And we have proven not only existence but also the uniqueness of this construction. Well, which means actually that the definition is pretty good. Definition of the line which is perpendicular to the plane is correct in some way. Because whatever we have defined exists and relatively unique as far as, you know, chosen points and lines. Well, that's it for today. Thank you very much. I do suggest you to go through notes for this lecture, read it again. And it probably would be very beneficial for you if you can prove all these existences and uniquenesses just by yourself as an exercise. That would be a very good exercise for you. Thanks very much and good luck.