 Hi, I'm Zor. Welcome to a new Zor education. Today's lecture will be about one particular kind of symmetry, symmetry in a three-dimensional world relative to an axis. So, symmetry about the axis. Now, we do know the symmetry about an axis in a two-dimensional world and, well, this lecture will be in a three-dimensional world. Well, this lecture is part of the Advanced Mathematics course for high school students presented on Unizor.com. So, the video is referenced from this website as well as a very detailed notes for this lecture, which I do recommend you to read before or after the lecture, just as a textbook, if you wish. And there are obviously other things on the web page, which allow you to basically control your educational process. Okay. What is three-dimensional symmetry around the axis? Well, actually it's exactly the same as far as the construction is concerned. As in two-dimensional world. Let me just remind you. Well, if if this board is two-dimensional space, we have a line and we have a point. We would like to construct a point which is symmetrical to this relatively to the line. How do we do it? Well, we drop the perpendicular to the line and continue on the same distance. So, A always equal to O A prime and the point A would be the image, the symmetrical image of point A relative to the symmetry line or axis, axis of symmetry S. Now, exactly the same thing in three-dimensional world. So, let's forget about this board. Let's just consider that we are talking about a line in a three-dimensional world. So, we have the board as, you know, one dimension, two dimensions, actually. And then we have whatever is on this side of the board and whatever is on that side of the board and the point A is not necessarily exactly on the board, but it can hang in the thin air, let's say. What can we do? Well, we still can drop the perpendicular to the plane, to the line. How do we do it? Well, very simply, if you have a point and you have a line, these two elements define a plane. So, there is a plane which goes through this line and through this point. And within this plane, we can drop a perpendicular using the plane geometry, right? So, that's just we can consider, for instance, this board to be exactly that plane which connects the line and the point. So, whatever that board position is, it contains both S, line S and point A. And within this plane, which connects the line and the point, we drop a perpendicular and continue it on the same distance. So, that's basically a construction, which is exactly the same as in two-dimensional world. Now, the first thing, which is very easy to prove, if A' is symmetrical to A relatively to the axis S, then A is symmetrical to A' relative to the same line S. Now, how can we prove it? Well, if we connect these points again with a straight line, it will intersect our axis. And obviously, since A' is a perpendicular to S, it means all A' which is a continuation is also perpendicular. So, the same line actually serves in both cases and the same distance actually, again, here and here. So, that's why it's symmetrical. Symmetry is symmetrical in some way, because symmetrical means as A relates to B, B relates to A. So, that's basically what it is. So, symmetry is symmetrical in this particular sense. Okay. Now, we can view symmetry around the axis in a more, I would say, dynamic view. And here it is. I would like to propose the view onto the symmetry as a rotation of the space by 180 degree around this axis. Now, let's just think about what is rotation of the space around the axis. Well, if this is my axis, how can the space be rotated around this axis? For instance, I would like to have a point somewhere in space, and I would like to rotate it by an angle phi, for instance. Okay? How can I do it? Well, the way to rotate the space and a point in this space in particular by a specific angle is the following. First, we draw a plane which is perpendicular to this axis and which goes through this point. So, in this particular case, let me do it vertically. It would be easier. So, this is my axis and this is my point A. So, what I do is the following. First, I have a plane which contains point A and it's perpendicular to axis S. This is an invisible part. So, it goes from here through the plane. Let's call it beta. And A is on that plane. Now, within this plane, I would like to, let's say this is point O. I would like to rotate this segment OA by angle phi within this plane beta. So, let's say I'm going to this way. And this will be my A prime. And this will be my angle phi. So, rotating is within the plane which is perpendicular to the axis and goes through my point. Now, within this plane, I can rotate. So, another way, well, let's consider this is my axis. This is my point. All right. This is my axis. This is my point. So, the plane beta would be actually the plane of this board, right? So, I'm turning the whole board and that would be around this axis. So, every point will retain its plane within which it is rotated. That's what the meaning of this. Now, obviously, if phi is equal to 180 degrees, then this point would be on this side with the same length, right? Since we are rotating this segment, it goes this way. So, basically, it goes into the position which is exactly prescribed by the relationship of symmetry relative to this axis. So, this rotation is just another view onto symmetry around the axis. So, first you just don't use any rotation. You just straight drop the perpendicular to the plane and, sorry, to the axis and continue it by the same distance. Another way of looking at this is having this plane which is perpendicular to S and rotating within this plane by 180 degrees. And obviously, every point during this rotation would be converted into symmetrical relative to the S. Wherever this point is, if point is here hanging in the air, then, again, my perpendicular plane would be this and I will rotate it to 180 behind the board on the same distance and that would be symmetrical point. All right. Now, just looking at this rotation, we can always say that we can view symmetry as operator. If you remember, operator is some kind of a transformation of one object into another within the same set of objects. Now, in this case, our set of objects is a space, three-dimensional space. Our elements are points. Now, if I have a certain axis, a straight line, I can always say that transformation of any point of a space into its symmetrical relative to this axis is a transformation of the entire space, which means this symmetry can be considered an operator, an operator of symmetry. So, this operator transforms every point into symmetrical relative to this axis. I mean, obviously, operator depends on the axis, but that's the only dependency because operator means I can use it to any point in space to get symmetrical. And this operator obviously is symmetrical because if A is symmetrical to A prime, A prime is symmetrical to A. Okay. Next. Next is a very important theorem. And it's not obvious. Like, it's not just one statement proof. The statement is the following. This transformation of symmetry preserves the distance between points. Now, how can I prove it? How can I prove that I will have, if I will have two points? So, let me make a slightly better picture. So, if I have two points, A and B, but what I will do first, first I will do one plane, which is perpendicular to my axis. And then another plane, which is perpendicular to my axis. Then I'll withdraw an axis. All right. Let's say, gamma and delta. One point would be, let's say, here. And another would be here. Now, they are in two different planes. They are in space. Don't forget that. So, obviously, the plane which goes through B perpendicular to my axis of symmetry is one plane. And if I have another point, the plane perpendicular to axis would be a completely different plane. So, I'm still interested in, okay. So, let's say, let me put it a little closer. Let's say this is my A. And on the same distance I will have A prime. Now, this is B. And on the same distance I will have B prime. Well, actually it's a solid line. And this is a solid line. So, what I would like to prove is that A B is equal to A prime B prime. Not very obvious statement, right? I mean, we feel that it should be this way, because we're turning the whole space, right? If we just think about, we're just turning the whole space. So, the relative position of A and B between themselves actually is not supposed to change, because since we are changing the whole space, then it's moving together, basically. So, whatever the distance between A and B was, we are rotating the whole space by 180 degree. Now, B goes into B prime, A goes to A prime, but the distance should be the same. I would like to prove it a little bit more rigorously. Okay, here's how we can do it. Now, these two planes are perpendicular to XSS. One goes through A, another goes through B. So, what I will do, I will project point B to, well, let's call it somehow differently. Let's call it C. And that will be projection from B prime. Now, you remember what projection actually is, right? It's just a perpendicular to plane delta going from point B and going from point B prime. So, these are two perpendicular to the same plane, which means they must be parallel. These issues were addressed before when I was talking about planes and lines, parallelism, perpendicularity, etc. All right, that's number one. Number two, these are two perpendiculars between two parallel lines. They are distances, actually. Their length is the distance between these two parallel lines. So, all perpendiculars between two parallel lines are equal to each other in size, which means that B, C, and B prime, D prime are lying within the same plane since they are parallel, being perpendicular to the same delta. And these segments are equal. So, what is this particular figure which is in the plane? It has two parallel sides. It's quadrilateral with two parallel and equal in length size. Well, parallelogram. We should remember that. Not only parallelogram, it's also a rectangle because these angles are 90 degrees. It's perpendicular, right? Perpendicular to the plane, which means perpendicular to the line on the plane. So, this is rectangle. B, C, D, B prime is a rectangle. And B, B prime is equal to C, D, obviously. Now, what I will do next is I will prove within the plane delta the equality between A, C, and A prime, D. This is I will going to prove. A, C is equal to A prime, D. Now, how can I prove that? Well, again, this is the plane geometry because obviously triangles, let's call it O, triangles A, O, C, and G, O, A prime are congruent. This is equal to this. This is equal to this because these are exactly equal parts. These are all. This is also equal to this. This is equal to this. Now, these are equal because they are symmetrical. So, these are symmetrical between themselves, obviously, as well. C and G, the angles are vertical. So, we have congruence of these triangles and that's why this is equal to that. So, A, C is equal to A prime, D. So, within the plane delta I have the equality of the distance between the point, between two points before and after the symmetrical transformation. Now, but I would like to prove that A, B is equal to A prime, B prime, which is a little bit more complex, but not much actually. Let's just think about it. A, B is this and if you consider A, C, B, it's a right triangle with A, C as one catheters and B, C another catheters. Now, what is A prime, B prime? It's this. It's part of the triangle A prime, D, B prime, which has also one catheters being B prime, D prime and another being A prime, D. But now, consider these two triangles. Again, triangle A, B, C and triangle A prime, B prime, D. Obviously, they are congruent because this catheters B prime, D prime, B prime, D is equal to this catheters B, C and this one A, C equal to A, G and A, G is equal to A, C. So, two catheters of one triangle are equal to two catheters of another right triangle. So, these two triangles are congruent, which means they are hypotenuses, which is A, B and A prime, B prime are equal to each other. So, I'm not saying it's a trivial theorem, but it's relatively easy. All I was doing, I was reducing my a little bit more complicated problem, the distance between A and B, to a little bit easier one, when this B is actually replaced with its projection onto the same plane where A is. Now, within the plane it's easy, and then going back to the original, I just, you know, built a couple of congruent right triangles. So, the transformation of symmetry around the axis preserves the distance between points. As a consequence, it obviously preserves the shape of any triangle, which means any angle also preserved. If you have an angle, you just convert it into a triangle, then you rotate it by 180 degree, or symmetrically transform into corresponding another triangle. And, well, that's it. I mean, triangles will be congruent, because every side of one would be preserved, as far as its distance is concerned, to side of another. So, triangles would be congruent by three sides. And therefore, the angles are the same. So, angles are preserved, distances are preserved, so it's invariant transformation. Symmetry around the axis is an invariant transformation. Well, actually, if you just think about it, if you consider symmetry around the axis as a rotation by 180 degree, which means everything in the space retains their relationship, the relative position is exactly the same before and after the rotation. And, obviously, you can consider that rotation by any angle, not only 180 degrees, which I was explicitly using here, actually, but any rotation by any degree around this axis would preserve the lengths of segments or distance between points and angles. So, I'm not going to prove it, but you just have to feel that this is just a rotation, and rotation is very kind of a static. I don't deform the space, I don't scale it, I don't reduce it in size, I'm just rotating without any change in dimensions. Alright, next theorem. Next theorem is that my straight line during this transformation of symmetry is transformed into a straight line. So, again, this sounds like a simple kind of a statement. Well, obviously, the symmetrical to line is supposed to be a line, right? Well, it's not that obvious. I mean, maybe the line will deform. Instead of straight line, it will be some kind of a curve after you transform every point of this line into its symmetrical counterpart. But let's prove that this is not true. So, this is my axis S, this is my line D, and somehow we are transforming into line D prime. Now, how can I prove that this is, if this is straight line, how can I prove that this is straight line? Well, that actually is easy. Let's just choose three points A, B, and C. So, that will be A prime, B prime, and C prime. So, every point on this line belongs to the line. I mean, point B belongs to the line which goes through A and C, right? So, what I would like to do is, I don't want to assume that this is a straight line. I just assume it may be a curve or whatever. So, I transform A to A prime, C to C prime, but I will prove right now that any point B taken on this line which goes from A to C would actually have an image lying on the line from A prime to C prime. It cannot be, this point, B prime cannot deviate from the line which goes through A prime, C prime. So, A prime, A and C, and correspondingly A prime and C prime, I choose any way I want to, right? That determines my line here and determines my line here. I don't know if this line is an image. And to prove it, I will take any point B on this line. Well, right now I'm just choosing in between, but that's the same thing outside. So, any point B would be converted into a point which is lying exactly between A and C prime on a straight line. How can I prove it? Well, very easily. In this case, what I can say is A B plus A plus B C equals A C, right? Because this is a straight line. That's how I choose point B in between A and C. Now, if B prime is not on a straight line from A prime to C prime, then A prime B prime plus B prime C prime would be greater than A prime C prime, right? If B is not on a straight line between A prime, if B prime is not on a straight line between A prime and C prime, then this is supposed to be a true statement. However, we all know that the symmetry preserves the distance. So, this is equal to this, this is equal to this, and this is equal to this, which means I must have equality here, not inequality. And to have an equality, I have to have B prime in between A prime and C prime. And that's basically the almost end of the proof. I mean, if I choose B outside, then I will just change the order of my transformation. I will choose A and B and then C in between, it will be exactly the same thing. All right? So, we'll have different like A C plus C B should be equal to A prime C prime plus B prime C prime. All right. Anyway, that's the end of this particular proof. So, the straight line is transformed into straight line. Okay. What else is important? Well, important is a plane, how the plane is transformed by this transformation of symmetry. So, let's say we have, instead of a line, we have a plane, which we would like to symmetrically transform into another plane. I mean, would it be a plane? Let's say this is my plane. If I will symmetrically transform every point on this plane, will it be a plane as a result? Well, the answer is yes. And here's how we do it. I will just choose three points on this plane, A, B and C, which define the plane completely, right? Three points have only one and only one plane, which goes through them. And I will reflect them here. So, it would be A prime, C prime and B prime. So, just take three points out of this. They are reflected into three points here. And now I will draw a plane through these three points. Now, how can I prove that this plane is the image of this? Well, I have to prove that any point which I will get from here, let's say point P, would be reflected to a point which is lying on the same plane. Now, how can I do it? Well, let's do it this way. Let's draw a line which crosses at least two sides of this triangle. Let's say M and M. Now, the image of M should be somewhere here on B prime, C prime, right? Because the line is transformed into line. So, M prime must be here somewhere. Now, M prime must be somewhere here. Because, again, every point on the line AC should be transformed into a corresponding point on A prime, C prime. It's all based on the previous theorem. When I was saying that the straight line is transformed into a straight line. And now, the whole line M and P must be here as a line, again, because line is transformed into line. But two points of these lines, this line, M prime and M prime, lie on this plane, right? Because they are lying on this segment and on this segment and segments are within the plane. Which means that any other point on the same line must lie within this plane. And that's the end of the proof. So, P prime is supposed to belong to this plane. So, the plane which I am just constructed based on three points actually contains all other points transformed using the transformation of symmetry. Okay, that's it. That's my three little theorems, which I wanted to prove. I do suggest you to read the notes for this lecture on unizor.com. It will probably give you just another nice picture of how the explanation actually can be provided. And, well, that's it for today. Thanks very much and good luck.