 Hi, I'm Zor. Welcome to Unizor Education. I would like to present you one very, very simple construction problem. It's related to lines and planes perpendicular to each other. Now, the problem is very simple, and I will present it first in a very, very short version, and then I will try to put some details into this explanation. I would also like to say that the lecture is presented on unizor.com website as part of the advanced mathematics for teenagers. I do suggest you to watch this lecture from this side because it has side notes which basically are identical to whatever I'm going to talk about right now, and it will be very beneficial for you if you will review all these notes before or after the lecture. So, the problem is, let's assume you have a line in space and you have a point outside of this line. What you have to do is to construct a plane which is perpendicular to this line and contains this point M. Now, what I will do first, I'll just very, very quickly present you the solution. You consider the plane which is defined by the line and the point, and in that plane you draw perpendicular from the point to the line. That's a plane geometry thing. Then what you might do is you have some other plane which goes through the same line. Any other plane which is different from the one which we were just talking about. Now, let's assume this is point P. Now, in that plane you construct a perpendicular to line A from the point P. Now, we have two lines, P M and P, let's say N, and through this intersecting line you can draw a plane which will be perpendicular to this line and it will pass through the point M as required. That's it. Very simple. How much time I spent? Like minute, two minutes? No more than that. And now I would like actually to present you something which I believe has certain educational value. How to approach this problem in a rigorous advanced format? Well, first of all, how did I guess that the problem is supposed to be solved this time, this way? Well, because I already did it before, right, obviously. But if you don't know how to solve this problem, if you did not really experience to deal with this problem before, how you can guess the proper way? Well, there is a stage of solving mathematical problems which is usually associated with the name analysis. So, let me just bring this step which is called analysis into this. So, you don't know anything about the problem, you don't know any methodology how to solve it. So, you're trying to analyze the problem in some way because maybe during this analysis it will prompt you to something which is important. So, that's how you should think actually solving this problem. Now, the analysis usually starts from the very end. So, assume that this particular construction has already been successfully accomplished. The plane exists which is perpendicular to our line and passes through this point M. So, let's assume that there is a plane which contains this line at this point M and it's perpendicular to the line A. Let's call it sigma. So, sigma is the plane which we actually have to construct. So, analysis usually starts from assumption. Okay, let's consider this plane is constructed. What's the properties? What's the interesting elements I can actually see in this particular construction which will help me to really construct it from scratch? Well, obviously within that plane, sigma, I know that since line A is perpendicular to all the lines which are crossing the point P, obviously I can think about two lines and by the way I do remember there was a theorem. If the line is intersecting the plane and is perpendicular to at least two lines which are crossing this point of intersection, then the line is perpendicular to entire plane. Well, and obviously if it's perpendicular to entire plane then there are two lines which are perpendicular to at least two lines. Obviously it's perpendicular to any other line which crosses the P. Now, obviously one of these two lines must be the one which connects our point which is supposed to belong to this plane sigma with the base of the perpendicular point P. And probably there is some other line which is also perpendicular to. So, A is perpendicular to Pm and it's perpendicular to Pm because it's perpendicular to entire sigma plane. Now, how can I construct these two lines? Because I understand that if I construct these lines my plane would be completely defined. Well, now I'm actually thinking that if I have a point here and I have this line so I can choose the point here, then in the plane which is connecting point M and line A, let's call it tau, in the plane tau my line M, P and A are perpendicular. And if I will draw the plane through the same line A and point N, let's call it this way, let's call it rho. So, in the plane rho this would be perpendicular. Now, going further in our analysis I can see that M, P I can definitely construct in the plane tau because I know the line, I know the point and I can draw perpendicular. Now, in the plane rho this one I also can construct this perpendicular because although I don't know the point N but I do know the base of the perpendicular. So, I can draw the perpendicular from the line A within this plane rho. So, that actually prompts me to basically the steps which I need to construct my plane sigma. So, here are a definitive steps of construction. Number one, draw a plane tau through line A and point M. That's my first step. Second step, within that plane tau I drop a perpendicular from M to line A getting point P. Then I choose any other plane which goes through the A which is not actually the same as tau. Let's call it rho. By the way, how can I do it? Well, if I know the plane tau I can choose any point N which doesn't belong to this plane and basically use N and the line A to draw another plane rho. Now, in that plane, forget about this point N from now on, we have the plane. Now, in that plane I use the point P which I already established by dropping perpendicular from M and draw a perpendicular to line A from this line from the point P. Now, I have two lines intersecting at point P which define plane sigma. So, that's basically all the steps. Incidentally, do you remember how to drop a perpendicular to a line or have a perpendicular to the line from a point located on it? Well, I do suggest you to, if you don't recall it, I do suggest you to go to the corresponding lectures. Well, actually I can do it very quickly here. If you have a line and you have a point, you have some radius which is greater than this distance and you do this circle from this point as a center which means these two are equal, right? Now, from these two points I can use, well, the same circle or another radius and I put here two circles and that's where they cross and this is the line which perpendicular to this one and passing through this. Now, if I have the base instead of this point and I would like to construct a perpendicular. So, if I have this base, how can I get this line? Well, I take again some radius and put the same distance in both sides then from these points, using some bigger radius, I do this, this, this and this and then connect. So, this is how to do it in plane geometry. All right. So, basically this is the construction. So, we have analyzed the problem and then we have established the steps to construct whatever we need to construct. Now, the next is proof that whatever you have constructed is indeed something which was needed. Well, let's just think about it. Which theorems we based upon our construction? Well, the theorem number one, we have the line and we have a point and we draw a plane. Well, there are either axioms or very elementary first theorems which we have already passed through which told us that this is possible. Point and line, point is outside of line, define one and only plane. So, the plane exists and it's unique which is passing through the line and the point. Okay. Next, the perpendicular, we drop the perpendicular. From the plane geometry, we understand that if you have a line on the plane and a point outside of this line on the same plane then you can drop the perpendicular, it exists and it's unique. That's all towards the theorems of the plane geometry. We can do that. Next is we have built a plane which is also passing through this line but is not the same as tau. How can we do it? Very simply, if there is a plane tau, we can choose any point outside of this plane and then use this point and the line to define plane row. Now, within plane row, we have constructed a perpendicular. So, again, from the plane geometry, there are a lot of theorems actually about perpendicular lines and basically we know that from every point on the line, we can construct the perpendicular to that line and it exists and it's unique. No problems about that. Finally, we have Pn and Pm both perpendicular to line A. So, the plane which goes through these, which also exists and is unique. We have already talked about this in our axioms in the first theorems. There is a plane which contains both intersecting lines and it's unique. So, that's how we built sigma. So, it looks like all our steps have certain bases underneath. So, it's not just we built it without any kind of consideration what we are building. We have built it based on certain theorems which we know lead to this particular result. Now, separate consideration should be given to uniqueness of the plane which is perpendicular to the line and passes through point M. Now, do not mix whatever we have just proved. And we actually did mention that on each step we had not only the existence but also the uniqueness of the element which we are building. What if we can choose some other set of steps and then it might actually bring us to a completely different plane which also perpendicular to line A and passes through point M. Is it possible? So, uniqueness is a separate issue which must be considered. So, the question is existence we have proven by actually constructing. The uniqueness is not actually proven yet because we can choose another methodology of constructing which would lead us to another plane. Is it possible? Well, the answer is now and we have to prove that this particular plane is unique. Okay, how do we usually prove the uniqueness? Well, we assume that there is another plane. Let's call it sigma prime. How can I draw it something like this? This is sigma prime. So, sigma prime also perpendicular. So, A is perpendicular to sigma and point M also contains in sigma. A is perpendicular to sigma prime and M is also contains in sigma prime. Now, since both sigma and sigma prime are perpendicular to line A we have two different bases for this perpendicular. So, whenever A intersects sigma it's point P and whenever A intersects sigma prime is P prime. So, this is P and this is P prime. So, P belongs to sigma, P prime belongs to sigma prime. So, let's assume that these two are different points. Now, both P and P prime belong to A. M belongs to both sigma and sigma prime. Which means that if I will connect P and M it's perpendicular to A. Now, M P prime is also perpendicular to A. So, we have from the same point M in space we have two perpendicular to the same line A. Now, it doesn't really matter that this is three-dimensional space because M, P and P prime all belong to the same plane which also contains line A. Right? Because three points by three points we can always consider a plane which contains these three points and since P and P prime belong to this plane the entire line A belongs to this. So, within that plane which contains M, P and P prime within that plane I have two perpendicular from the same point outside of the line A to line A which is impossible in plane geometry as we know. Which means what? Which means P and P prime should be one and the same. So, these two planes maybe they are different but the base of the perpendicular must be the same. Okay, fine. So, can we still have two different planes? So, we have line A, we have two perpendicular planes to line A which are going through the same base P and the same line at the same point M. So, basically if this is one plane this can be another plane. So, they are just turning around M P but still being perpendicular to A. Is this possible? Well, again the answer is no and what I suggest is just to draw another plane which is supposed to cross sigma and sigma prime somewhere. Now, again within this plane if this is the line where this plane intersects sigma and this is the line where it intersects sigma prime within this plane it looks like we have two different perpendicular to the line A at the same point P which is on this line A which is also impossible because again within this plane we have a plane geometry and we have two perpendicular to the same line A which means that again two intersections also should be one and the same. So, these two planes sigma and sigma prime not only have this line in common but also this line also must be in common and basically any line whatever I am it doesn't really matter what kind of a plane this one I choose I will have certain line where and on this line they are supposed to be the same. So, I have two lines right now where sigma and sigma prime are identical they are the same but two lines determine uniquely the plane which means the entire plane sigma and sigma prime should be the same. Well, you see how much time I spent to try to solve this problem construction problem in a way which is supposed to be considered rigorous. So, I was trying to prove the existence of whatever I am going to construct using exact construction steps and I also proved that this construction actually results in a unique result, unique plane which is perpendicular to line A and passes through M which means any other set of steps which I would choose would actually lead to the same final result, the same final plane sigma. So, I think it's very good if you always think about all these rigorous steps like analysis of the problem, like assuming that the problem is solved what can I actually pick from this to help me to construct whatever I want to construct. Then very detailed steps to accomplish whatever you are constructing and finally you have to prove that whatever you have constructed is really what is required and preferably unique because some other person might choose different set of steps and come up with different results. That's usually a no-no in mathematics. If there is a problem to construct something, whatever way you construct must lead exactly to the same result. Alright, so basically that's all I wanted to present to you today. It would be great if whatever, why don't you read the notes to this lecture on Unisor.com and what I suggest you if you understand everything completely close the computer, have a piece of paper and try to write your own analysis, step of constructions and the proof in relatively the same way as I put it on the web because I was trying to put all these in a very detailed fashion. I think if you will be able to do this, it will discipline your mind and it will definitely be extremely beneficial for development of your analytical thinking and logic. And that's what the whole course is about. It's not about how to construct the plane or anything like that. I'm sure you will never need to construct a plane which is perpendicular to a line which is passing through a point. It's not a practical kind of a problem. I mean we have some construction problems but we have a lot of computer programs which help, etc. That's not the purpose of this course and it's not the purpose of this particular problem and any other problem. The purpose is to help you in your development of your analytical thinking because the next time you will see the problem which you don't know how to solve your analysis, your detailed description of the steps and proof that you are right actually would be extremely helpful. That's it for today. Thank you very much and good luck.