 Hi, I'm Zor. Welcome to New Zor education. Today we will spend some time solving problems, my favorite kind of thing. Now, I hope I will be able to solve these problems. They're supposed to be very simple. It's all about lines and planes in three-dimensional space. Some of them actually we did address before, so maybe it's a little bit repetitious. Anyway, this is part of the advanced mathematics course for teenagers. Present it on unizor.com where I suggest you to watch this lecture from because it has nodes with the conditions of these problems. And I do suggest you to spend some time before you watch this lecture. Try to solve these problems yourself. They're not difficult at all, so Well, let's just try to do it. Okay, first I wanted to remind you three major axioms. Basically, everything in solid geometry is based upon. It's not a complete set of axioms. Complete set of axioms is a very big one, but these are major ones which you will be using like always and all the proofs which you make. So the first axiom says that if you have two points that belong to the same line, a and b belong to the same line, a, lowercase a, and these two lines also belong to the same plane gamma. Therefore, an entire line belongs to the plane gamma. So if you have a plane and you have two points of a line belong to this plane, then every point on the line which connects them also is part of this plane. So that's one thing. Next thing is if you have two planes intersecting with each other and there is some point a which belongs to this intersection, then there is a line, straight line, which actually is there. I should say it belongs to not equal to. Then there is a line, straight line, which is their intersection, which contains this point a. So if you have one plane and then another plane and they intersect somewhere and there is a point a which belongs to this intersection. So there is a line, straight line, which is their intersection. It belongs to both of those. So it's actually is an intersection is a line, which contains this point. So that's the second axiom and the third axiom is very simple. If you have three points in space, which do not lie on the same line, then there is one and only one plane which passes through these three points. So three points not on the same line uniquely define a plane, which contains them. Okay, so I just wanted to remind you these main axioms because you will always be eventually using it somewhere. All right, so now let's go to the problem. The first problem is construct a plane which intersects three given planes. Well, it's actually a very simple thing because you can pick one point on each of the plane of the three given planes and three points uniquely or not uniquely define the plane. If they are not on the same line, they uniquely identify the plane and since these three points belong to three different planes, let's say you have one plane, you have another plane, somehow here and the third plane. So you pick three points and since three points define the plane, so we can always draw some kind of a plane which goes through these three points and obviously it will intersect this plane, this plane and this plane because the points belong to these three planes, right? Now, if they are on the same line just by accident, it's still possible to draw a plane through this line, so through any line we can draw as many planes as we want to. So again, we can construct any plane which intersects these three. So there is no problem with this. Okay, this is easy. Well, they are all easy. Next is, I believe we have already addressed it before, but as a repetition, I would like to mention it right now. If you have two angles in space formed by two pairs of half lines and the sides of the angles are correspondingly parallel, then the angles are congruent. So here it is, you have one, let's say A, B and point P and then you have another A' B' P'. So A parallel to A' B parallel to B' in space, in three-dimensional space. Then I'm saying that these two angles are congruent. Now, how can we prove it? Well, we can do it this way. Let's take some segment and put this segment here, here, here and here. Let's call it A, B, A' B'. So P A is equal to P B is equal to P' A' equals to P' B'. Okay, now let's connect P to P' A to A' and B to B'. So what can we say about P A A' P'. Well, P A and P' A' are parallel, right? Because lines A and A' are parallel. Which means they lie in the same plane and do not intersect, right? Okay, also P A is equal to P' A' by lengths. So in this plane P A A' P' in this plane which contains these two parallel lines and these two basically segments are equal. We're talking about parallelogram because it's a quadrilateral, quadrilateral which has two opposite sides parallel and equal in lengths to each other. That means this is the parallelogram. Which means P P P' and A A' are parallel and equal. So P P' parallel and equal to A A' Now similarly P B B' P' is also parallelogram because P B is parallel to P' B' and equal in lengths. So therefore we can say that P P' is parallel and equal to B B'. So all three of them P P' A A' and B B' they are all equal to each other in lengths and parallel. Which means that if I will move this figure along this direction on the same distance my point P will go into point P' E goes to A' and B goes to B' which proves congruence of these two angles. That's it. Now this is very important because this particular theorem will be used in others. So again let me just repeat it. If two angles have correspondingly parallel sides in space then the angles are congruent. Next. Next I have a plane and two lines A and B. I know that they are parallel and I also know that line B is perpendicular to the plane gamma. I have to prove that the second line is also perpendicular. So one is perpendicular and another is parallel to a perpendicular. These are bases. So how can I prove perpendicularity of A to the plane gamma? Well you remember that there is a very important characteristic of perpendicularity. Perpendicular is defined as the line which is perpendicular to all lines which goes through this base. But there is a sufficient condition, it's sufficient to be perpendicular to two lines. So what I'm saying is the following. I will draw from this base of the perpendicular B one line and another which connects to this base of the A. And from this base I will draw a parallel to this one. So what happens is the following. This angle is straight, I mean right, sorry straight. It's a right angle because B is perpendicular. But now since this line is parallel to this and this is parallel to this, this is also perpendicular. Now same thing with this angle. This is the right angle because B is perpendicular. Now A is also making the same angle because again A is parallel to B and this is one in the same line so it's parallel to itself so this angle is also right angle. So here we have two lines, the A is perpendicular to this one and this one. That's a sufficient condition for A to be perpendicular to gamma. So if you have two parallel lines, one is perpendicular and other is also perpendicular to the same plane. Next, well in a way it's basically a reverse problem. You have two lines again, A perpendicular to gamma and B perpendicular to gamma. So perpendicularity is given now and I have to prove the parallelism. Now the previous problem was the parallelism was given and one of them is perpendicular then another is perpendicular. Now I have another one is also perpendicular and I have to prove the parallel. So two perpendicular to the same plane are parallel to each other. So we know both of them are perpendicular. The question is to prove that they are parallel. Alright, well let's do a trick which is actually using the previous theorem. Let me from the point A draw a line parallel to B, A prime. Now by previous theorem I know that this line is supposed to be perpendicular to the plane gamma. So if A is perpendicular and A prime is perpendicular I have one point from which two perpendicular are going to the same plane. And when we were actually discussing this issue in one of the previous lectures we have proven that if there is a plane and a point on this plane there is one and only one perpendicular to this plane which means that A and A prime are supposed to be one and the same which means A is parallel to B as well as A prime is parallel by construction. So we are basically referring to the previous theorem and the fact of the uniqueness of the perpendicular at point. Okay, the next is the theorem about the following. If you have a plane and a line parallel to this plane then the distance from all points to the plane is exactly the same. So let's pick up two points A and B. Well actually in my notes I call them M and N. So let's call it M and N. Now you remember that the distance from a point to a plane is measured by the perpendicular to the plane. You have to make a projection. This is P and this is Q. Okay, so basically I have to prove that M P equals to N Q. Now M P is a perpendicular and Q is perpendicular. By previous problem these two are supposed to be parallel to each other which means they belong to the same plane. So M N Q and P belong to the same plane. Okay, that's good. Well, what is P Q? P Q is intersection of this plane M and Q P with gamma. With plane gamma, right? Because if two points belong to a plane the line in between belongs to one of the axioms, right? So what I can say is that M N Q P plane is intersecting plane gamma along the line P Q. Alright, now there was another theorem that if you have a line parallel to a plane and any other plane which goes through this line then intersection of that other plane would be parallel to the line itself. So that was another theorem which we have learned in the lecture where I introduced the parallelism between lines and planes. Which means that P Q is parallel to M N. Now before that I mentioned that M P is parallel to M Q as two perpendicular to the same plane. So within the plane M N Q P, M N Q P is parallelogram because the opposite sides are correspondingly parallel to each other. And obviously the consequence of this is this, that M P is equal to M Q. So two perpendicular from two different points on the line which is parallel to the plane. The perpendicular to this plane are equal in lengths which means that the plane and line are equidistant. Wherever you make the measure of the distance between the point here to the plane it will be exactly the same. Which is obvious, I mean if you are line parallel to the plane obviously you are moving on the same distance from the plane all the time. But this is the proof. Ok, now my last problem is if you have three planes and two of them, let's say gamma, delta and rho. If you have two of them separately parallel to the third then they are supposed to be parallel to each other. So if this parallel to this and this parallel to this they must be parallel to each other. Again, obviously it's obvious, obviously it's obvious. Yes, I mean visually it's obvious. Alright, how can we prove it? Well, here is what we can do. Well, let's assume that they are not parallel, which means they intersect somewhere. Something like this. This is their interception. Just for chance. Ok, now let's pick point here and any point here or two points here and draw a plane. Now this plane would intersect all three planes along some lines. Now, there was another theorem before we were proving when we were talking about planes, parallel planes and traversal. That if you have two parallel planes and traversal then the intersection lines are parallel to each other. So, we have some kind of a line of intersection of our new plane here, here and here. So, these two lines are supposed to be parallel to each other. Because it's the result of the intersection of the plane which goes through these three points with this and with this. Now, these are, we have to prove that these are parallel. But what we can do is that if they are not parallel it means that they are actually intersecting at this point. Now, within this new plane which contains this line, well and obviously this line, this point, this line, this line and this line. Within this fourth plane this is supposed to be parallel to this, right? Because gamma and rho are parallel to each other. This is the condition of our problem. And we have cut these two parallel planes with this fourth plane. So, this line and this line are supposed to be parallel to each other. Now, this line and this line are also supposed to be parallel to each other because delta and rho are parallel to each other. So, within this fourth plane we have an interesting situation. We have this line parallel to this and this line parallel to this. But nevertheless, these two intersect at some point. Now, this is a pure plane geometry and we know that this is impossible because if two lines are separately parallel to the third one, they are supposed to be parallel to each other and to do not intersect. So, that's the contradiction. So, what we have done, we have reduced the three dimensional problem into two dimensional by cutting through some point which lies on presumable intersection of these two planes. So, we presumed that there is an intersection and we came to the incorrect conclusion that in the plane geometry is basically impossible. The two lines which are separately parallel to the third one intersect, which is impossible. So, that proves that our initial consideration that these two planes somewhere intersect is wrong. So, they do not intersect, which means they are parallel. Well, that completes this set of problems. Very easy problems. As you see, I maybe made just one extra drawing, one extra construction if you wish to prove these statements. These are easy problems and what I would suggest you to do is go to the website Unizor.com, look at the conditions of these problems, look at whatever they are stating and try to either replicate or come up with a different proof of whatever I just did. It would probably better educate you in relatively strict logic which we are trying to achieve right now in these conditions. Alright, that's it. Thank you very much and good luck.