 Hi, I'm Zor. Welcome to Inuzor Education. Today I will continue talking about lines and planes and about their parallelism. We have already defined all the different elements we will be dealing with in solid geometry and now we are going through different properties, characteristics and theorems, which is very important. Actually, that's my kind of favorite part of the course when I'm trying to prove the theorems, solve the problems, because that's exactly what is required from the young student's mind to be developing. So that's the purpose of the whole course to teach you to think logically, analytically, creatively, etc. And that's why I would actually recommend you to try to prove all these theorems, which I'm going to present today just by yourself. Go to theunizor.com where this lecture is presented, and in the notes you have all the theorems and the proofs as well. So you can just look at the proofs, think about how you would prove it yourself, then listen to the lecture, then try it again, and you will actually be comfortable with the logic of this. Now logic is very important here because we are very close to the axioms. We don't really have a big set of theorems which we can base upon. There are actually practically none. We just have a couple of theorems and a couple of axioms which we have already passed through, which I will repeat actually. So it's very interesting how you build the logic in such a way that you base your logical conclusions only on those axioms and the theorems which you have already proven before. So that's why you have the logical stepping forward. Okay, so today we're talking about parallelism between lines and planes. Let me just very briefly remind you the axiom, the main axioms. I mean, there are many axioms. Hilbert actually was the guy who put the set of like 20-something axioms into geometry, which definitely encompassed Euclid's postulates. So let me just remind you. Okay, axiom number one, if two points belong to a plane, then the line which connects these two points also belongs to the plane. That's an axiom and it seems to be quite obvious. Now the second one is if a point belongs to the intersection between two planes, then there is a whole line, a straight line obviously, where these planes intersect and obviously this line contains the point. And the third axiom is if you have three points in space which are not lying on the same line, there is one and only one plane which goes through them. Again, an obvious thing. Now as a very brief consequence of this, we have proven a couple of theorems and one of them which is probably more important for this particular lecture is that if you have a straight line and the point outside, there is only one plane which there is one and only one plane which goes through the line and the point. So the line and the point outside of it define the plane. So these are axioms and the trivial consequence which I will be using right now. So let me just go straight to the definition. Now the previous lecture was about the parallelism of two lines. Let me just remind you that two lines are parallel in the three-dimensional space if they, well obviously number one do not have common points, but number two they should be in the same plane. Now what about the line and the plane? Well the parallelism between the line and the plane is even simpler. Just no common points, that's it. As long as there is no, the intersection between the line and the plane is empty, these are parallel to each other. So let me just write it down. If there is a line A and there is a plane gamma and their intersection is empty, therefore A is parallel to plane gamma by definition. So this is a definition, this is the definition of the parallelism. Now I will be using different symbols as I'm going through and primarily my lower case legend letters will be lines, upper case legend letters will be points and the Greek letters will be planes, lower case Greek letters. So I'll try to adhere to the standard throughout the course. Now we are, now we have five different theorems which we would like to prove one after another and that's what I'm going to do. And I hope I will be able to do it. Okay, theorem number one. Okay, if you have a plane, let's call it gamma, you have a line B on the plane and line A outside of the plane. And I know that A parallel B. So these two lines, so one of these two lines is inside the plane and this is outside. So the theorem states that if this is a situation when line outside is parallel to the line inside, then it's a sufficient condition for this line A to be parallel to the entire plane. Okay, so how can we prove it? Well, let's just think about it. What is a definition of the parallel lines? The parallel lines are those lines which number one line the same plane and number two do not have common points, right? Now, if I'm stating that these two lines are parallel, it means there is some kind of a plane they both belong to and in that plane they do not have any common points, right? All right, okay, so if you assume that line A intersects somewhere the plane gamma, so let's say that there is some kind of intersection. Now, at the same time, line A completely belongs to some kind of a plane, well, let's call it delta, with line B, since they are parallel. So since line A completely belongs to beta, this point which is a continuation of line A also belongs to plane beta, sorry, delta. So these two lines within the plane delta are supposed to be parallel, which means it cannot actually happen any intersection between these two lines. But if, however, I assume that line A somewhere intersects gamma, it inevitably means that these two lines are intersecting because these two lines belong to the delta. So everything happens within this plane delta and that contradicts basically the parallelism of the lines A and B, that they are not actually intersecting to each other. So it looks like any continuation of line A, it cannot hit the plane gamma because these two are inevitably will be intersecting within this plane delta. Because this line is completely belongs to the delta and this line completely belongs to the delta, that's why they are intersecting within the plane delta. And that contradicts the two-dimensional parallelism. This is basically the fifth postulate of Euclid, that the parallel lines are not actually intersecting. So that basically proves that this is impossible, the intersection between line A and the plane gamma is impossible, which means that they are parallel, A and gamma. Okay, that's the first theorem. Actually all of these theorems are relatively easy. It's like basically one logical step from something which we have already learned to this particular theorem. Theorem number two. Okay, now let's assume that we have a plane gamma. Again. And we have a line A which is parallel. Now we draw a plane which is crossing. So this plane delta is containing the line A and crossing the plane gamma. So the theorem states that intersection between gamma and delta, which is this line, let's call it B, is parallel to the A. In some way it's a reverse theorem to the previous one. So in this case we know that the line A is parallel to gamma, to the plane gamma, and it intersects, the plane delta contains the line A and intersects the plane gamma. So the intersection will be parallel to this. Okay. How can we prove it? This is the easiest of all these theorems. Now the parallelism between this and this means that these two lines are supposed to be in the same plane, but they are in the same plane by construction of delta, right? Since B is an intersection between gamma and delta, it obviously belongs to delta. So A and B both belong to delta. So they do belong to the same plane. Now the second condition is that they should not really intersect. These two lines, if we want them to be parallel in three-dimensional space. Well, they cannot intersect because B completely is part of the gamma. So if A and B are intersecting, it means A and gamma are intersecting. So somewhere if there is some kind of a point where they intersect, this point belongs to gamma because the B belongs to gamma. So, and that contradicts the parallelism between A and gamma, which means that these two lines A and B must be parallel. One plane they belong to and no intersection. Very simple. Next. Next is, okay, if you have a line in space and some kind of point outside it, outside it. Question is how many lines parallel to A you can draw through A? Well, the obvious answer is one. Now we have to prove it. So that's the theorem. So we have to prove that through a point lying outside of a line A in three-dimensional space, you can construct only one, one and only one line parallel to line A. Well, again, let's assume that there is another one. So we have two lines. Both are going through this point A and both are parallel to line A. Okay, how can we prove that B and C is actually one in the same line? Let's just think about it. Well, since A and B are parallel, we obviously can construct a plane which contains both of them and in that plane they do not have any common points. All right, so A is parallel to B. All right. Now what? Well, now let's try to have another plane which connects A and C through A and C. So we have another plane. And again, it's possible because A and C supposedly are parallel, right? Okay. Now, so what do we have now? What do we have now? So here's an interesting point. Let's call the first gamma and delta. Okay, so let's consider the plane delta, which is the plane which connects A and C. Now B is parallel to this plane delta because B is parallel to one line which is A which belongs to delta and that's our first theorem, right? So according to the first theorem B is parallel to delta. On the other hand, B has a common point with C and therefore with delta. So it looks like B should be parallel to delta but at the same time it has a common point. Well, that's a contradiction because the parallel line to a plane cannot have any common points. So that's why B and C must be one in the same line. It cannot be two different lines. Well, that's it. So again, we consider delta and since B is parallel to A, B is parallel to entire delta and it cannot have a common point. All right, that's it. Short and sweet. Next. Okay, if you have two different planes, one plane and another plane. Well, let me just continue it from this point. That would be easier. All right, so two planes. And I know that the line A is parallel to both mu and mu. So if line A is parallel to two intersecting planes, then the theorem states that it's supposed to be parallel to their intersection. So if A parallel to the plane mu and A parallel to plane mu, therefore it's supposed to be parallel to their intersection. So B is intersection between mu and mu. Okay, let's try to prove that. Well, again, what does it mean to be parallel between two lines? It means it's supposed to be in one plane, both supposed to be in the same plane and no intersection. Well, no intersection is obvious in this case because if A intersects B, it means it intersects with both of these planes, which is impossible since it's parallel to them. So no intersection is fine. So the question is whether it belongs to the same plane. All right, what we will do is the following. Let's just take one particular point on the B and draw a plane based on the line and the point, which is one and only plane possible which we were proving in some other theorem. So this is our plane which is constructed from line A and point B. Well, now this plane, let's call it gamma. So this plane gamma intersects with both mu and let's call this line of intersection m. Now m is a line in mu which goes through p and what's also important, since A parallel to plane mu, this intersection should be parallel to the A. That's the second theorem, right, which we have proven before. Now, analogously, this plane gamma, if I intersect it with this plane mu, it will be some kind of line m. And also, sorry, parallel to A. So what do we have? We have both m and n parallel to A and both m and n both contain point P. Now, the theorem number 3 says that from this point outside of A we can draw only one line which is parallel to the given line. So from the point P we can have only one line which is parallel to A in space, which means that m and n are one and the same line. So it's one and the same line. Now, m belongs to mu and which is the same line belongs to mu. So one and the same line belongs to both mu and mu. And therefore, m or n, whatever you want to call it, belongs to their intersection. And intersection between two planes is one and only line B. So m is B. Well, same thing is n. So that's the end of the proof that B and m and n are all one and the same line is parallel to A. And that's what we needed to be proven. Okay, that's it. And the last theorem, the fifth one for today is the following. If you have two lines parallel to one and the same line, they are parallel to each other. Something quite obvious. If two different lines are parallel to the third one, they must be parallel to themselves. In two-dimensional, we know about this theorem, now about the three-dimensional thing. All right, we have to prove it. So let's just think. First of all, m and n cannot have common points. Because if they do, if they intersect somewhere in space, it would look like from the point P, we have drawn two different lines parallel to the same line A. And we have already proven it in theorem number... Which one? Three, right? Yes, that's the theorem number three. Which we have proven that this is impossible. There's only one line parallel to another, which you can draw from one particular point. So no common points between m, m and n. So what we have to prove is only that they belong to the same plane. All right? Okay, so let's just try to prove that they belong to the same plane. Let's choose the point, let's say P on line n. And we will draw the following planes. This plane, m and a, are parallel by the condition of the theorem, which means we can draw the plane. It belongs to, let's call it nu, and they do not intersect. n and a are also parallel, so we can draw a plane, n, nu. And a is the intersection between nu and nu, obviously, because a belongs to both of them. Now we will draw the third plane between the line m and one particular point on the line n, which I call P, right? Okay, so it's something like this. Now I do not know that the whole line n actually is part of this plane. Let's call it gamma, let's say. I do not know that. That's what I have to prove. If I know that n belongs to this plane gamma, then it means that m and n are in the same plane, and that's what the parallelism means, because they don't have common points. All I know is that only one point, p of n, belongs to this plane. But now let's just think about it. m is parallel to a, and therefore m is parallel to nu. The entire plane nu, which contains a, and m is parallel to a, therefore m is parallel to the entire plane, right? That's the first theorem we have proven. Okay, now let's use the second theorem. Since line m is parallel to plane nu, it means that intersection between gamma and nu is parallel to m. So we have a plane, which is nu, and we have intersected with gamma. And m is parallel to the plane. The second theorem says that in this case the intersection would be parallel to the m. So this intersection is parallel to m. Now let me ask you, can this intersection be anything but the line n? Well, if you assume that there is some other line instead of n and prime, then it looks like m prime and n both are going through the same point p parallel to m, which again we have proven before the third theorem that this is impossible. At this point we can draw only one line parallel to another line. So basically the intersection between plane gamma and plane nu is this line n, which means that n and m lie both in the same plane gamma. And that's a sufficient condition for them being parallel because there is no common points. So basically, as you saw, it just takes one or two logical steps from the beginning, from the axioms and one of these theorems to prove another one. But it's very important to make them right because it's very easy to make some kind of a logical loop and start proving a from b and then b from a. And that's basically a distortion of the entire logical construction which we are trying to build. So try to be very careful with this. And I definitely urge you to go through the notes for this lecture again and follow the proof. Try it yourself as well. It's very important that from the very beginning, when everything is really simple and requires only one or two logical steps, that you do it yourself. So ultimately, I would like you to be able to basically stand like I'm standing here and explain it to somebody else based on certain axioms and certain very primitive theorems to make some logical steps to prove a little bit more complicated theorems. Well, that's it for today. Thank you very much and good luck.