 Hi, I'm Zor. Welcome to a new Zor education. We will solve a few problems today, problems about planes and lines, parallel, perpendicular, but general solid geometry stuff. This lecture is part of the course of Advanced Mathematics for Teenagers, which is presented on unizor.com. That's where I suggest you to watch this lecture from because it contains basically the conditions of these problems. No solutions, just whatever the problem states. And I do recommend you to try to solve these problems yourself before you listen to this lecture. Well, these problems are not difficult at all. Most of them are in basically like one or two logical statements and references to the previous problems or axioms, so they should not present much difficulties for you. It's very important that you try to do it yourself. Even if you don't come up with a good solution, it's still important that you think about this. You're trying to make some logical connections between components of each problem and only then listen to whatever I'm suggesting is my solution, which obviously is not necessarily the only solution. You might come up with your own, and by the way, you're welcome to send me an email about your own solution. I will gladly include it into the website with reference to you. All right, so problems. So I have specified them in a short hand notation. So here they are. Problem first. You have two planes parallel to each other, one plane and another plane. Now the problem states that the distance between these planes is constant. Now what is the distance between planes? Well, the distance from any point on this to this plane is basically the length of the perpendicular. And if you have another point and have its projection to this plane, it's supposed to be the same length of these perpendiculars. So we're talking about two randomly chosen points on the plane gamma, which is parallel to the plane delta. And we have two projections, which means two perpendiculars from the corresponding points to the plane delta. And I have to prove that the length of AB is equal to the length of A prime D prime. If planes gamma and delta are parallel. All right, so how can we prove that? Well, let's think about it. First of all, we have already proven before, that was one of the problems. I think it was a previous lecture actually, that two perpendiculars to the same plane are parallel to each other. Okay, so these are parallel to each other, which means that they are in the same plane, right? Okay, now since they are in the same plane, what we can do right now is we can prove that this is a rectangle. Now, why is this a rectangle? Let's just think about it. If we will draw a line from A parallel to BB prime in this plane, it must coincide with A A prime. Why? Well, because let's say it's not A double prime or second A second. So A A second within this plane is parallel to BB prime. Okay, now since they are parallel, then A A second B prime B is a rectangle, right? Because this is parallel to this. This is parallel to this. Now these are right angles because A A B and A prime, B prime are perpendicular. So it's a parallelogram with the right angle, which means it's a rectangle. And A A second is equal to BB prime. All right. So all I have to do right now is prove that A A second and A A prime are one and the same line. Now, first of all, they do lie in the same plane. That's number one. Number two, A A prime is parallel, but A A second is parallel by construction. And A A prime cannot intersect with BB prime in this plane. I'm talking about A A prime, B prime, B plane. So within that plane, A A prime and BB prime cannot intersect because otherwise the planes would not be parallel, right? So if they're not intersect, A A prime is also parallel to BB prime. So A A second and A A prime are one in the same line. And therefore, A A prime, B prime, B is a rectangle. So we can just wipe out this line. It coincides with this one. So if I draw a parallel from A parallel to BB prime, it will completely, within the plane, A A prime, B prime, B. It will completely coincide with A A prime. All right, that's it. So it's a rectangle. And since it's a rectangle, A A B and A prime, B prime are equal in lengths, which means that the distance between two parallel planes measured along the common perpendicular is always constant. All right, that's it. Basically, all of these problems are on this level of difficulty. You just have to have maybe one extra construction or something like this. Next one. You have again two parallel planes. So that's number one. Gamma is parallel to delta, gamma delta. Then you have a point A on plane gamma and a line A, which contains this plane, this point A. Also is known that line A parallel to delta. So the planes are parallel and this line is parallel. Now, what I have to prove is that this line A completely inside the plane gamma. So basically, what it says that if you have a point on a plane parallel to this plane, any line which passes through this point parallel to this should be within this plane. So all lines parallel to the delta are supposed to be within this plane gamma. They are horizontally rotating around the point A but always within plane gamma. So that's the pointer. Okay, so how can we prove that? Actually, you know what we can probably use the previous problem. Since planes are parallel, what I will do is the following. I will drop a perpendicular from A to B and I know it has a certain fixed length, right? Now, what I will do next is let's assume my line A passes through point A on the plane gamma but it's intersecting gamma at a certain angle. It's not within plane gamma. And what I will do is, within the plane gamma, I will draw a line which is parallel to this plane. Now, how can I draw it? There are many different lines, right? So what I will do is the following. I will draw a plane through A and the perpendicular H. So A is somewhere there and I assume it intersects gamma at a certain angle. And there is a perpendicular H. So if I will draw a plane through A and H, it will intersect my plane gamma at a certain line A prime. Okay, so now A prime is parallel to this plane delta. Why? Because it's very easy because the same plane which I draw, which is going through A and H, it intersects this plane delta at a certain line B, right? So this same plane, so the A prime and B are within the same plane and they're not intersecting because the planes are parallel, right? So the line A prime is parallel to B and therefore it's parallel to delta. Okay, fine. So I have my line A which is outside of the gamma by assumption, right? And line A prime which is within plane gamma. And they're all within the same plane. But now look at it this way. If my two lines A and A prime both are parallel to delta, now A is not intersecting with B. Otherwise A would not be parallel to delta, right? And A prime is not intersecting to B. And everything is occurring within the same plane which is the plane formed by line A and altitude H. Because A prime is intersection of this plane with gamma and B is intersection of the same plane with delta. So it looks like within the same plane I have line B and two lines A prime and A parallel to B. Now, and both are actually going through the same point A. Now from the plane geometry we know that this is impossible to have two different lines parallel to this one which is passing through the same point, right? So A and A prime must coincide. So our assumption that A is outside of the plane gamma is just wrong. It must be inside. So A and A prime are supposed to be the same lines. So again the key to this problem is to draw a plane through this line and H which is just a projection from point A to delta. And that gives me two lines and the parallelism is obvious. Okay, fine. Number three. I have two intersecting planes. I have two intersecting planes. Let me try to draw it this way. This is one and this is another. Gamma, delta. C is the edge of this dihedral angle if you wish. Now we have two lines. Now within gamma we have line A parallel to delta. No, sorry. It's not parallel to delta. We have line B which belongs to delta and these two lines are parallel to each other. That's what happens. So the planes are at the angle to each other but the lines are parallel. Now my statement is that in this case both lines A and B must be parallel to the intersection between these two planes. Okay, here's how we can solve it. If A is parallel to B then A is parallel to entire delta. Right? So A is parallel to delta. Now if B is parallel to gamma, sorry, to A then B is parallel to gamma. Because if the line parallel another line on the plane that's enough to say that the line is parallel to an entire plane. Okay, now look at it this way. We have a plane delta. We have a line which is parallel to this plane and we have another plane gamma which goes through A and intersects delta. Now there was a theorem that if you have, let me just draw it differently so you would recognize it better, so if you have a line in this case A parallel to delta and you have a plane which intersects delta and goes through A then the line of intersection, by the way this is gamma, plane gamma, then the line of intersection which in this case C is parallel to the line itself. So from this we know that A is parallel to C. Now symmetrically we have a line B which is parallel to plane gamma and we have a cutting intersecting plane delta going through the B and the theorem stated that the line of intersection which is C actually would be parallel to B. So similarly B is parallel to C. So all you have to do right now is to consider this problem as a continuation basically or a very close relative to the theorem which has been proven before that if you have a parallel line to a plane and then the plane which intersects this plane passing through this line then the intersection will be parallel to the line. Same thing we just applied it twice for two different cases and we have two different parallelism between A to C and B to C separately. Okay done that next next so we have a plane gamma we have line A within it now we have a line B which is not part of gamma but it is parallel to A. Okay now we have some plane which goes through B and intersects plane delta which intersects gamma at line C. So what I have to prove is that A is parallel to C. So C is gamma intersect delta. Alright so it's absolutely similar consideration as in the previous case right. You have this line parallel to the plane and then we have a plane which goes through this line and intersects the gamma at C. So B is parallel to C from that theorem which I was referring to in the previous problem but A is parallel to B that's the condition of our problem. So there is a transitivity here A is parallel to B, B is parallel to C and from this we follow that A parallel to C and again I can refer to another problem which we have solved before. If two lines separately are parallel to the third line then they are parallel to each other. We did it before in one of the lectures. That's quite interesting actually that in mathematics if you have proven something you can use it to basically shorten the whole proof for the next problem that's what I just did. I immediately referred to some theorem and the problem which we have already solved and that's sufficient as a proof obviously. Okay next that was fast. Okay now we have two skew lines A and B which means they are not intersecting and they're not parallel to each other. Okay I will probably have to draw it slightly differently. Let me start from the planes these two lines belong to. So there is a plane gamma which belongs which A belongs to and then there is a plane delta and B belongs to. So I draw these planes parallel to each other because that's what I'm going to prove actually that they are parallel but the condition is so A is parallel to delta, A is parallel to delta and B is parallel to gamma. B is parallel to gamma and by themselves A and B are basically completely random so they can be skewed. Not parallel, obviously not intersecting according to this drawing at least. So what I would like to prove is that gamma and delta are parallel lines so if one line on one plane is parallel to another plane and the line on that other plane is parallel to the first plane then the planes must be parallel. This is actually quite interesting I mean you might not really expect this type of thing this is one of the more interesting problems which I would say. All right so what can we do about this? Here is what I suggest. Let's just take a point on B and draw a line parallel to A, A prime and take a point here and draw a line parallel to B. Now these lines are parallel and these lines are parallel to each other. Now my point right now is to prove that the line A prime is completely within delta and the line B prime is completely within gamma. How can I prove that? Well let me do it this way. I'm not taking any point here. What I will do is I will project line A onto the plane delta. So it means that these are perpendicular. So from each point on line A I draw a perpendicular down to B basically forming a projection of the line A onto plane delta. Now projection of a line is a line we were talking about this before that's number one. Number two all these perpendicular are parallel to each other so it's basically they form a plane if you wish which cuts across. So A and A prime are parallel to each other. Why? Because A prime belongs to this plane delta, A belongs to plane gamma and I know that A is parallel to delta which means it cannot intersect. So if A prime and A are not parallel while lying in the same plane I would have that line A intersect somewhere plane delta which is impossible since it's parallel. Now similarly it's not just any B prime which is parallel to B. I will project my B onto plane gamma getting the line B prime and again it's supposed to be parallel to this one because B is parallel to the plane gamma and the line and its projection are lying in the same plane so the parallelism is obvious they cannot intersect because otherwise B would intersect gamma. So what I have done right now I have proven that line A prime which lies within delta because it's a projection of A onto delta and line B which is also that's the condition of the problem it lies within delta. They form some angle and these lines also both lying within gamma they also form an angle and sides of these two angles are parallel to each other and as we know again there was a theorem before that for parallelism of two planes it's sufficient to have some two lines at angle to each other any angle as long as they're not parallel to be correspondingly parallel to the lines on another plane. So that's the major theorem about parallel planes so it's sufficient to have two lines on one correspondingly parallel to two lines on another as long as the lines are intersecting that's sufficient to have planes parallel to each other that was one of the theorems. Okay so that's the proof of the parallelism next next so we have a plane gamma we have a line B which is perpendicular to plane gamma and then we have line A which is perpendicular to B to line B now I did not draw it this way obviously I could because this line A can be anywhere within this within this plane which is parallel to gamma so basically I have to prove that line A is parallel to gamma given that A is perpendicular to B and B is perpendicular to gamma. All right so what should we do in this case well what we can do is and it's actually similar to whatever we have done in in one of the previous problems we will construct a plane defined by A and B it will be just a cutting plane through this is a plane okay now let's put some points here A, B, C and D so B is just any point on A and C becomes its projection it's perpendicular to gamma so everything occurs within A, B, C, D plane so what do I know about this now I know that these two are parallel now this angle is 90 degrees because B is perpendicular to gamma and therefore it's perpendicular to C and this is also 90 degree because I just dropped the perpendicular within this plane I dropped the perpendicular to C, D what else this is also right angle so we have basically a parallel ground here obviously because all angles are at least three angles are 90 degrees right so that's what that's why the fourth angle is also 90 degrees so this one is 90 degree because that's how A was defined relative to B this is 90 degree because this is perpendicular to the plane and these two are parallel to each other because I dropped a perpendicular from B within this plane onto the line C, D so it's a parallelogram and from this being parallelogram follows that A, B is parallel of C, D which means A is parallel to one line on the plane gamma which means it's parallel to the entire plane gamma all right these are all the problems and what I suggest you to do right now is try to go through the same problems again on Unisor.com and and solve them again that would be a good exercise to basically firm put it into your mind that's it thanks very much and good luck