 Hi. I'm Zor. Welcome to Inusor Education. I would like to present to you a few construction problems in three-dimensional space. I do recommend you to try to solve these problems yourself. Just go to Unizor.com and this is problems number 7 within the topic lines and planes. Try to do it yourself. It's very important actually that you try to solve the problems yourself first. Even if you would not succeed, that's still very useful exercise. Now, as far as construction problems, obviously it's kind of difficult to do the real construction. For instance, in two-dimensional case we can just draw something on a paper or a board. In three-dimensional case it's kind of difficult. We assume certain things basically as given and among them are basically the things about planes. Since we can't really draw the planes in space, there are certain assumptions which we are making. Now, assumptions are that the plane is fully defined and considered as constructed if you know the position of, let's say, three points, not lying on the same line in space which belong to this plane. So, three points uniquely determine the plane. Or, for instance, you have a line and a point outside of this line. Again, this line and a point define the plane which contains them, uniquely defined. Similarly, if you have two intersecting planes, they define the plane which contains them. Or two parallel lines, also that's a sufficient condition for uniquely define the plane. So, if anything of this is given, then we consider that the plane which contains these elements is also given without physically drawing the plane in space. I also consider as known facts some elementary construction problems which we have solved before. For instance, how to drop a perpendicular to a plane from a point outside of this plane. Or how to draw a line parallel to another line in space. I mean, these similar and similar construction problems were considered before in prior lectures and I'll just use them basically as given. Now, let's talk about the problems. Problem number one. We have two different lines in space and a point. Lines are considered to be general speaking skew lines probably. I mean, they're not parallel, they're not perpendicular, they're not intersecting, nothing is known about these lines. So, you can consider that the general case is there are skew lines. Now, what you have to do is you have to draw a plane through this point M which is parallel to both lines A and B. So, this plane gamma should contain M and it should be parallel to A and it should be parallel to B. These are conditions of this plane gamma. How can we do it? Alright, now, that's actually quite easy because through the M we can draw the line which is parallel to A, call it A' and draw the line parallel to B, call it B'. Now, these two intersecting at point M lines, A' and B', define the plane as we have just mentioned. Now, this plane is parallel to A because it contains the line A' on it which is parallel to A. And it's parallel to B because it contains the line B' on it which is parallel to B. Which means that this plane is parallel to both A and B. Now, there are certain cases when probably it's impossible to do. And one of the cases is when A and B intersect at certain angle, for instance. You cannot make the plane which is parallel to both of them, I guess. Well, it depends on the position of the point. I mean, sometimes you can, sometimes you cannot. So, it all depends. I mean, there are certain obscure cases when you can't really do it. Okay, that's it. And there are some cases when the problem actually is not sufficiently defined because, for instance, if A and B already are parallel, then there are infinite number of planes going through the point M which are parallel to these lines. So, it's not uniquely defined. But generally speaking, in a skew case, it's uniquely defined. Okay, next. Okay, so we have the plane. We have a point and a line. What we have to do is we have to draw a line through M which is intersecting A and is parallel to gamma. Something like this. So, B is supposed to intersect with A. Their intersection not empty. And B is supposed to contain the point M and B is supposed to be parallel to plane gamma. That's our conditions for line B which we have to meet. Now, how can we do it? Well, I think here is the way how I would suggest to do it. Through the point M, we can always draw the plane parallel to gamma. So, any line which goes through M which is parallel to gamma is supposed to lie within this plane. So, it can be this way or this way or this way. Considering gamma is horizontal, then this horizontal plane which goes through the M contains all the lines which are parallel to gamma. Now, A supposedly intersects this plane at some point. Well, again, unless A is parallel, but let's not consider that. It's a special case. A general case intersects somewhere, right? And then, if you connect the point of intersection A with this plane and point M, this line would be still within this plane which I have constructed. And it will intersect, obviously, A and that's what is required. Now, obviously, if A is parallel to the plane gamma, it will not intersect this auxiliary plane which I have built and there will be no solution. Also, if A goes through M, then there will be an infinite number of solutions. But these are not really general cases. We are talking about a general case and that's what it is. Next. Next, we have three lines in space. I have to draw a line which intersects two of them and is parallel to the third one. Now, I have drawn them on the surface of this board, but in theory, they are all skew lines. They are all somewhere in space. All right. So, how can we do that? So, H is supposed to be parallel to C and it's supposed to intersect A and B. Not empty and not empty. Okay. Here is what I might suggest to do in this case. Let's choose any point A on the line A. Now, any line which is parallel to C and goes through point A would be within the plane which is going through the A. So, what I would do, I would draw line C prime parallel to C and draw a plane through A and C prime. Now, what is this plane about? Well, first of all, it's parallel to C. The whole plane is parallel to C because C is parallel to C prime. I would also say that there is somewhere where this particular plane, I don't know how to draw it, intersects my line B. Let's say I don't have this line yet. I have to construct it. Let's consider that this plane defined by A and C prime which is parallel to C. Let's consider it actually intersects line B at some point. Call it B. Capital B. Now, this point B is within this plane. Call it gamma. So, A and C prime define the plane gamma and B is intersection of gamma with line B. Obviously, within gamma as well. Now, within that gamma, I can draw a line parallel to this one, to C prime. It will intersect here somewhere. Now, this line is parallel to this one and this is parallel to this, so these lines are parallel. Also, this line intersects both lines A and B. So, that's the one which we really need. So, this is the line H. So, again, I have this C prime parallel to C and therefore I have the plane which contains, which is actually parallel to C. And then, from intersection of this plane with the line B, I draw another line which is parallel to C prime within this plane. Yeah, and that would be the solution. Okay, looks okay. Next, number four. I have two lines A and B and point H. What I have to do is I have to construct a line which is perpendicular to A and B and goes through the H and goes through the point H. So, line H which contains capital H and it's perpendicular to A and it's perpendicular to B. That's my conditions. Now, this line doesn't really have to intersect A and B. It can be somewhere in space. But the perpendicularity is assumed to be as angle which is equal to 90 degrees, right angle. In a sense, as it was explained in the lecture about angle between two skew lines, you basically bring them to the same origin with a parallel shift and that's the angle. So, that's exactly what I'm going to do. I will bring A parallel to A, I put A prime, parallel to B, I put B prime. Both of them now are going through this point H. Now, how can I build a perpendicular line to both of them? Well, they define the plane. So, I just, okay, the plane. So, I just build the perpendicular to this plane from the point, uniquely defined. And that perpendicular to the plane will be perpendicular to this line. And since it's parallel to this, this will be perpendicular to this as well. And since it's perpendicular to plane, it's perpendicular to B prime. And since B prime is parallel to B, then H will be perpendicular to B as well. Okay, so that's it. That's number 15. Next, 5. Okay, we have a plane and a line which is parallel to this plane. What I have to do, I have to build another plane, something like this, which intersects this plane gamma at certain dihedral angle. So, there is an angle 5, which is given. So, the line is given, the plane gamma is given, and there is an angle. And I have to build the plane through the line A, which intersects plane gamma at this certain dihedral angle. Now, well, obviously it's similar to the plane geometry. If you have a line and you have a point, and you have to construct an angle which is basically given somewhere. Now, in a plane, it's kind of easy. You basically have a triangle, and then you build an equal triangle from this. So, you basically have all the elements of this triangle, and you can build it at certain angle. All right? So, the simplest thing is to have a right triangle, for instance. That would be even easier. So, you built a perpendicular here. So, you have this triangle. Now, how to build a triangle here? You build it here, somewhere, and draw a line parallel to this one. That would be an angle, right? So, we will use this technique, basically. Because what is a dihedral angle? Well, the dihedral angle is very much defined by its corresponding linear angle, right? So, if this is the intersection between these two planes, what I do is I have to build a linear angle which basically represents this dihedral angle by building two perpendicular to the line of intersection. You know, I would probably draw it slightly differently. So, it would be more obvious. Let me just draw it from the side. It would be better. So, if you have something like this plane, and the line would be somewhere here, above it. So, I have to put something like this. So, this is my line A, and this is a dihedral angle. It would probably be even better if I would open it up. I would put A on this side. It would be like a book. This would be one. And this would be my A line. Now, I'm looking into this angle which is supposed to be equal to phi, right? So, this is perpendicular to one. This is perpendicular to another. And this is an angle which is supposed to be phi. This is a linear angle. Now, this is perpendicular and this is perpendicular. We just look at them from a side. So, now the question is if line A is given, how can I basically build this thing? Alright. If I will... Now, A is parallel to the plane, right? Now, if this plane, let's call it delta, which I have to construct, intersects along line B, my plane gamma, then A and B are obviously parallel to each other, right? Now, this is perpendicular. So, this is also perpendicular. This is also 90 degrees. So, what I will do, I will pick any point on line A and draw a plane which is perpendicular to A. Obviously, it will be perpendicular to my potential B line, right? Which I don't know yet how to construct. However, that would give me the intersection which is this, right? Now, within this plane which is perpendicular to A, and by the way, it's perpendicular to gamma s-gold, but it doesn't matter right now. I can build this angle phi without any problems, right? So, I don't know this line B yet. It does not exist. But what I do know is that if I construct a plane perpendicular to A, it will intersect my gamma at certain point. And within that plane, I know this line, because that's the line of intersection between my auxiliary plane, let's call it rho. So, rho is this plane. So, I know that within this plane I have a point, that's where I have chosen that point, and I have a line, because this is the line of intersection with gamma. So, I can build an angle, this one, from point goes to this line which contains this angle, which gives me this point, let's call it B. Now, this point is on intersection between gamma and rho. Now, and A, B, call it C, gives me my angle phi, because that's how I constructed it. Now, since I know this point B, I can draw a plane through A, line A, and this point D, and that will be my plane gamma, sorry, delta. So, again, from A, I construct a perpendicular to A plane, I just cut everything. And that gives me this line on plane rho. Rho is my perpendicular to the line A. Then I'm dropping a line from A to this intersection of the rho and gamma to get point B such that this angle is equal to phi. And as I was saying, we know how to do that in plane geometry. And having the point B, I can draw a plane. Now, there is no freedom, because apparently I randomly chose A, but if I choose something else, A prime. Now, my plane, which is perpendicular to line A, would be different. It would be lower, it would be something like this, rho prime. But again, I will get this angle and I will get another point B prime. But the plane will be exactly the same, whether it's B or B prime, whether it's A or A prime, the plane defined by line A and point B or B prime is exactly the same plane, and it will intersect at certain line B in both ways, it will be the same intersection. Okay, that's it. Now, next problem. Next problem, we have a plane and we have two points above it, on the same side of the space. So, a plane apparently divides the space into two halves, right? So, they're both in the same half. I have to find a point M in such a way that A M plus M B is minimum, minimum possible. I mean, obviously it should be minimum, because if M goes to infinity here or there, the sum would go to infinity as well. So, there is somewhere the minimum of this sum of these two distances. So, where is this minimum? The way how this problem is solved is very similar to an analogous plane geometry problem. In the plane geometry, you have the same kind of thing. You have line and you have two points A and B, and you have to find point M such that some of these two distances is the shortest. Now, how to do that? You just reflect B relative to this line. So, these two are the same on the perpendicular B prime, and you connect A and B prime. Now, obviously this A M plus M B would be exactly the same as A M plus M B prime, since B and B prime are symmetrical relatively to this line, right? Because this is equal to this. But now, A M plus M B prime would be smaller if it's a straight line, rather than two segments which are not on the straight line, right? So, we will use exactly the same technique here. For instance, we are reflecting through perpendicular to B prime. Now, obviously sum of these is equal to this plus this. So, all I have to do is basically reflect the B relatively to this plane through a perpendicular, drop a perpendicular down to the same distance below this plane, and then just connect A and B prime with a line, and this line will intersect gamma at certain point M, which is shorter than any other, because any other would be not a straight line, and A and B prime is a straight line. Okay, so that's my last problem for today. I will suggest you to go through these problems again and try to solve them again. Maybe you forgot whatever I said, then go back to the lecture, or maybe you can come up with your own solution, which is fine too. But it's very important that you repeat it again, maybe even more than once. So, all these logical compositions of the steps which you know how to do, like elementary steps in construction, you can combine together into more complex procedure. That's it for today. Thank you very much and good luck.