 Hi, I'm Zor. Welcome to Unizor Education. We continue talking about angles and straight lines and planes, and in particular this lecture is about an angle between a line and a plane. Well, this is obviously part of the course presented on Unizor.com. It's advanced mathematics for teenagers. I suggest you actually to watch this lecture from the Unizor.com website because it contains notes, very important notes, which you can just read before the lecture or after the lecture. And one other detail is if, for instance, I'm presenting a theorem in the lecture, first try to prove it yourself. I mean, you read the notes before the lecture, try to prove the theorem yourself. It's really very easy theorems, which I usually include into this theoretical material, but it's a very good exercise for your logical thinking, your analytical abilities, etc. Okay, so an angle between a line and a plane. First of all, let me just talk about some philosophy of math. The concept of an angle is very much similar to the concept of a distance. The distance between two points, it's basically how much you have to cover moving from one to another. It's a measure of your movement from one to another in certain direction. An angle, it's also a measure, it's a measure of rotation from one direction to another. So in this particular regard, these two concepts are similar. Both are measures, both are additive measures, which means if you move from point A to B and then from B to C, the measure from E to C, if it's along the same direction, is added together and measure from rotation from one direction to another and then from that other to the third one, again, the sum of two measures is basically the measure of an entire movement. So there is a similarity. Now, let's move the similarity a little bit further. If you're talking about a distance not between a point and a point, but let's say between a point and a set of points, is there a concept of a distance between point and a set of points? Well, yes, there is. Traditionally, we choose the point which is nearest to this one and we're saying that the distance from this point to a set is actually the distance to the nearest point. Well, there might be cases when there is no nearest point. For instance, if this particular area on this surface is without the border, but we're not considering these cases, these are really some obscure cases which we don't really need right now. What's important is that in a relatively general case, we are choosing the nearest point, so the minimum of the distance between this point and any of these is the distance from the point to the set. Now, it is possible that there are two points which are in the same distance. Let's say if you have something like this picture, then there are two points actually, this one and this one, which can be on the same distance. So, the distance between the point and the set is still the same, that's the smallest, but not necessarily the one to the nearest because there is no nearest, there are two different nearest points, right? But in any case, what's important is that there is some kind of a minimum distance from a point to this particular set. So, out of all the different distances, we choose the minimum. Can we expand this particular concept to an angle between a line and a plane? Well, the answer is yes because you can always consider an angle, let's say this is your plane and this is your line which intersects the plane. Now, you can always consider different lines on the plane and choose the angle between this line and any of these and try to find the minimum. So, if it can be found, then this is a good definition to an angle between the line and the plane. So, it's a minimum of all possible angles between this line and all the different lines which basically comprise together a plane. All right, well, as a philosophy it's fine, but mathematically we have to really prove a couple of things. Number one, we have to prove that this minimum exists, that there is a particular line or maybe lines on the plane which are actually those where this minimum is achieved. It would be really great if we have this particular line existent and it would be even better if it's a single line, not like I was just describing two different nearest points. It would be much better, it would be much nicer from the mathematical standpoint if we have one and only one line on the plane which forms the minimum angle with our given line. So, this is some kind of a preamble to whatever I'm just talking, will be talking about just to explain that it's not really just obtained from the thin air, whatever I'm going to talk about. It's obtained from the purpose which I would like to achieve. I would like to show that there is one and only one line on the plane which forms the angle which is smaller than any other line forms with this given line. That's my goal today. Now, how I am going to achieve it? Well, first of all, let me introduce a very simple concept, a concept of projection. Now, if I have a plane and a point outside, projection of the point onto the plane is the base of the perpendicular which I can drop from this point to a plane. Now, perpendicular is one and only, we were talking about this before, which means that the base point is also one and only. So, for any point outside of the plane, projection is the point, one and only point, which is the base of the perpendicular. That's easy. Now, let's consider we have a line, straight line. Now, what is a projection of the straight line? Well, we would like actually to have this line as a projection, which connects, let's say, one of the bases of the perpendicular from any point on the plane and point of intersection because obviously, if you have a point not above the plane, but let's say on the plane, what's the projection of this point? Well, obviously, it's itself, right? So, what I'm going to prove is the following theorem, that the projection of the line onto a plane is a straight line which goes through point of intersection of the line and the plane and one of the bases of the perpendicular dropped from any point because if you take any other point and drop the perpendicular, it also goes to the same line. That's what I'm going to prove. Well, this is again one of the theorems which you can try to prove yourself. So, if you want, you can just stop the video if you didn't do it before and try to prove it. It's really very simple. Here is how. Again, pick any point on the line, drop a perpendicular and now we can consider three points. This, this and this. So, any point on the line outside of the plane, point of intersection of the line and the plane and the base of the perpendicular from this. Well, let's put some letters on it. A, B, C. All right? Now, three points define the plane. So, let's draw a plane through these A, B and C. Well, since the plane contains A and C, it contains an entire line, obviously, and it contains an entire line A, B, obviously, and it contains the line B, C. Now, any plane which goes through a perpendicular A, B to a plane gamma is perpendicular to gamma. That's one of the theorems which we have proved before in one of the lectures. So, the plane A, B, C is perpendicular to gamma. That's one thing. Another thing is that if I have two planes perpendicular to each other, plane A, B, C and plane gamma, then if I drop a perpendicular from any point on this plane A, B, C, it will be completely within the plane A, B, C. Which means that any other point here, if I drop a perpendicular, it will drop within plane A, B, C. Now, since I'm dropping the perpendicular, the base belongs to the gamma. So, the whole perpendicular belongs to the A, B, C, but base belongs to both actually A, B, C and gamma. So, it belongs to intersection. Now, intersection between plane A, B, C and gamma is the line which connects B and C. So, that's why we have proven that the projection of any point on the line lies somewhere on the line B, C. So, the projection of the straight line is a straight line which connects, again, the point of intersection and you can take any base of the perpendicular dropped from any point on this given line to the plane gamma. So, that's what the projection of the line is. Now, by the way, I did not really talk about this, but projection of any other curve, I should say, not necessarily a straight line, is basically something on this plane gamma which is formed by all the perpendicular. So, I didn't really mention this, but it's kind of assumed and that's why I suggested that the projection of the line is something on the plane gamma which happens to be a straight line. All right, so, what else I have to mention here? Here is another interesting detail. You see, if this line intersects gamma, then projection is, as I was saying, B, C. Now, A, C and B, C have common point C. They are not skewed, they have common point, right? So, the straight line and its projection belong to the same plane which is plane ABC, right? Now, if the line is not intersecting C, there is no point C, there is no intersection between line and gamma, which means if the line is parallel to the gamma, right? What is a projection in this case? Well, basically the projection is already defined. It's the basis of all perpendiculars, right? But if you drop two perpendiculars from a straight line, okay, I think I have to redraw this picture, if you have plane and the line parallel to the plane, then if you drop two perpendiculars, now, two perpendiculars to the same plane are parallel to each other. So, these four points belong to the same plane, right? Since A, D and B, C are parallel. And within the same plane, obviously, ABC, D is a parallel, and a rectangle because these are perpendiculars, right? So, these are all perpendiculars. So, if the line is perpendicular to the plane and this line is parallel to the plane, this is also perpendicular. We have already proved this particular point. So, in which case, again, the line and its projection belong to the same plane. They are not skewed because this is a rectangle. All right. So, we have proven that straight line is projected onto a plane as a straight line. Well, in this case, it's a straight line which is basically parallel to the one which we have in this case. It's still a straight line which intersects, obviously, with ours. Okay. The only thing which I really have to prove right now to go back to my concept of an angle between the line and the plane, remember, I was always trying to reach the point when some angle with some line on the plane is less than others. So, that's exactly what I'm going to prove right now. If I have a line and a projection of this line, this is perpendicular and this is projection, then I would like to prove that this particular angle is smaller than an angle which a C forms with any other line CD. Okay. In a way, it's kind of obvious from the visual standpoint because let's say my plane is this board and my line is this one. Then, obviously, this is projection. It's smaller than this one. Visually, it's kind of obvious, right? Okay. So, let's prove it. And by all means, try to prove it yourself first. Okay. So, what I'm going to do is I will drop a perpendicular from A to the line CD. So, this angle is straight angle, right angle. So, this is the right angle because AB perpendicular to the whole plane and that's Y to be C. AE is not perpendicular to the plane but it is perpendicular to line CD. So, triangle ACE is the right triangle and triangle ACB is the right triangle. Now, they have common hypotenuse and the characters AE in the first triangle is obviously greater than the characters AB of the second triangle. Now, why is it? Well, again, if you consider another triangle, ABE, it's also right triangle because AB is perpendicular to BE since AB is perpendicular to the plane, right? And this is the characters. This is perpendicular to this one and this is hypotenuse. So, hypotenuse is longer. So, we actually have that AE is greater than AB. So, forget about this line, we don't need it anymore. We have proven this. And now let's consider a plane geometry problem. If you have two triangles, ACE and ACB, both are right triangles and they have common hypotenuse. But in one triangle, the characters opposite to this angle is smaller than in another triangle. This character is bigger than this. So, I'm actually stating that the smaller characters corresponds to a smaller angle. Now, I would like to flatten this picture. Let's say this is my hypotenuse, this is my AC. This is my AB and this is my AE. So, AE is greater than AB. Now, why did I draw a circle? Well, because, obviously, the locus of all the points, all the vertices of right triangles with the same hypotenuse are lying on the same circle, right? So, we know it from the plane geometry. But now let's think about it. So, if AE is greater than AB, this is the chord, this is the chord and the chord, bigger chord corresponds to the bigger arc, right? So, arc AE is bigger than arc, AE is bigger than arc AB. Arc AE is bigger than arc AB. But arc is actually a measure of either central angle or vertical angle because vertical angle is half, this one is basically half of a central angle which is supported by this, which is this one. And this angle is half of this central angle. Now, central angles are obviously corresponding to arcs, the bigger arc, the bigger central angle. Now, these are halves of the central angles because these are inscribed angles into a circle. So, that's why basically the bigger arc corresponds to a bigger central angle and therefore the bigger half of the central angle. So, that's kind of a very easy plane geometry problem. And that proves that my angle, ACE, is greater than ACB. So, ACB, the angle between the line and its projection, is smaller than anything else. So, what we have proven right now is, number one, the projection of the line is a straight line. It exists, we basically constructed it. And the second, what we have proven is that the angle between the line and its projection is smaller than any other angle with any other line. Well, that actually kind of proves that my initial philosophy about the distance or measurement actually, either linear measurement or angular measurement, between an element and a set, can be actually considered as a minimum among all different distances or angles between that element and elements of the set. In this case, my set is a plane which is a set of all lines and there is a minimum angle. So, it all makes sense. I mean, I could probably start from just defining projection and defining the angle, et cetera, but I think it would be better if I explained why I did all this. And the reason why is to direct the whole thing into a general philosophical concept of a measure, of a distance measure or angular measure between something and something. So, that completes this particular lecture. I suggest you to read the notes again if you didn't do it before and again try to prove these couple of theorems which I have in the notes just by yourself. Obviously, I always recommend you to sign in to register with Unizor.com which opens you for certain procedural mechanisms which are built in this particular website. You can enroll into certain courses or topics. You can take exams. You can take unlimited number of exams. I mean, you can just repeat the same exam again and again until you are really perfect in something. And the site is free, by the way, completely free. Alright, thanks very much and good luck.