 Hi, I'm Zor. Welcome to Unisor Education. This lecture is about a couple of problems, very simple problems related to vector and vector lengths. Well, I strongly suggest you to try to solve these problems yourself first. They are presented on Unisor.com website. It's a chapter dedicated to vectors, and a particular topic is lengths. I think it's problem number two. So, two easy problems related to vectors. In this case, these are vectors in three-dimensional space. The first problem is the following. You have a vector represented as a tuple. Remember what tuple is? It's just three numbers in three-dimensional system. Which are coordinates of the endpoint of the vector. This is x, this is y, this is z, and this is the vector which has coordinates a, b, c. Which means that its projection on three-axis are correspondingly projection on the x, it's a, which is the same as this. Projection on the y is b, which is this, which is the same as this. And projection on the z is something like this, right? So this is c, and this is c, obviously. So, try to imagine in three-dimensional space. So these two, x and y, are horizontal plane, z goes up. Now, this is the vector which has coordinates a, b, c. And these are projections, basically. This is projection on the x, y plane. And from this point, we have projections to x and y. This is projection on the z-axis, all right? So a, b, and c are given. What's necessary to find? Okay, number one, lengths. Lengths of the vector a, b, c. Okay, l, lengths. All right, so how can we find the lengths? Well, let's consider triangle, all p, q. Now, p, q is a projection on the x, y-axis, which means it's perpendicular to all different lines, including all q. So this is the right angle, 90 degrees. And it's a vertical right triangle. So all p is a hypotenuse. p, q is one calculus and all q is another calculus. We do know p, q. p, q is projection on the z-axis, right? Because that's the projection on the, that's the segment which constitutes the distance from the point to projection on the x, y plane. Which means that the length of the segment is projection to the z-axis. So we know p, q. Now, do we know all q? Because if we do, we can just use the Pythagorean theorem, right? Well, we don't know all q easily. But we do know that all q is a hypotenuse with a and b. This is also the right angle, right? Because from the q, this line is perpendicular to all x. If you imagine this is x, y is a horizontal plane, we just look at this from the angle. But if you look at this from the top, that would be perpendicular, right? So this is the right triangle, all q, r. And in this right triangle, we do know all q, because this is a hypotenuse with two characters, a and b. So all q square equals to a square plus b square. Now, all p square, which is the square of the lengths, which we are looking, lengths of the vector, is all q square plus pq square, which is equal to all q square is a square plus b square. pq is c, so we have a square plus b square plus c square. So the lengths of p l is equal to square root of a square plus b square plus c square, which is absolutely similar to the lengths of the vector in two-dimensional space. Because if you have only two dimensions, let's say this and this. So forget about the bottom of the picture. Then obviously, the lengths is equal to, length square is equal to some of these two squares, right? Because these are two projections, and these are two coordinates. So for two-dimensional case, it will be a square plus b square, three-dimensional a square plus b square plus c square. And actually, if we will go deeper into the higher levels of mathematics, the vectors in n-dimensional space, which have coordinates like a1, a2, a2, a3, an, they do have the lengths, which is defined. Now it's a definition as a1 square plus a2 square plus a3 square, et cetera, et cetera. OK, so that's the lengths. Now next, angle with x's, x, y, and z. Well, these are, again, spatial angles. So try to bear with me. It's angle from vector op, for instance, to vector oz, or ox, or oy. So it's these 1, 2, and 3, these angles. And again, try to imagine it in three-dimensional space. This vector op sticks right in the middle of these three x's. The three x's are perpendicular to each other. Well, what do we know about these triangles? Well, look at this way. If you will connect, what do we have? We have more, let's say, m and n. If you connect p to point r, or p to point m, or p to point n, which we have already connected. Now, it's very easy to understand that these are perpendicular to corresponding x's. So these are basically projections from the point p onto the corresponding x's. q is a projection on the plane x or y. m is projection of the point p onto the y, because that's exactly what coagulants actually are. It's a projection of the point on the corresponding x's. And n is a projection to the z-axis. So what do we know about each of these triangles? Let's say OPR, for instance. Well, we know that this is L is a hypotenuse. And the catatose is 8, because PR is perpendicular to OR. It's projection from the p to the x's. Same thing here. What do we know about triangle? Let's say OPM. Again, OPM is a hypotenuse, the same hypotenuse. And this is one catatose, PM, and another catatose is V, in this case. So all these triangles, we have three triangles. Their properties are, their hypotenuse is L. And one of the catatose is corresponding to A, B, or C. Now, we are looking for an angle which is adjacent to this particular catatose, or an angle adjacent to this particular catatose, or angle adjacent to this particular catatose. So let's go to the two-dimensional picture. So if you have a hypotenuse, which is always L, and one of the catatose, which is A, or B, or C, and you need to know angle 5, which is adjacent, how would you calculate the angle 5? Well, assuming that this particular point is, has positive co-ordinates, A, B, and C, just for simplicity, we can say that these angles are all acute and cosine of phi is equal to A over L, capital L. So from this, we derive very simply that angles, let's call them phi 1, phi 2, and phi 3. So phi 1 is equal to arc cosine of A over L, phi 2 is equal to arc cosine of B over L and C over L. So these are angles between the vector and the corresponding axis. And again, we derive it by projecting the endpoint of the vector onto each of the axes. And the projection point has the lengths A, B, and C. And lengths of the vector, we have already calculated, which is this. So just substitute L into this formula and you will get one angle. And then instead of A put B, and you will have another angle with a y-axis and C for z-axis. And the last one about this particular problem is let's consider, I think we have too many different things on this picture. So let's redo it. Now we need an angle between the vector and its projections not onto the axis, but projections to the planes, including planes. So let's just consider one of them. OK, this is the one. We project the point P to point Q. So this is L. This is A. And this is A. This is B and B. This is the right angle. Now this is projection, so this is also the right angle. And we need an angle between OP and OQ. Basically, it's very similar. What do we know about triangle OPQ? We know the hypotenuse OP. Now this is the right angle. We know one coordinate, which is PQ, which is a C coordinate, the z-axis projection. And then we don't know this particular catatose, but we don't really need it because we have this opposite catatose. So the angle can be calculated as sine of phi is equal to C over L. If this is phi, the opposite catatose towards hypotenuse. And we know L because we have already calculated it. So in this case, well, the only thing is let's use, instead of phi, let's use psi, which means psi 1 is equal to arc sine. In this case, sorry about this, arc sine. Now in this case, it's C over L. And in other cases, it's B over L and A over L. So the angle between vector OP and this projection on xy is arc sine C over L, where L is this. Now the angle between the vector and this projection on, let's say, xoz plane would be the B over L, in this case. And projection on yoz plane would be arc sine of A over L. Well, that completes this problem. So I think all you need here, the problem is very simple. As long as you understand this spatial relationship, which angles are right angles, which angles are acute angles, et cetera. And again, it's easier for me to solve this problem with all positive A, B, and C. So my angle lies in the main octampal, whatever. The opposite quadrant in two-dimensional case, we have octant, I guess, octant. So this is the main octant with all positive coordinates in the three-dimensional space. So we have to understand how this projection actually is working. I don't have these three-dimensional models to show you. So that's why I try to display it on the whiteboard. But basically, that's what it is. Just imagine everything is happening in three-dimensional space. Next problem. Also easy and also requiring certain dimension analogy of your vision. All right, so what do we know? We know about the vector. Well, it's basically kind of opposite problem. In this problem, in the previous problem, we knew the top of representation, A, B, and C. Now we know something else. We know the length of the vector. We know L. And we know these angles between the vector and its projection on each of the plane. So this is projection onto x, y. So let's call this angle gamma. Now projection on this particular plane, it should be something like this, I guess, at this point. So we know this point. And the vector projection is this. So we know this angle. Now this is xz, so we will call it beta. And then gamma, similarly. So what do we know to find out? Well, first of all, we have to know A, B, and C. Well, let's think about it. Again, it's very simple. What is z-coordinate in this particular case? It's the length of this particular catheters, because this is at the right angle. It's perpendicular down to the plane. And this is one of the lines from this point. So the projection of OP would be OQ, we know L, we know gamma. So we know this opposite catheters, right? This PQ, which is z-coordinate, right? Which is C in our representation, is equal to, so, catheters, if we know hipotons, and opposite angle, right? So it's L times sine of gamma. And absolutely similarly, we will get that B is equal to L sine of beta, and A is equal to L sine of alpha. That's it, it's very simple. So all you have to understand is how this spatial arrangement is working. All right, next. And what are the lengths of three projections on the planes? OK, so now it's asking, what are these lengths of projections? OQ, for instance. Well, but again, we know everything about this. We know this catheters and we know this catheters. So the projection OQ is equal to L cosine of gamma, right? This is catheters, this is hipotons, this is an angle adjacent. So the cosine of this angle is this over this. So this particular line, this particular segment, catheters is equal to hipotons times the cosine. And similarly, all other, the projection, the lengths of the projection on the, let's say, xz would be L cosine beta and the correspondingly L cosine alpha. So are these problems difficult? No, but let me repeat again. You really have to understand how it's all arranged in three-dimensional space. Well, other than that, I do suggest you to go back to the notes for this lecture. And again, they are unizord.com in the chapter topic called lengths. And just do it yourself. Try to draw your own picture of illustration of these problems and get the same results. I think it's a very useful exercise. And it helps you to visualize how everything is done in three-dimensional space. That's it for today. Thank you very much, and good luck.