 Hi, I'm Zor. Welcome to a new Zor education. We continue talking about vectors. We just recently covered the topic lengths of the vector, and we also visited the different representations of the vector, like a tuple or like a geometric some kind of a segment with an arrow. I mean, that's also representation. So after that, let's just solve a few problems related to different representation of the vector and its lengths and how it's positioned in the space where it acts. So these few problems are very, very simple. Obviously, I do encourage you to go to Unisor.com, go to Vectors topic and in the lengths lecture, under the lengths lecture, there are these problems listed. So try to do it yourself first and then listen to this lecture. Okay. So problems are really simple. So don't think it will be any challenge or anything like that. But still, it needs to be done just to make sure that you perfectly understand all these simple concepts. Okay. Problem number one. It's just a conversion from one representation to another. So let's consider you have two-dimensional plane with coordinates, and you have a vector originated at the beginning of the coordinate origin point. It has a length L and the angle phi with the positive direction of the x-axis. So this is a geometric representation. You have a length and you have a direction relative to some chosen direction. Now, what is required is to convert this particular representation into a table representation. Now, table representation is basically the coordinates of the endpoint of this particular vector in this coordinate system. So all I need to do is I have to find coordinates of the endpoint of this vector, and that will be my table representation of the vector. Obviously, coordinates are this is a, this is b, this is b, the same thing. All we need to do right now is to use the plane trigonometry of the right triangle. The right triangle is obviously this one. The vector itself plays the role of a patternus and coordinates a and b are basically two catechic. So a is equal to L cosine of phi and b is equal to L sine of phi. So this is a conversion from the geometric representation when the vector is represented by its length and direction relative to some chosen direction angle from the chosen direction. And a and b are the symbolic representation in the table form which is related obviously to coordinates. And in this particular case, what's important is that the horizontal x-axis, which is used as the one from which we actually tried to deviate in the directional movement, and it coincides actually with the x-axis of the coordinate system. So that's explicitly used because if for instance this angle phi is angle from some other direction, then we have to do some angle manipulation and conversion. But this is a simple case. That's all it's needed. Now, second problem is very much related to this. So let's consider that again, we have a two-dimensional plane and there is some point a, b where an object is located at certain time. But it's not just sits in this position. It actually moves with a constant speed and the speed is again equals to L which is represented by the length of this vector. And there is also a direction between horizontal line and the direction of the movement which is phi. So my task is to represent this vector as a tuple. Now, I would like to recall the fact that the vectors are characterized by two things, lengths and direction, which means this vector which also has a length phi and which is an angle phi and the length L. It's exactly the same vector. These two vectors are equal, congruent or whatever you want to say. Which means that initial position, the beginning of the vector has absolutely nothing to do with its tuple representation. Tuple representation of this vector and this vector are exactly the same because these are two vectors which have exactly the same lengths and exactly the same direction. So all vectors, and anywhere else, wherever I put, as long as they have the same lengths and the same direction relative to horizontal line, these all vectors are exactly the same. They are equal to each other, they're congruent to each other, and their tuple representation is exactly the same, which is position of the endpoint of this vector if it's originated at zero-zero. We know this position is equal to L cosine phi and L sine of phi, as we did in the previous problem. So these two, L cosine phi and L sine phi, are the coordinates of the endpoint and the tuple representation of this particular vector regardless of the beginning. So this is extraneous condition which we don't really need. It doesn't matter whether the point is at A, B, or anywhere else on the plane. All vectors with the same direction and the same lengths are the same. Three. Okay. Given vector in its tuple representation, minus 6, 6. So this is an endpoint of the vector originated at zero-zero. Going towards this direction. So question is, what are its lengths and angle with the positive direction of the x-axis? Which means this end. Okay. Now, that's easy. The lengths, you can use this right triangle. Now, minus 6 and 6, it means that basically the absolute value of these two lengths of these two categories are the same. This is 45 degrees. 45 degrees. So this is 135 degrees, which is what? 3 pi over 4, right? And 45 is pi over 4. So we know the direction which is this. This is an angle, pi. And the length is basically right triangle. If you go in theory, this is 6 and this is 6. So you have 6 square plus 6 square, which is 6 square root of 2, right? So this is L and this is pi, end of story. So mind you that I'm using the angle from the positive direction of the x-axis towards the direction of the angle. I mean, that's basically something which is sufficient to define the direction of this vector, right? I can probably start from any other point. I can start from the vertical point, in which case it's 45. Doesn't really matter. Traditionally, we do it from the x-axis. As you see, all problems are very, very simple and very basic. They are just based on definition of the vector and its length, et cetera. Now, similar problem, different coordinates. So it's 6 minus 6. This is the vector. So what is the angle? It's either this one or this one. This is positive direction. This is negative direction. Well, obviously, it's easier to use negative direction. It's obviously minus 45 degrees. So phi is equal to minus 45 degrees. Or minus phi over 4 in radians, right? This angle, phi. So negative angle minus phi over 4 is exactly the same as this one. And as far as the direction, now, as far as the length is concerned, it's exactly the same length as in the previous case. Because the lengths of these two cateches are 6 and 6. So you have L is equal to 6 square root of 2. Easy. Let's say we have two points. One point is A, B. Another point is C, B. Now, we have an object which moves from here to there with a constant speed, S. So I have S and I have A, B, and C, D. Let's assume for simplicity that A is less than C, B is less than D, which means both coordinates are increasing. That's easier this way. So I don't have to worry about negative things. So question is, what exactly are the characteristics of this particular vector of speed or velocity or other vector of velocity, which is the same in every point on its way, because I said that the S is constant. I hope I said it. If I didn't, I'm saying it right now. So S is constant. So it moves with a constant speed from this point to this point. Now, let's refer back to the first problem. I gave the lengths of the vector. Well, S is basically the lengths of the vector. That's the absolute value of the speed. And I was also using the angle, the direction. And from these two, L was the length. Phi was an angle. I derived L cosine phi and L sine phi as tuple representation. So this is the geometric representation. This is a tuple representation. And that's how they're related. Now, I don't have exactly the same as in that particular problem. I do have lengths of the vector, which is, in this case, letter S. But I don't have a direction. Well, I don't have direction in explicit form, but I have two points which basically define the direction. So how can I determine this angle? Well, it's easy because this is A, this is B, this is C, and this is D. So what do I know about this particular triangle? Well, I know this and this. These two catch-a-tree, I do know because this vertical one is the difference between D and B. And that's where I'm using the D is greater than B. So this is positive. It's easier for me this way. Now, this catechus of this right triangle is obviously C minus A. Tangent of this angle is the ratio between the opposite catechus towards the adjacent one. The opposite is D minus B. Adjacent is this. Now, considering these are all positive, and as I was saying, I'm looking for this particular case, I can resolve this as this. So if tangent of some angle, in this case, it's an acute angle. The tangent of the acute angle is this. Then angle itself is the inverse to tangent function called r-tangent, which I'm sure you remember from the trigonometry course. If you don't, just go back to Unisor.com. There is a trigonometry course where I explain what arc tangent is. So knowing this, I know the phi. I know the angle. And from the angle, I can define cosine and sine. And if you want, you can obviously make some kind of a transformation from cosine of the angle tangent of which is equal to this. You can find what is the cosine and sine using very elementary trigonometry formulas. Considering the angle is acute, you will not have any problems with square roots, et cetera, et cetera. So basically, that's it. All I wanted in this particular problem, in addition to whatever I did before, was to define the angle phi if I have end points of the direction. Because these end points do characterize the angle fully through its tangent in this case. Well, that's it for today. I do recommend you to do exactly the same problems just by yourself if you didn't do it before. Problems are very easy. Go to Unisor.com. And I do encourage you to get registered as a student, have somebody else registered as your supervisor or your parent, and enroll you in classes. And that will enable you to take exams. You can take any number. You can take exam any number of times until you will get it perfect, actually. So that's the purpose. And try to solve all these problems again yourself. That's very important. The whole course is oriented towards solving problems rather than just explaining certain simple things. So these problems are very important just to understand different representations of the vector. They're not difficult to solve. Thanks very much, and good luck.