 Let's talk a little bit about vector components. I have a vector. I have represented graphically on a piece of paper. Actually, it's over here on this piece of paper, but I'm not going to show it to you quite yet. It has a magnitude and a direction, and I want you to draw my same vector on a piece of paper of your own. Well, right now you're saying, well, wait a minute. How do I draw your same vector? I need some information. Well, I'll give you a little more information. It's on a piece of paper that's in two dimensions, so this is a two-dimensional vector, so I will need to give you two pieces of information. So let me give you a direction. The direction of the vector is 30 degrees north of east, and the magnitude of the direction of the vector is 5 miles. So 30 degrees north of east, and the magnitude is 5 miles. Go ahead and draw your vector. So let me see. Let's see if your vector looks like mine. Is this the vector you drew? Probably not. It's unlikely that you drew a vector that looks exactly like mine, because we probably didn't choose the same basis. I drew my vector by establishing a basis. I decided that east was going to be in this direction for my particular picture, and I realized that north and east, if this is east, then that direction must be north. So I drew 30 degrees north of east. So I established a basis that was up vertically on my piece of paper, and this angle is being north. I decided it was going to be counterclockwise from this direction. And then we set a magnitude of 5 miles. Well, how do I represent 5 miles? I haven't really established what 5 miles is in here. Looks like I might need to go ahead and draw them on my basises. Let's see here. I probably need to figure out about how long this is. There's about halfway, and divide it into five roughly equal parts. One, two, one, two, three, four, and five, where each of those parts is going to represent one mile. Notice yours might be a little bit longer or a little bit shorter than mine. However, as long as we've established what our basis is, this is a mile, this is the direction I'm measuring off of, the axis I'm measuring off of, we're effectively representing the same thing. So yes, your vector is the same as mine. No matter how you drew it, as long as you give me the information about what the basis is, how you got the direction from an established direction, and how you got the magnitude based on some sort of scale of units. Let's sort of reconsider this vector. Well, let's go ahead and write. I'm going to turn this here, and I'm going to write down what the basis was. So my basis, in this case, was an angle from east, whatever we define as being east, and a distance. That was my basis. And my components ended up being some angle, and r will use as representing a radius or a distance. My components for this particular example were 30 degrees angle from whatever I define as east, and 5 miles is my distance. Well, that's one way that we can represent a vector. However, if we want to do math with vectors, we actually have to use a different form of basis. Let's consider this same vector in an orthogonal basis or a Cartesian basis. In other words, where our basis is two perpendicular number lines. For example, we could use the number lines east and north. Let me try that. There's east, and there's north. And we know we still have the same information we had before, where this was 5 miles in magnitude, and we have 30 degrees. But now, instead of me identifying it with losing those two pieces of information, instead, I want to identify it in terms of my basis of distance east and distance north. I'm going to go ahead and label those. I'm going to use the algebra convention. We're going to call this our x-axis and this my y-axis. And if I do so, I can start, or I can recognize that I have a right angle here, which means I have a right triangle. That's a right angle, and this is a right triangle. I can start using some of the information I remember from a trigonometry class to relate this distance x here and the distance y here to my original components, theta, the angle, and r, which is the distance of that hypotenuse there. So if you remember our trigonometric identities, if you remember, let's see here. What was the mnemonic that people are often using? Recalling, s-o-ca-to-a. Hopefully, you've seen that in a trigonometry class. The sine of an angle is the ratio between the opposite side and the hypotenuse. The cosine is the adjacent side and the hypotenuse. And the tangent of the angle is the ratio between the opposite side and the adjacent side. So if I use that information, I know that the sine of my angle, and in this case, we're going to call there, here's my angle theta. The sine of the angle has to be related to the opposite side over the hypotenuse, which we were calling r. So my sine of my angle is equal to y over r. If I do a little math, multiply both sides by r, I can also say that my component y is equal to r, that's my hypotenuse, times the sine of the angle. Now I have a relationship that tells me how to get y if I have r and theta, which I do. Similarly, the cosine of theta is the adjacent side divided by the hypotenuse. So the adjacent side is component x over r. Again, doing some algebra, x is equal to r cosine of theta. And because I don't want the tangent to be lonely here, let's go ahead and use this relationship between the tangent. The tangent of the angle is equal to the opposite side, there's the opposite side over the adjacent side, the opposite side being y, the adjacent side being x. Well, in this particular case, that doesn't help us find either y or x, because they're both in the equation together. But it does give us a relationship between the angle and the two components, that if we take the inverse tangent function of this y over x, if we happen to have y and x, we could then find the angle theta from that y and that x. So we have a relationship that allows us to find y, given the two components in our original basis. We have a relationship that allows us to find x, given those two components. And we have a relationship that allows us to move in the other direction that would allow us to find theta given y and x. There's one more relationship that would be useful. If we use the Pythagorean theorem, Pythagorean, we can notice that x squared plus y squared equals r squared, or a squared plus b squared equals c squared. Again, c is the hypotenuse. x squared plus y squared equals r squared. If I do a little algebra there by taking the square to both sides, I can say that r distance r, the magnitude of our vector r, is equal to the square root of x squared plus y squared. So now I have four relationships, two of them that allow me to take my current basis and find the new basis, and two of them that would allow me to go in the other direction, that if I had a new basis, I could find something in my current sort of basis. So my new orthogonal slash Cartesian basis, my new basis is distance due east and my distance due north. And in that case, x and y are my components. And if we use the relationship that we just talked about here, we know that that would be r cosine theta and r sine of theta, or 5 miles times the cosine of 30 degrees and 5 miles times the sine of 30 degrees. If I do a little math there, I get 4.33 miles and 2.5 miles. In other words, I can locate this same point by going out 4.33 miles and up 2.5 miles. Those two pieces of information, along with the basis, with me drawing the east and north, will allow me to again replicate that same vector. Let's try this in just one more basis. Notice there are multiple ways I can describe this vector because there are multiple choices that I can use for my basis. Let me trace the vector here. But this time, instead of my basis being east and north, I'm going to let my basis be southeast, which is a 45-degree angle from east and northeast, which is also a 45-degree angle. And if we remember, our angle itself was 30 degrees north of east. Well, if we think about this new sort of system, we can rotate it and think about it. If we look, we can still see a nice right angle. However, we have a slightly different angle to use for our relationships. In this case, the angle we have here is the combination of 30 plus 45. We have an angle of 75 degrees. The magnitude, however, has not changed. We still have the magnitude of 5 miles. Well, in this new basis, if our basis instead of east and north happens to be distance southeast and distance northeast, then our new components are going to have an x in the southeast direction and y being the northeast direction. I'm just putting a little label on there. Would be r cosine theta. But this is a different angle. So I'm going to just give this a number, we'll call this number 2, and r sine of theta 2, where theta 2 is this 75 degrees. It's a different orientation because our basis is oriented differently. If I plug those numbers in, 5 miles times the cosine of 75 degrees and 5 miles times the sine of 75 degrees, I get 1.2 miles and 4.82 miles. So if I instead choose my basis to be along these lines, I have a different set of components. The vector itself has to be represented by both the basis, a description of what you're using, and the components in that basis. Now you might be asking yourself, well, why in the world would we ever want to orient it in a basis that's strange like this? Well, many of you live in different cities, and the cities will often be oriented in something like a grid. But there's nothing that says that that grid of roads and streets is always going to be perfectly oriented north and east. Some cities are, but many times the grid might be determined by the location of a river that's running through something. And they would, well, more likely, they would align it with something like the river itself. Let's say the river is running through the city in a northeast to southwest direction. Well, then it's likely that the grid of the city would follow the river as opposed to being somewhat arbitrarily located in a north-south direction. So now you have sort of a reason for using your trigonometry. It's our primary means of converting between components of vectors in different bases.