>> Here's how vectors are given, are written a lot of times. A vector can be written in its component form, in its form of horizontal and vertical components, you see, like this. Now, notice the notation of this little symbol on either side here. And the horizontal component is R cosine theta. The vertical component is R sine theta. And for our problem, it would be 25, 43.3. Now, notice the horizontal and vertical idea. Gee, this corresponds really well with our notion of ordered pairs and our coordinate plane and all of that kind of thing. So it's very consistent with that and easy to recall. So a vector, in its component form, would be kind of an XY pair here, given by mostly this R cosine theta R sine theta notion. Let's talk about a problem on the coordinate plane. Now, we want to express U in terms of horizontal and vertical components. Now, just looking at our coordinate plane here, we can set up a right triangle configuration and make the calculation very quickly. That is, all we have to do in terms of horizontal and vertical is do this kind of thing, you see? And we have a right triangle formed. And it's real easy just to count here, if we want to, and count here for horizontal and vertical components. Now, the calculations, though, need to be made in a rather special way. We have direction associated with problems like this. And so when we make the calculations of--now, notice for the horizontal value, it's horizontal distance minus this horizontal distance to give us this horizontal distance. So it's 7 minus 2. And it's important to perform the operation in a particular order. We actually use the components of, I shouldn't say components because we use that same word elsewhere, but the coordinates of our terminal point, minus the coordinates of our initial point, you see, in this calculation. So 7 minus 2 for the X value and 5 minus 3, the 5 coming from here and the 3 from here, as the vertical value. So 5, 2. Now, the magnitude--now, let's stay right here for just a second--the magnitude of our result in vector can be found rather easily because we have this right triangle configuration, and we've determined that this side is 5 and this side is 2. Gee, we can use the Pythagorean relationship, you see, to find the distance here. And it would be, let's see, C squared is 5 squared plus 2 squared or C is equal to the square root of 5 squared plus 2 squared. Now, let's generalize that idea. We're dealing with vectors here, and we're talking about the magnitude then of U. So the magnitude of U has this notation. This is the magnitude for U. And it's the square root of the horizontal magnitude, that H we were talking about, squared plus V squared. And in this problem, it's 5 squared plus 2 squared or the square root of 29. Now, what about the angle involved? Well, when we talk about a coordinate plane and we get away from the notion of navigation where we think about angles off of a vertical or a north, south line, we think about angles from the horizontal. And remember that this just corresponds with our study of trigonometry from the beginning where we had angles in standard position and all of that kind of thing. So we're just looking for this angle, you see? And we're thinking of it generated like that. Okay, so we're looking for that angle. Hmm. We have information about opposite side and adjacent side. Now, we calculated the hypotenuse. But gee, with just the given information, we can make the calculation for theta using the tangent ratio. So it is that the tangent of theta is the vertical component over the horizontal component, the Y over X, you see. And tangent theta then for our problem is 2/5. Theta turns out to be approximately 21.8 degrees.