Hello, class, Professor Anderson here. Let's talk about adding vectors using their components, and see if this is a perhaps simpler way to attack some of these problems. I think you'll find that it is. So let's say we have the following. Let's say we have vector A. We can write that as A sub X I hat, plus A sub Y J hat. Now, we have vector B, which is B sub X I hat, plus B sub Y J hat. Those are some numbers that we would put in there, but we don't have those numbers yet. And let's calculate C, where C is the sum of those two. How do I deal with this mathematically? Well, we can just plug these things in, right? C is equal to A, which is this thing, AX I hat, plus AY J hat, put some parentheses around them, plus the vector B, which is this thing, BX I hat, plus BY J hat. And now those parentheses are irrelevant, right? We don't need the parentheses anymore, and so we can rewrite this as the following, AX I hat, plus BX I hat, plus AY J hat, plus BY J hat. And now look what happens, we have common terms in the first two, we have common terms in the second two. And so we can pull them together and we get the following, AX plus BX I hat, plus AY plus BY J hat. Okay? And this is the great thing about vector addition, all you have to do is figure out the components, and then you just add up all of the components, add up all of the X components, keep it separate, add up all the Y components, keep it separate. And in fact, you could do this for any number of vectors. We just did two here, but you could have three, or ten, or 100 vectors, and all you have to really worry about is the components. It's a very nice approach. So let's take a look at two things here. Let's ask the following question, "What is C equal to, and what is the angle theta equal to?" Right; if we are in polar coordinates, we want to know the magnitude, which is C, and we want to know the angle relative to the X axis, which is theta. So how do I do it? Any thoughts? How would we get C? Yes. >> (student speaking) Well, since C's the length, wouldn't we do Pythagorean? >> We would do Pythagorean Theorem, absolutely. What do I want a square to add up? >> (student speaking) You want would to square I hat and J hat. >> Okay. I hat squared is just one, J hat squared is just one. So we don't really have to worry about the I hat and the J hat, but what we do have to worry about is the stuff that's in front of the I hat and the J hat, okay? So let me ask you a question, is this the right answer? Is that the right answer? Is that the magnitude of vector C? Show of hands. Who thinks it is the right answer? Who thinks it's the wrong answer? Who is still awake out there in that dark room? I can barely see you guys. Okay. It's the wrong answer. Why is it the wrong answer? What have I done wrong? Yes; what do you think? >> (student speaking) Add those components before you square them. >> Add them before we square them, absolutely right. The X component for C is AX plus BX. So I need to add those first, and then square it. The Y component for C is AY plus BY. I need to add those first, and then square it, okay? And then, you know, you're going to foil it out, and you're going to get some extra terms in the middle, all right? So you've got to be careful about adding them first, and then squaring them. All right; and now we got it right, we can take off our little question mark. What about the angle theta? How am I going to calculate theta for this vector C? Somebody that hasn't answered yet. Raise your hand and tell me what you think. Whether it's right or wrong, don't even worry about it. How should I calculate theta? Anyone? Yes. >> (student speaking) Using sine or cosine? >> Using sine or cosine; all right. So let's say we use sine, right, what should I put inside the argument? What will go in there? >> (student speaking) Opposite over hypotenuse? >> Yes; opposite over hypotenuse. What is the opposite? Yes. >> (student speaking) Four? >> Four. >> (student speaking) No, wait, no sorry. I was totally looking at something else. Never mind. >> It could be four. I haven't given you any numbers yet, but it could totally be four. [ Inaudible Comment ] There's this old joke in physics, the answer is three, you just have to pick the appropriate units. >> (student speaking) Yes. [Laughs] >> Yes, I know, physics humor, can't get enough of it; I understand. All right. So R sine of something opposite. What are we missing here? What we're missing is a picture. Right; we don't really have a very good picture of this thing. So let's put a picture together, okay? We don't know exactly what A is, we don't know exactly what B is, but what we do know is that C is going to be some vector in this XY space. And that vector is going to have a CY component and a CX component. And it's going to be at an angle theta. The CX component is just this stuff right here, AX plus BX. The CY component is going to be other stuff right here, AY plus BY. So what is the sine of theta? You said sine of theta is opposite, which would be this, right, C sub Y over hypotenuse, which is regular old C. Right? And now you know what that stuff is; you can plug it in. CY is this stuff right here, AY plus BY. C is this big long thing here, the square root of all this stuff. Okay; so at that point now, if you have numbers, you can plug it in, and maybe you get four or maybe you get something else. Who knows? All right. I don't think that was clear at all, but if it's really bad, come see me in office hours. Otherwise, cheers.