Trigonometry: Finding the Components of a Vector

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Finding the Components of a Vector
Let's see if we can actually find the components of a vector if you're just given in some sense the directional angle and its magnitude, so let's go backwards. Before we gave the coordinates or the component of the vector and we found the magnitude and the directional angle, but let's now do it the other way just to see that in fact you can go back and forth, back and forth. Not a problem.
So suppose we have a vector that comes out like this. This angle here is 30° and the length of the vector, the magnitude, is 20. How would I write this vector? It's , let's say. It equals <>. What are and ? Well, is just this length, and is this length. So how would I find that? Well, I'd use some sort of trig stuff. Now you can see another application of all this trig stuff in the vectors things. So for example, if I wanted to find , what would I do? Well, that's the adjacent to this 30° angle, so that would be "adjacent." And so if I look at, for example, , that would equal adjacent, , over hypotenuse. Now, what's the cosine of 30°? Well, I happen to know that's . So what I see is . Well, now I can solve for . Multiply through by 20 and I see . So there's the component.
What about the component? Same business but now I use "opposite" so I should use sine. So is opposite, which is , over the hypotenuse, which remains 20. And what is ? Well, that's , and so what do I see? I see that that forces to equal 10. So what's this vector ? Just knowing its angle and its magnitude, I know I can write it as <>. So that's the answer. By the way, just to sort of muck up the works a little bit, notice that's just the scalar 10 multiplied by the vector <>. Just a tiny little aside; no big deal. Just to remind you how scalar multiplication works, I'd multiply the 10 by both terms and I would get this. This answer is preferable. I'm just trying to illustrate how scalar multiplication works here.
Okay, let's take a look at one more example just to see--well, maybe I'll look at more than one. Maybe I'll look at five, ha! You think, "Maybe he's going to do one more lecture," but you don't know. I could do as many as I want. Don't try to second-guess me! Well, you can try, but don't get mad if you're wrong.
All right, here we go. How about this vector? It goes way around, way around, and it comes out. It has a length of 5, and this angle here is 225°. Name that vector. Well, let's call that vector . It has an and a . Now in fact, if this is really going to be correct, the and should both be negative. I think you can see that because I go and then I go down . So now, what's the reference angle here? Well, this whole thing is 180, and then this entire angle is 225, so what's this little reference angle? Well, that reference angle is actually 45°. So if it's 45°, then where am I? Well then I have actually an isosceles right triangle and I could either use a little bit of trig stuff or I could just use a little bit of the Pythagorean Theorem. Let's use the Pythagorean Theorem and see what goes on here.
So if this side is s, then this side would be s because both angles here are 45°, which means both these sides are equal. And now I use the Pythagorean Theorem. I know the hypotenuse, so I see . Well, that's . , which is 25. So that means that s would be--and s is going to be this positive length--it will be the positive square root of , which is . So that's s. So therefore, what are the coordinates of ? Well, there's , and notice is in this negative direction, so it's , comma. And is also negative, so . Notice that we have to be careful of those negative signs because that angle tells me whether I should be negative or positive. So in this case we see that this completely expresses this vector, in vector form.
Okay, now I'll do the last one. There was a little more, you see. How about this one? Just a little angle, no big deal. This angle here is 20°, modest. And the length of this, the magnitude, is 10. So we know that we're going to have this as , and is right here. And so the vector is <>. That's how we write it.
Now I'm going to use a little trig to figure this out. How could I figure out the ? Well, I'll use cosine. So I'll see that . So I see that . Now that 20° is not one that I know, even though it sounds like, "Oh, 20°, we should know that one." Well, I don't. Maybe you do but I don't know that one off the top of my head. So I type in 20 and make sure it's in degrees, and multiply by 10, and I see that . All right. And now how do I find ? Well, I'll use now sine, since that's the opposite side. So should equal opposite, , over hypotenuse, 10, and so I see that . And so what does that mean? That means that equals--and you could do the same thing here. You take 10 and you multiply it by and you get--oh, hold on a second, folks. I made a huge typo here. I typed in sine on the calculator, so in fact this answer is 3.4202. Here if you type in cosine, let's do that again; 10 times cosine--sorry about that; I typed the wrong button--20--ah! This turns out to be 9.396.
And actually, we would have figured that out for ourselves, because notice that this is 20° and this is a right angle, so what's the angle of this? Well, that's got to be 70° because the whole thing has to add up to 180. So 70°, this should be pretty long and the 3 was too small. This should be 9, which sounds right, and this smaller angle of 20 is opposite the small side of 3, so that makes sense. So in fact the answer is, what's the vector? The vector is <9.3,3.4>. So in fact you can either use a calculator to find out how to express this in standard form as a vector, and you can do that using a little bit of trig. Not a big deal. Okay, try these and see what you think. I'll see you soon.