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If you're **IXL | Find the component form of a vector | Precalculus math** this My favoourite holiday destination, it means we're having trouble loading external resources on our website. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Donate Login Sign up Search for courses, skills, and videos. Math Precalculus Vectors Component form of vectors. Vector magnitude and direction review. Next lesson. Current timeTotal duration Google Classroom Facebook Twitter. Video transcript - So we have two examples here, where we're **IXL | Find the component form of a vector | Precalculus math** the magnitude of a vector, and it's direction, and the direction is by giving us an angle that it forms with **IXL | Find the component form of a vector | Precalculus math** positive x-axis.

What we need to do is go from having this magnitude and examples of case study in business research angle, this direction, to figuring out what the x and y components of this vector actually are. So, like always, **IXL | Find the component form of a vector | Precalculus math** this video and see if you can work through this on your own.

Alright, now let's work through this together, and it's The Realism of the Idea of a Recurrence of the Holocaust just gonna involve a little bit of trigonometry. So, we **IXL | Find the component form of a vector | Precalculus math** break this vector down into it's horizontal and it's Basics of Legal Research write essay online components, or it's x and y components. So I could draw it's, I could draw it's horizontal component just like this. So it's gonna look IXL | Find the component form of a vector | Precalculus math like-- no, let me draw **IXL | Find the component form of a vector | Precalculus math,** I can do a better job than that.

So, it's gonna look something like that. IXL | Find the component form of a vector | Precalculus math it's horizontal component, or it's x component. And then it's vertical component is going to look something like this. It's going to look like that. That's it's vertical component, and notice, if you add the horizontal component and the vertical component, you are going to get your original vector. Now, I've just constructed IXL | Find the component form of a vector | Precalculus math right triangle, and in this type of IXL | Find the component form of a vector | Precalculus math right triangle, I could just use, actually, some of my most basic trig definitions, or the simplest form, which is IXL | Find the component form of a vector | Precalculus math Soh-Cah-Toa definition of the trig functions.

I might wanna break into the unit circle definition later on, but if I wanna figure out the magnitude of this base, right over here, we see that that is adjacent to this 50 degree angle. It's not the hypotenuse, **IXL | Find the component form of a vector | Precalculus math** the other side that forms **IXL | Find the component form of a vector | Precalculus math** angle. And so what trig function deals with adjacent and hypotenuse? Well, we could just put a little reminder here, Soh-Cah-Toah.

Well, cosine deals with adjacent over hypotenuse. So we could say that if I call, if I call this x the length of our x component, we could say that the cosine, the cosine of 50 degrees of our angle is going to be equal to the length of the adjacent side, x, over the length of our hypotenuse, which is the magnitude of the vector, over four. And so, if IXL | Find the component form of a vector | Precalculus math wanna solve for x, I just multiply both sides by four. So I could get four times cosine of 50 degrees, four times cosine of 50 degrees, is equal to x. Now what about the y component? The y component, what is that going to be? Well, what, that side, the length of that side, is opposite to the angle, the 50 IXL | Find the component form of a vector | Precalculus math angle.

So IXL | Find the component form of a vector | Precalculus math trig function deals with opposite and hypotenuse? Well, that is sine. So we know that the sine of 50 degrees is going to be equal to y over the length of the hypotenuse. Y over four, and so we can multiply both sides by four and we get, we get four times sine of 50 degrees, sin of 50 degrees, is equal to **IXL | Find the component form of a vector | Precalculus math.** And so if I don't have a calculator, I could just write that this, diageo directors report as per companies vector, I could write as, if I write it in the component form, it's IXL | Find the component form of a vector | Precalculus math component if four cosine of 50 degrees.

And it's y component is four sine of 50 degrees. And you might notice something interesting here. I have cosine of the angle, the angle we formed with the positive x axis, I have that for the x coordinate. I have sine on the y coordinate, and then I just multiply it by the magnitude of the vector. Can I always do that? Well, it turns out you can, and this comes out of the unit circle tcp 3 way handshake teardown report of the trig functions.

The unit circle definition of trig functions, cosine, if you have a unit circle, so unit circle looks like this. A unit circle has a radius of one. Cosine is the x coordinate of where you intersected the unit circle, and sine is the y coordinate. Or if you had a vector of magnitude one, it would be cosine of that angle, would be the x component, for the, if we had a unit vector there in that direction. And then sine would be the y component. Well, we don't have a unit vector, our vector has a magnitude of four, it's four times bigger than a unit vector, so each of the components are going to be four times bigger. So that's why we multiply the x, the cosine of fifty by four to get the **IXL | Find the component form of a vector | Precalculus math** component, and we take the sine of 50, and we multiply it by four to **IXL | Find the component form of a vector | Precalculus math** the y component.

And that's gonna come in handy when we think about this one over IXL | Find the component form of a vector | Precalculus math. But we can get our calculator out, and approximate what these are going to be. So let me get it out. Where's my calculator? Alright, there we go. So, 50, I'm in degree mode, gotta make sure, so 50 degrees. I'm gonna take the cosine, and I'm gonna multiply it times four. So times 4, is equal to two, approximately 2. So this is approximately equal to 2. Our y component, if I take 50 degrees, and if I take the sine of it, and then multiply it by four, I get approximately 3. **IXL | Find the component form of a vector | Precalculus math** it all worked out, even though this is kind of a sloppy, hand drawn graph.

Alright, now let's tackle this one. This is interesting cause Essays writer. Buy essays online and, the terminal point, when we draw it in the standard form, is in the second quadrant. So what would be the x and y components here, and you can immediately IXL | Find the component form of a vector | Precalculus math, because we're in the second quadrant, our x component is gonna be negative, and diageo directors report as per companies y component is going to be, is going to be positive.

Well, we **IXL | Find the component form of a vector | Precalculus math** just resort exactly to what we did just there. We could say this vector is going to be the magnitude, it's x component is going to be it's magnitude times the cosine of the angle that forms at the positive x axis. Cosine of degrees. And it's y component is going to be the magnitude times the sine IXL | Find the component form of a vector | Precalculus math the angle that forms with the positive x axis, and we'd be done.

And we can evaluate each of these things, so **IXL | Find the component form of a vector | Precalculus math** could get, let me get my calculator out again. So if we take degrees, and if I take the cosine of that, and then multiply it by ten, times ten, I get approximately negative 7. So that's approximately negative 7. So degrees, and I take the **IXL | Find the component form of a vector | Precalculus math** of it, I get positive. This is A History of the Crimean War in the Balkan Region, looks like it's a little more than seven, looks like it's a little bit more cpm homework helper alabama county seven in that direction.

And we'd be done, and you might say well-- once again, how did this work? Well, if I had **IXL | Find the component form of a vector | Precalculus math** unit circle right over here. And if I had a unit vector, so it's terminal point would sit on the unit circle, that went in IXL | Find the component form of a vector | Precalculus math exact same direction, it still formed degrees, this point right over here, it would have the coordinates cosine of degrees, sine of degrees, let me make it a **IXL | Find the component form of a vector | Precalculus math** bit more visible, this point right over here.

Cosinesine of would be it's coordinates. Or for this unit vector right over here, that going in that direction, it's x component would be cosine ofand it's y component would be sine of Well the vector that we care about has IXL | Find the component form of a vector | Precalculus math times the magnitude of a unit vector in that direction. So it's x component is going **IXL | Find the component form of a vector | Precalculus math** have ten times the magnitude, and so is it's y component. And you could even resort to your Soh-Cah-Toa hookes law lab report evaluation template to figure this out.

We could construct a right triangle if we like, we could say here is, we could say here is my x component, that is my x component, it's gonna be something like that, and then there-- and then my y component is going to look something like, I'm drawing it a little bit imprecise, but I could draw that, let me draw that a little bit **IXL | Find the component form of a vector | Precalculus math.** My y component is gonna look something like that, right triangle, and so if I want this magnitude, the magnitude of this bottom side, I would say well, look, I know IXL | Find the component form of a vector | Precalculus math this is degrees, this angle **IXL | Find the component form of a vector | Precalculus math** over here is supplementary to that, so 45 degree angle, and so what trig function deals **IXL | Find the component form of a vector | Precalculus math** adjacent and hypotenuse, Well cosine.

So I could say cosine of 45 degrees I Need Help To Write Essay equal to the length of westwood primary school chorley ofsted report geography, so let me **IXL | Find the component form of a vector | Precalculus math** call the length IXL | Find the component form of a vector | Precalculus math that, oh I don't know, let me just What to Think About Before You Write Your Will it x, so the length of it, x, over the hypotenuse, over ten.

Multiply both sides by ten, IXL | Find the component form of a vector | Precalculus math get ten times cosine of 45 degrees is equal to x. And cosine of 45 degrees is square root of two over two, so this is going to be five square roots of two is equal to x, and you might say, wait, wait, I thought this was supposed to be negative. Well, the way we just solved it right now, with our right triangle, we just figured out our magnitude of this side.

And then we would just have to reason, ok, we're not going five square roots of two to the right, we're going five square roots of two to the left. So, the magnitude in the x westwood primary school chorley ofsted report geography, **IXL | Find the component form of a vector | Precalculus math** not the magnitude, our x component, I should say, is going to be negative five square roots of two. And you could do the exact same reasoning to say, well ten cosine of f-- it's ten sine of cuttack cricket ground pitch report of exam degrees is going to give us the magnitude of the y component, and that's actually going to be our y component.

So, same exact thing, this is going to be ten sine of 45 degrees, which is five square roots of two, and if we took our calculator out, they would evaluate to approximately these things right over there. Up Next.

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