Find the vector v having length |v|=8 and direction theta=2pi/3. Write answer in component form.

Question

ipwnbunnies · Answer

$$x-coordinate = |v|*\cos(\theta)$$
$$y = |v|*\sin(\theta)$$

emilyjones284 · Answer

So y=8 times sin(2pi/3)?

johnweldon1993 · Answer

^ Correct

emilyjones284 · Answer

And then y=8 times sin square root of three/2?

whpalmer4 · Answer

$$y = 8\sin(\frac{2\pi}{3}) = 8*\frac{\sqrt{3}}2=$$

whpalmer4 · Answer

$$x= 8\cos(\frac{2\pi}3) = 8*(-\frac{1}{2}) = $$

emilyjones284 · Answer

So 8 square root of three/2 and -4?

whpalmer4 · Answer

You can simplify the first one a bit more, can't you?

emilyjones284 · Answer

would it be $$4\sqrt{3}?$$

whpalmer4 · Answer

Yes.

emilyjones284 · Answer

Oh alright got it... and then how would I put all of that into component form?

whpalmer4 · Answer

Beats me, but I'm sure your book has an example of something in component form, doesn't it?

Your two components are $4\sqrt{3},-4$, that's all I know for sure.

ipwnbunnies · Answer

It's the x and y component of a coordinate. (x,y)

whpalmer4 · Answer

I'd hate to be responsible for you getting a problem that you've correctly solved marked wrong because I told you the wrong way to format the correct answer...

ipwnbunnies · Answer

If you graph that point, and the vector at the same angle, it'll be at the same spot.

emilyjones284 · Answer

okay no problem thank you both!

whpalmer4 · Answer

I think, but do not promise, that it is just <x,y>

whpalmer4 · Answer

$$<4\sqrt{3},-4>$$

emilyjones284 · Answer

Okay yeah I think you're right, thanks

ipwnbunnies · Answer

Oh, right. You're writing the vector, not the point. My bad.