Question

Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j.
absolute value of v =50, theta=120 degrees

Answer 1

x component : \(r \cos\theta\)
y component : \(r \sin\theta\)

Answer 2

\(r\) is the magnitude of vector

Answer 3

wait so how do you get the answer?

Answer 4

is that the answer?

Answer 5

I don't know why my \(\theta\)'s came out like Q's in the drawing.

Answer 6

so is that the answer or...

Answer 7

pleease help me

Answer 8

Here's how you get the x and y components of any vector:

x component is \(r \cos \theta\)
y component is \(r \sin \theta\)

In your case, they've referred to \(r\) as \(v\). Don't be fooled :-)

Answer 9

Note that if the angle is 0, the vector points to the right along the x-axis, and the x component is the entire vector length, because you are multiplying by \(\cos 0 = 1\). The y component is 0, because you are multiplying by \(\sin 0 = 0\). If \(\theta\) is \(90^\circ\) or \(\pi/2\) radians, the vector is entirely y component, with no x component. Usually, you'll end up with an angle other than a multiple of \(90^\circ\), and there will be both x and y components.

Answer 10

thank you!

Answer 11

That is very helpful (:

Answer 12

You're welcome! It's really pretty simple, perhaps that's what confused you :-)

Answer 13

Yeah I think too much about some things haha

Answer 14

It might help if you think about the unit circle and the values of cos and sin as theta varies. When your vector points to the right, down the x axis, all of the magnitude goes into the x component. As the angle opens, less of it goes to the right and more goes up in the y component, until at the top of the unit circle, there is no x component, it's all y component.

Also, if you've done parametric equations, the parametric equations for a circle should remind you of the decomposition of a vector, and vice versa.

http://www.mathopenref.com/coordparamcircle.html

Answer 15

okay thank you (: