SOLUTION: Hello, I do not understand completely how to find the direction of vectors. I understand that you are suppose to use this formula: θ=arctan(y/x). What I don't understand is

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Question 755269: Hello,
I do not understand completely how to find the direction of vectors. I understand that you are suppose to use this formula: θ=arctan(y/x). What I don't understand is what to do if you have negative y's and/or x's, and how to solve for the different quadrants. My problem that I am really confused on is this one:
What is the magnitude and direction angle of the vector <2,-sqrt(3)>?
Here is how I tried to solve it:
Magnitude =
Magnitude =
Magnitude = =
Direction angle = arctan(y/x)
=
How would I finish solving this? Do I include the negative sign for all vectors?
Thanks!

Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
Think of your vector as the arrow from point O to point A:

Point A is in the fourth quadrant.
It is at a distance from O.
The coordinates of A are


Your direction angle is

The angle does not help you figure out the quadrant because <-2,sqrt(3)> would give you the same angle.
The signs of {{x}}} and tell you the quadrant.
<(2,-sqrt(3)> and <(-2,sqrt(3)> are mirror images of each other.
<(-2,sqrt(3)> forms a with the positive x-axis.

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