# Vector direction

1. Sep 16, 2007

### klm

vector E= (125 i - 250 j) what are the magnitude and direction?
ok so what i did was found out that vector E should be in Quad. 4. and that the magnitude will be 279.51 b/c you do the square root of 125^2 + 250^2 = 279.51. but i am having trouble understanding how to get the direction. i think my problem is trying to see how to draw the picture.
i know that you go +125 on the x axis and then down -250 on the y axis, so the vector will be in quad 4. but what angle are you trying to find? what i did was tan^-1 (125/250) = 26 and then i added 270 and got 296 as my angle. is this the correct way to do this problem

2. Sep 16, 2007

### Staff: Mentor

Nothing wrong with what you did. You found the angle with the -y-axis, then added 270. Sounds good to me! You are following standard practice of measuring the angle with respect to the +x axis using counterclockwise as +.

You can also, of course, start by finding the angle with respect to the x-axis using tan^-1(250/125) = 63.4 degrees below the x-axis. Subtract that from 360 and you get back to 296.6 degrees (same as you did).

3. Sep 16, 2007

### G01

Yes, that is a correct way to denote the direction of that vector. You could have also said, "26 degrees from the negative y axis" or "64 degrees from the positive x axis." That would have also been a correct representation of the direction.

4. Sep 16, 2007

### klm

thank you both very much!