# Vector direction

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

Doc Al
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).

G01
Homework Helper
Gold Member
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.

thank you both very much!