- #1
klm
- 165
- 0
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
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