Vector direction

  • Thread starter klm
  • Start date
  • #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
 

Answers and Replies

  • #2
Doc Al
Mentor
45,137
1,433
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
G01
Homework Helper
Gold Member
2,682
16
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
klm
165
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thank you both very much!
 

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