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Direction of the vector

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A vector has x component of (-25.0) units and y component of (40.0) units. Find the magnitude and direction of this vector.

    2. Relevant equations



    3. The attempt at a solution

    Magnitude = [tex]\sqrt{}(-25.0)^2+(40.0)^2[/tex]
    = 47.17 units.

    Direction = tan x = | 40.0 | / |-25.0|
    = 1.6
    x = tan^-1 (1.6)
    = 58
    since the vector is located in the IInd Quadrant theta = 180 -58 = 122 degrees.

    But text book answer is 58 degrees. I couldn't get it :confused:

    Any suggestions ?? Thank you...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 10, 2010 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    hi thushanthan! :wink:
    (which book is it?)

    you're right, the text-book is wrong :smile:

    (as you know, if the direction was 58º, both x and y would be positive)
     
  4. Jun 10, 2010 #3
    Thank you :smile:
     
  5. Jun 10, 2010 #4

    HallsofIvy

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    Staff Emeritus
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    What, exactly, did your textbook say? Anything like "58 degrees west of north" or "north 58 degrees west"?
     
  6. Jun 10, 2010 #5
    No there is no direction mentioned. Only the value of theta is given, which is 58 degrees. :frown:
     
  7. Jun 11, 2010 #6
    It's painful when the textbook is wrong and you've spent hours on one question you KNOW you did correctly.... but there is some sense of victory when you find out they are wrong and you are right :)
     
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