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Drawing a simple vector field issue

  1. Mar 22, 2007 #1
    Hello everyone i'm not sure if this is right or not...

    If i have

    F(x,y,z) = zj; where j is the vector, j hat.

    Would that be all vectors are going to be pointing up if you assume z is up, and are in the y plane?


    If the coordinate sytem is, z is up, y is to the right, and x is pointing at you.

    If i had F(x,y,z) = yj; the answer is, No vectors emanate from the xz plane since y = 0 there. In each plane y = b, all the vectors are identical.
     
  2. jcsd
  3. Mar 22, 2007 #2

    Dick

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    No. j is a unit vector pointing in the same direction as the +y axis. Which you seem to want to call 'right'. The length of the vectors is proportional to their z coordinate. What is the 'y plane'? What does 'in' a plane mean? Parallel to?
     
  4. Mar 22, 2007 #3
    I worked it out and it seems the vectors are pointing parrallel to the +y axis which is pointing right, for values > 0, and for values < 0 its pointing to the left or in the -y direction.
     
  5. Mar 22, 2007 #4

    Dick

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    That's the right picture (where 'values' means z, right).
     
  6. Mar 22, 2007 #5

    HallsofIvy

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    z is a number, j is a vector pointing in the positive y direction. Your vector field consists of vector pointing in the positive y direction, longer as z increases. (And pointing in the negative y direction for z negative.)
     
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