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How do i find the orthogonal projection of a curve?

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    curve S is the intersections of two surfaces, i have to find the curve obtained as the orthogonal projection of the curve S in the yz-plane


    2. Relevant equations
    how do i find the orthogonal projection of curve S??


    3. The attempt at a solution
    i found the equation of curve S to be (y-1)^2+(z+2)^2=5
    and i know that orthab=b-projab, where a and b are vectors
     
  2. jcsd
  3. Oct 25, 2009 #2

    Dick

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    (y-1)^2+(z+2)^2=5 doesn't describe a curve. It's cylindrical surface. How did you get that by intersecting two other surfaces?
     
  4. Oct 25, 2009 #3
    the two surfaces are x=y^2+z^2 and x-2y+4z=0, i substituted x in the second equation.. is that correct?
     
  5. Oct 25, 2009 #4

    Dick

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    Not really, if you want to get a curve. f(x,y,z)=C doesn't generally describe a curve. It's still a surface. On the other hand you did the right thing. Projecting an intersection of two surfaces to the yz plane just means eliminating x. Your expression in terms of y and z is already the correct curve in the yz plane.
     
  6. Oct 25, 2009 #5
    do you know how i can find the curve obtained as the orthogonal projection of the curve S in the yz-plane? or did i already find the answer?
     
  7. Oct 25, 2009 #6

    Dick

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    You already found the answer. If you have an (x,y,z) point on the curve then the projection to the yz plane is (y,z). That just means you take your two surface equations and eliminate x. What could be wrong with that?
     
  8. Oct 25, 2009 #7
    Oh i see, thanks for your help! ^_^
     
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