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Parametric Equations

  1. Feb 27, 2014 #1
    1. The problem statement, all variables and given/known

    Find the equation of a line passing through the point P=(-1,2,3) that is parallel to the line of intersection of the planes 3x-2y+z=4 and x+2y+3z=5 . Express your answer in parametric equations .

    2. Relevant equations

    Cross product of normal vectors
    N1 X N2

    3. The attempt at a solution
    Normally all other problems we've had to use cross product of the normals vectors of planes.
    n1= <3,-2,1>
    n2= <1,2,3>

    Cross product I get < -8,-8,8>

    Idk what my next step is since I am trying to find a line passing through point P PARALLEL to line of intersection?
  2. jcsd
  3. Feb 27, 2014 #2
    The vector you get from the cross product gives you your direction vector for the line -- it points in the direction of the line. Two lines are parallel if they have the same direction vectors (or really if their direction vectors are parallel). To find your line, assign it the vector of <-8, -8, 8> (or <-1, -1, 1> if you want to reduce it)
  4. Feb 27, 2014 #3


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    Can't you use your cross product vector for a direction vector for your line?
  5. Feb 27, 2014 #4


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    hi yazz912! :smile:
    looks fine so far :smile:

    ok, you're only interested in the direction, so you may as well call that < -1,-1,1>, or even slightly better still < 1,1,-1> …

    so what is the parametric equation of the line through P parallel to <1,1,-1> ? :wink:
  6. Feb 27, 2014 #5
    So using my point P=(-1,2,3)
    And my direction vector of <-1,-1,1>

    Would my parametric equation be

    x= -1t -1
    y= -1t +2
    z= 1t +3
  7. Feb 27, 2014 #6


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  8. Feb 27, 2014 #7
    Ok thanks!!:)
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