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Finding an equation of the plane (Linear Algebra)

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Find an equation of the plane that has y-intercept -5 and is parallel to the plane containing the points P(3, -1, 2), Q(0, 2, 1) and R(5, 2, 0)


    2. Relevant equations

    ax + by + cz + d = 0

    3. The attempt at a solution

    I got two directional vectors
    u = PQ = (-3, 3, -1)
    v = PR = (2, 3, -2)

    n = u x v = (-3, -8, -15) which is also equal to (3, 8, 15) because it's a nonzero multiple of n

    So putting P and the norm together, I get
    3(x-3) + 8(y+1) + 15(z-2) = 0
    3x + 8y + 15z - 31 = 0

    Now I'm stuck. I have no idea what to do with y-intercept of -5. A push in the right direction would be nice.
     
  2. jcsd
  3. Oct 8, 2012 #2
    The point P is not in your plane. It is in a parallel plane. The y-intercept (0, -5, 0) is the only point that is presented as being in your plane.
     
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