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Finding the scalar equation of a plane

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the plane that goes through points P, Q and R. P = (3, -1, 2), Q = (8, 2, 4) and R = (-1, -2, -3)

    2. Relevant equations
    Eq of plane
    0 = a(x - x0) + b(y - y0) + c(z - z0)

    3. The attempt at a solution
    In order to find vector normal to the plane, my teacher took the cross product of PQ X PR. Would I still get the correct normal vector if I take the cross product of PQ X QR? Finally, when I plug in the numbers to the equation in "2. Relevant equations", does it matter which point I choose (P, Q or R) for x0, y0, or z0? Thank you.
     
  2. jcsd
  3. Feb 16, 2015 #2

    SammyS

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    Yes, you can use PQ×QR or, for that matter, PR×QR . Each gives a vector which is normal to the plane.

    No, it doesn't matter which point you use. Try more than one and compare results.
     
  4. Feb 16, 2015 #3
    Thank you
     
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