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Math Help - Equation of a Plane

  1. #1
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    Equation of a Plane

    Three points are given: P = (2,1,0) , Q = (1,0,1) , R = (3,2,-1).
    I'm supposed to find the plane that goes through P, Q, R in the form Ax + By + Cz = D.

    My attempt:

    M : (x,y,z) = P + t*PQ + s*PR , s,t R.

    PQ
    = Q - P = (-1,-1,1) , PR = R - P = (1,1,-1).

    This gives me a system of equaton where I use the elimination method to find out what the coefficients A, B, C and D are equal to, but I'm stuck at the second step:
    x = 2 - t + s
    y = 1 - t + s
    z = + t - s

    <=>

    x = 2 - t + s
    -x + y = - 1
    x + z = 2

    Did I do something wrong? / How do I proceed from here?
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  2. #2
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    Re: Equation of a Plane

    Quote Originally Posted by Cinnaman View Post
    Three points are given: P = (2,1,0) , Q = (1,0,1) , R = (3,2,-1). I'm supposed to find the plane that goes through P, Q, R in the form Ax + By + Cz = D.
    Here a different way.
    Let N = \overrightarrow {PQ}  \times \overrightarrow {PR} then the plane is N\cdot(<x,y,z>-P)=0
    Thanks from Cinnaman
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  3. #3
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    Re: Equation of a Plane

    We haven't reached cross product yet, I need to solve it using the method above.

    I just don't know what happens when both t and s disappear in the third equation at step 2 of the elimination process.
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  4. #4
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    Re: Equation of a Plane

    So my lecturer just sent out an email saying there was an error in the question itself, now it makes sense -_-
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