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Finding cartesian equation of plane

  1. Jun 17, 2014 #1
    1. The problem statement, all variables and given/known data
    determine the Cartesian equation of the plane through the points (3,0,1) and (0,1,-1) and perpendicular to the plane with equation x-y-z+1 = 0



    2. Relevant equations




    3. The attempt at a solution
    Well I know the normal of the plane (a,b,c) dotted with (1,-1,-1) = 0 which gives me a-b-c=0
    but other than that Im stuck
     
  2. jcsd
  3. Jun 17, 2014 #2

    Simon Bridge

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    You need to write out the description in terms of the maths.
    You know the standard form for the cartesian equation of the plane (or the vetor form - whichever you are most comfortable with). So write it out.

    That equation has to satisfy some conditions - write out what those mean.

    i.e. point (3,0,1) is on the plane - so if you put x=3, y=0, z=1 into the equation, how does it come out?

    The rest is simultaneous equations.
     
  4. Jun 18, 2014 #3

    HallsofIvy

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    (a, b, c).(0, 1,-1)= b- c= 0 so you know b= c. The equation must be of the form a(x- 3)+ by+ b(z- 1)= 0 and you know that x= 0, y= 1, z= -1 must satisfy that: -3a+ b- 2b= 0. Of course, any multiple of (a, b, c) will give the same plane so you can take either a or b to be whatever you like.
     
  5. Jun 18, 2014 #4
    Ok, thanks
     
  6. Jun 18, 2014 #5

    Simon Bridge

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    Lucky person you - HoI does not normally do people's homework for them...
     
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