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Equation of a plane

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the equation of a plane with distance 3 units from the origin and perpendicular to the line through P(1,2,3) and Q(-2,4,1).


    2. Relevant equations

    [tex]\vec{n}[/tex] = [tex]\frac{\vec{PQ}}\left|{\vec{PQ}}\left|[/tex]

    Plane Equation: a(x-x0) + b(y-y0) + c(z-z0)


    3. The attempt at a solution

    Ok so I think I can solve this all the way up until the end.

    We have [tex]\vec{PQ}[/tex] = <-3, -2, 2>

    so [tex]\vec{n}[/tex] = [tex]\frac{1}{\sqrt{17}}[/tex]<-3,-2,2>

    Now if I scale that vector by 3 or -3 I can get a point on the plane that I am looking for, I need to put this into the form of a plane so I use the equation above and end up getting:
    [tex]\frac{-3}{\sqrt{17}}[/tex]x + [tex]\frac{2}{\sqrt{17}}[/tex]y - [tex]\frac{2}{\sqrt{}17}[/tex]z = [tex]\frac{-27}{17}[/tex] + [tex]\frac{18}{17}[/tex] + [tex]\frac{18}{17}[/tex]

    This can be simplified further but this is where mine and my professors work differs. He gets the following on the right side of the equation = [tex]\frac{27}{17}[/tex] + [tex]\frac{12}{17}[/tex] + [tex]\frac{12}{17}[/tex]

    Any ideas?
     
  2. jcsd
  3. Feb 24, 2010 #2
    nevermind I figured it out.
     
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