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Intersection of line and plane

  1. May 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane which passes through the line [itex]\frac{x-2}{3}=\frac{y-3}{1}=\frac{z+1}{2}[/itex]
    and it is normal to the plane x+4yy-3z+7=0

    2. Relevant equations



    3. The attempt at a solution

    I find the intersection point of the plane and the line. It is M(2,3,-1). Also I got the condition (A,B,C)(3,1,2)=0

    [tex]\left\{\begin{matrix}
    2A+3B-C+D=0 & \\
    A+4B-3C=0 &
    \end{matrix}\right.[/tex]

    What is the third condition ?
     
  2. jcsd
  3. May 22, 2008 #2
    Sorry I must fix something. Up in there is x+4y-3z+7=0, and (A,B,C)(1,4,-3)=0.
     
  4. May 24, 2008 #3
    Is (A,B,C)(3,1,2)=0 the third condition?
     
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