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Calc: finding an equation

  1. Sep 17, 2005 #1
    how do i work this?

    find an equation for the plane that passes through the point (1,3,1) and contains the line x=t,y=t,z=-2+t

    would the line equation be -2i+t(i+j+k)? and then would that mean that the direction or normal vector for the plane be i+j+k? and then what do you do? i have the answer and i can't get it....
     
  2. jcsd
  3. Sep 17, 2005 #2
    to find the equation of a plane you need to have TWO distinct directional vectors and one point the plane passes through

    you have been told that it contains the line - what is the directional vector of that line
    you need one more directional vector - a vector between the line and the point will give you this. Pick a point on the given line and find the vector between this point and the given point.
    the equation of the plane will look like this

    [tex] P = (x,y,z) + sd_{1} + td_{2} [/tex]
    where s and t are scalar mutliples, d1 and d2 the directional vectors and x y z is the point

    p.s. there are many possible answers
     
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