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Math Help - Cartesian equation of plane parallel

  1. #1
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    Question Cartesian equation of plane parallel

    Question states "Find the Cartesian equation of the plane parallel to the plane x+y+2z-5=0 and passing through the point (1,2,-1)".

    I approached it like this, vector normal is <1, 1, 2>

    then using
    <br />
\begin{array}{l}<br />
 n \bullet (r - r_0 ) = 0 \\ <br />
  < n_1 ,n_2 ,n_3  >  \bullet ( < x,y,z >  -  < x_0 ,y_0 ,z_0  > ) = 0 \\ <br />
 \end{array}<br />

    I got this
    <br />
\begin{array}{l}<br />
 n \bullet (r - r_0 ) = 0 \\ <br />
  < n_1 ,n_2 ,n_3  >  \bullet ( < x,y,z >  -  < x_0 ,y_0 ,z_0  > ) = 0 \\ <br />
  < 1,1,2 >  \bullet  < x - 1,y - 1,z - 2 >  = 0 \\ <br />
 x - 1 + y - 1 + 2z - 4 = 0 \\ <br />
 x + y + 2z = 6 \\ <br />
 \end{array}<br />

    However the answer is x+y+2z=1 Was I suppose to do something with the 5 in the equation given within question?
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  2. #2
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    First i noticed that the required plane must have the form x+y-2z=c for some constant c. Substitute in the given point to find c=1.

    EDIT: You answer went wrong when you substituted in (x_{0},y_{0},z_{0})=(1,1,2) which is not a point on the plane. If you substitute in (x_{0},y_{0},z_{0})=(1,2,-1) you get the correct answer
    Last edited by whipflip15; October 25th 2008 at 12:36 AM.
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  3. #3
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    Quote Originally Posted by Craka View Post
    Question states "Find the Cartesian equation of the plane parallel to the plane x+y+2z-5=0 and passing through the point (1,2,-1)".

    I approached it like this, vector normal is <1, 1, 2>

    then using
    <br />
\begin{array}{l}<br />
 n \bullet (r - r_0 ) = 0 \\ <br />
  < n_1 ,n_2 ,n_3  >  \bullet ( < x,y,z >  -  < x_0 ,y_0 ,z_0  > ) = 0 \\ <br />
 \end{array}<br />

    I got this
    <br />
\begin{array}{l}<br />
 n \bullet (r - r_0 ) = 0 \\ <br />
  < n_1 ,n_2 ,n_3  >  \bullet ( < x,y,z >  -  < x_0 ,y_0 ,z_0  > ) = 0 \\ <br />
  < 1,1,2 >  \bullet  < x - 1,y - 1,z - 2 >  = 0 \\ <br />
 x - 1 + y - 1 + 2z - 4 = 0 \\ <br />
 x + y + 2z = 6 \\ <br />
 \end{array}<br />

    However the answer is x+y+2z=1 Was I suppose to do something with the 5 in the equation given within question?
    Remember < x_0 ,y_0 ,z_0  > = (1,2,-1)

    EDIT: Too late ...
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  4. #4
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    Ok think I realise now. x+y+2x-5=0
    The co-efficients here give a point normal vector and so for a plane to be parallel to this is will also have the same point normal vector, right?

    That is why it is just x+y+2z= (constant)

    and so just than sub in the point to get the constant, right?
    ie 1+2+2(-1) =1

    giving x+y+2z=1
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