# Thread: Cartesian equation of plane parallel

1. ## 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
$
\begin{array}{l}
n \bullet (r - r_0 ) = 0 \\
< n_1 ,n_2 ,n_3 > \bullet ( < x,y,z > - < x_0 ,y_0 ,z_0 > ) = 0 \\
\end{array}
$

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

However the answer is x+y+2z=1 Was I suppose to do something with the 5 in the equation given within question?

2. 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

3. Originally Posted by Craka
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
$
\begin{array}{l}
n \bullet (r - r_0 ) = 0 \\
< n_1 ,n_2 ,n_3 > \bullet ( < x,y,z > - < x_0 ,y_0 ,z_0 > ) = 0 \\
\end{array}
$

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

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 ...

4. 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