# Complex Equation of a Plane Problem

• Jun 9th 2008, 08:58 PM
Morphayne
Complex Equation of a Plane Problem
Problem:

Find the equation of the plane that passes through the line of intersection of the planes $x-y+2z+5=0$ and $2x+3y-z-1=0$, and is perpendicular to the plane $x+2y-2z=0$.

$4x+y+3z+9=0$

I'm stuck on tis one big time. Please grant me you help.(Crying)
• Jun 10th 2008, 12:34 AM
TheEmptySet
I hope this is clear
Quote:

Originally Posted by Morphayne
Problem:

Find the equation of the plane that passes through the line of intersection of the planes $x-y+2z+5=0$ and $2x+3y-z-1=0$, and is perpendicular to the plane $x+2y-2z=0$.

$4x+y+3z+9=0$

I'm stuck on tis one big time. Please grant me you help.(Crying)

First we need to find the equation of the line of the intersection of the two planes.

multiplying the 1st by -2 and adding it to the 2nd gives
$5y-5z=11$ Now we can parameterize the line. Let $z=t$ then $y=t+\frac{11}{5}$ and finally x is

$x-y+2z+5=0 \iff x=y-2z-5=t+\frac{11}{5}-2t-5=-t-\frac{14}{5}$

so we get $<-t-\frac{14}{5},t+\frac{11}{5},t>=<-\frac{14}{5},\frac{11}{5},0>+t<-1,1,1>$

We can now find the normal vector from the plane to be $<1,2,-2>$

Now if we cross these two vectors we will get a vector that is perpendicular to both of them so it is parallel to the plane.

$\begin{vmatrix} i & j & k \\ -1 & 1 & 1 \\ 1 & 2 & -2 \\ \end{vmatrix}=<-4,-1,-3>$

From our work before we know the point $(-\frac{14}{5},\frac{11}{5},0)$ is in the plane and let (x,y,z) be any other point in the plane then the vector
$\vec v= $ is in the plane.

Now If we dot v with the vector from the cross product and set it equal to zero we will get the equation of the plane perpendicular to the original plane.

$<-4,-1,-3> \cdot =0$

$-4x-\frac{56}{5}-y+\frac{11}{5}-3z=0 \iff -4x-y-3z-\frac{45}{5}=0 \iff 4x+y+3z+9=0$

Yeah!!!(Clapping)