# Math Help - easy intersection of planes question:(confusion!)

1. ## easy intersection of planes question:(confusion!)

Find the equation of the plane passing through the intersection of $2x+y+2z=9$ and $4x-5y-4z=1$ and passing through $(3,2,-1)$

I forgot how these sums are done:Here's what I did:
solving the two equations we get $x=\frac{23-3z}{7}$ and $y=\frac{17-8z}{7}$
hence we get the parametric equation: $<\frac{23-3t}{7},\frac{17-8t}{7},t>$
hence the direction vectors= $<\frac{-3}{7},\frac{-8}{7},1>$
as the line is perpendicular to the intersection, it is also the normal vector of the 3rd plane:
equation of required plane: $-\frac{3}{7}x-\frac{8}{7}y+z=D$
putting $(x,y,z)=(3,2,-1)$ we get $D=\frac{32}{7}$
new simplified equation: $3x+8y-7z=32$

but the answer given in the book is: $11x-5y-z=24$

What am I doing wrong here?

2. ## Re: easy intersection of planes question:(confusion!)

The line is not perpendicular to the intersection, the line IS the intersection...

3. ## Re: easy intersection of planes question:(confusion!)

Originally Posted by Prove It
The line is not perpendicular to the intersection, the line IS the intersection...
how do we get the normal vector to the vector: $<-\frac{3}{7}, -\frac{8}{7},1>$

4. ## Re: easy intersection of planes question:(confusion!)

You don't need to. Just choose two points that you know lie on the line, and use the point you are told lies on the plane, to substitute into the plane equation ax + by + cz = d. You will get three equations in three unknowns to solve simultaneously.

5. ## Re: easy intersection of planes question:(confusion!)

Originally Posted by earthboy
how do we get the normal vector to the vector: $<-\frac{3}{7}, -\frac{8}{7},1>$
Vectors do not have normal vectors, planes have normal vectors.