Math Is Fun Forum

djbip

Member

Registered: 2008-03-27

Posts: 13

problem involving 2 planes

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 that is perpendicular to the plane
x+2y-2z=0

Anyone? please...

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: problem involving 2 planes

The normal vector of x-y+2z+5=0 is  <1,-1,2>
The normal vector of 2x+3y-z-1=0 is <2,3,-1>
The cross product of <1,-1,2> and <2,3,-1> would be the directional vector of the line
So the direction vector of the line would be <-5,5,5>
Then we shall find a point which lies on the line
solving the group of equations of two planes
2x+3y-z-1=0
x-y+2z+5=0
make x=0, y=-3/5 , z=-14/5
The line would be  r= <0,-3/5,-14/5>+t<-5,5,5>
The plane which include the line would be
n*r=0
n must be perpendicular to the plane's(x+2y-2z=0) normal vector <1,2,-2>
n must be perpendicular to the line's directional vector <-5,5,5>
<x,y,z>*<1,2,-2>=0
<x,y,z>*<-5,5,5>=0
make y =1, get
n=<4,1,3>
<4,1,3>*<x,y+3/5,z+14/5>= 4x+y+3/5+3z+42/5=0
So the plane would be 4x+y+3z+9=0

Too much calculation... hope I didnt get it wrong
btw, I am bad at calculating

Last edited by Dragonshade (2008-04-15 17:09:21)

djbip

Member

Registered: 2008-03-27

Posts: 13

Re: problem involving 2 planes

The normal vector of x-y+2z+5=0 is  <1,-1,2>
The normal vector of 2x+3y-z-1=0 is <2,3,-1>
The cross product of <1,-1,2> and <2,3,-1> would be the directional vector of the line
So the direction vector of the line would be <-5,5,5>
Then we shall find a point which lies on the line
solving the group of equations of two planes
2x+3y-z-1=0
x-y+2z+5=0
make x=0, y=-3/5 , z=-14/5
The line would be  r= <0,-3/5,-14/5>+t<-5,5,5>
The plane which include the line would be
n*r=0
n must be perpendicular to the plane's(x+2y-2z=0) normal vector <1,2,-2>
n must be perpendicular to the line's directional vector <-5,5,5>
<x,y,z>*<1,2,-2>=0
<x,y,z>*<-5,5,5>=0
make y =1, get
n=<4,1,3>
<4,1,3>*<x,y+3/5,z+14/5>= 4x+y+3/5+3z+42/5=0
So the plane would be 4x+y+3z+9=0

Too much calculation... hope I didnt get it wrong
btw, I am bad at calculating

k...thanks! you got it right, no calculation mistakes.

listen...
if the two planes intersect in a line then the following equation would represent another plane that contains this line:

Ax + By + Cz + D + k(A'x + B'y +C'z +D')=0

can this formula be used to solve the question? it might shorten it..

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: problem involving 2 planes

I believe this is called rectilinear system, I am not sure
You might use it, I am not familiar with it tho, I heard that it's used to solve for question like this
And there's another system called elliptical system, used to solve for intersection of ellipse

Last edited by Dragonshade (2008-04-15 17:37:27)

djbip

Member

Registered: 2008-03-27

Posts: 13

Re: problem involving 2 planes

you solved that stuff and you're not familiar with this??

Ax + By + Cz +D=0 is the usual equations of a plane... where normal vector n=[A,B,C] and (x,y,z) is a point on the plane

Ax + By + Cz + D + k(A'x + B'y +C'z +D')=0 is called a " linear combinations of equations of planes"

still can't relate anything?

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: problem involving 2 planes

I've looked into that before (linear combinations of equations of planes), but I soon forgot
now I just stick to using normal vector to get the equation for plane

I think linear algebra is concerned , I havent studied it yet

Last edited by Dragonshade (2008-04-15 17:45:57)

djbip

Member

Registered: 2008-03-27

Posts: 13

Re: problem involving 2 planes

OK....can you please help me with one more. I would be forever grateful

Determine a vector equation of the line of intersection of the two planes

p1: 3x-y+4z-2=0 and p2: x+6y+10z+8=0

Thank you

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: problem involving 2 planes

First, the normal vectors for the two planes are <3,-1,4> , <1,6,10>
cross product of the two normals gives a vector perpendicular to the two normals
<3,-1,4>x<1,6,10>= <-34, -26,19>

Then you just need to find a point which lies on the line
x+6y+10z+8=0
3x-y+4z-2=0

make z=0
x=4/19 , y=-26/19
r= <4/19 ,-26/19 ,0> + t<-34, -26,19>