Parametric Equation of a Line from the intersection of two planes

[Aneeshrege]

Homework Statement

Find the parametric equation for a line of intersection of these two planes

x+2y+3z=0
4x+5y+6z=5

Homework Equations

Normal to plane 1= <1,2,3>
Normal to plane 2= <4,5,6>

The Attempt at a Solution

I know the way to do this problem is to take cross product of two normals etc etc,
but i want to know if the way i did this is correct also.

I already turned the work in so there's nothing i can do to change it but the curiosity is killing me.

First, i set the two planes equal to each other

4x+5y+6z-5=0
x+2y+3z=0

=> x+2y+3z=4x+5y+6z-5 (please correct me my thinking process is wrong,im winging it)
=> 3x+3y+3z=5
=> x+y+z=5/3 (now I am thinking this is the equation of the intersection of the two planes but this isn't the equation of a line, it looks like a plane, or is it?)

so i took a point on this set, (5/3,0,0) and two other points (0,5/3,0) and (0,0,5/3)

i did <0,0,5/3> - <0,5/3,0> = <0,-5/3,5/3> as a directional vector.

so L(t)= (5/3,0,0) + <0,-5/3,5/3>t

x=5/3
y=-5/3t
z=5/3t

IS any of this wrong?!

 

[Dick]

Try it out. Put your x,y and z into the two original plane equations. At t=0 you get (5/3,0,0). That's not on either of the original planes, is it?

 

[Aneeshrege]

Ah i do notice that, but

x+2y+3z=4x+5y+6z-5

if i plug them into that those points solve the equation, which is the x,y,z such that those two planes are equal, or is that me failing at winging a problem?

 

[Dick]

If you mix up two planes you get a third plane. A line in that plane may or may not be in either of the two planes you started with. I admire your spirit of winging it, but when you scamble the two planes, you loose information.

 

[Aneeshrege]

so i just needed to restrict that plane to a set of values that lie in both planes. so I am just wondering what exactly did i do by setting the two planes equal to each other? i notice i get the same result if i subtract equation 1 from equation 2, so did i just subtract the two planes by setting them equal? I am confused as to what that actually does

 

[vela]

Yeah, you effectively just subtracted one equation from the other. This gives you a new plane in which the line of intersection lies, so you're just finding the equation of a plane rotated about that line.

 

[Dick]

so i just needed to restrict that plane to a set of values that lie in both planes. so I am just wondering what exactly did i do by setting the two planes equal to each other? i notice i get the same result if i subtract equation 1 from equation 2, so did i just subtract the two planes by setting them equal? I am confused as to what that actually does

Try a simpler example. Take the planes x=0 and z=0. They intersect along the line (0,t,0). Subtract them and get x-z=0. That's a different plane. Still contains (0,t,0) though. Adding and subtracting the plane equations isn't getting you any closer to finding the intersection. I think you'd better stick with the normal and cross product method.

 

[Aneeshrege]

okay thanks guys for all your help. i understand this a lot better now