Finding a third plane that has a dihedral angle to two other planes.

1. Mar 28, 2013

user8899

1. The problem statement, all variables and given/known data

The acute angle between two planes is called the dihedral angle. Plane x−3y+2z=0 and plane 3x−2y−z+3=0 intersect in a line and form a dihedral angle θ . Find a third plane (in point-normal, i.e. component, form) through the point (-6/7,0,3/7) that has dihedral angle θ/2 with each of the original planes. Do the three planes intersect at a point or in a line? Explain all steps carefully.

2. Relevant equations

cosθ=|n1 (dot product) n2| /|n1||n2|,

3. The attempt at a solution

I found the dihedral angle of the first two planes to be 1/2, but then I'm not sure what to do after that.

2. Mar 28, 2013

LCKurtz

Don't you mean you found the cosine of the dihedral angle to be 1/2? Wouldn't your new plane bisect the dihedral angle? Can you get its normal from the given normals?

3. Mar 29, 2013

user8899

Yes, that is what I meant. Do we add the two normals to get the third normal? I am really confused.

4. Mar 29, 2013

LCKurtz

Think about it and draw some pictures. Say you have two vectors with their tails at the same place. Will their sum bisect the angle between them or do you have to have some condition on the vectors to make it happen?

5. Mar 31, 2013

user8899

No because the vectors have to be tip to tail to add up?

6. Apr 1, 2013

LCKurtz

No they don't. Think about the parallelogram rule for addition of vectors. The sum is the diagonal of the parallelogram. What has to be true about the two original vectors for that sum vector to bisect the angle between them?