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

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SUMMARY

The discussion focuses on finding a third plane that forms a dihedral angle of θ/2 with two given planes: x−3y+2z=0 and 3x−2y−z+3=0. The cosine of the dihedral angle between these two planes is determined to be 1/2. Participants emphasize the importance of understanding vector addition and the conditions required for a vector to bisect the angle between two others. The discussion concludes that the normals of the planes must be manipulated correctly to derive the normal of the third plane.

PREREQUISITES
  • Understanding of dihedral angles in geometry
  • Knowledge of vector operations, specifically dot products
  • Familiarity with the concept of normal vectors of planes
  • Ability to apply the parallelogram rule for vector addition
NEXT STEPS
  • Study the properties of dihedral angles in three-dimensional geometry
  • Learn about vector addition and the conditions for angle bisectors
  • Explore the derivation of normals from given planes
  • Practice problems involving the intersection of three planes in space
USEFUL FOR

Students studying geometry, particularly those focusing on three-dimensional plane intersections, as well as educators looking for examples of dihedral angles and vector manipulation.

user8899
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Homework Statement



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.

Homework Equations



cosθ=|n1 (dot product) n2| /|n1||n2|,

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.
 
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user8899 said:

Homework Statement



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.

Homework Equations



cosθ=|n1 (dot product) n2| /|n1||n2|,

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.

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?
 
LCKurtz said:
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?

Yes, that is what I meant. Do we add the two normals to get the third normal? I am really confused.
 
user8899 said:
Yes, that is what I meant. Do we add the two normals to get the third normal? I am really confused.

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?
 
LCKurtz said:
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?

No because the vectors have to be tip to tail to add up?
 
user8899 said:
No because the vectors have to be tip to tail to add up?

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?
 

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