(FB/Math/BM – Problem by Souradeep P.)
If three non-coplanar lines passing through the same point in space make angles A, B, C with each other, what are the angles between the three planes defined by the three lines two at a time?
A plane not through the given point is inserted so that it has one of the lines as normal. This creates a tetrahedron and the angle in the plane (v) is by definition the angle between two of the planes defined by the lines.
I assume (for the time being…) that A, B, and C are acute angles.
From two rightangled triangles
(1) tan A = a/h , tan C = c/h.
Using Pythagoras’ theorem and the cosine theorem
(2) b² = a² + c² + 2h² – 2√(a² + h²)·√(c² + h²)·cos B
(3) b² = a² + c² – 2ac cos v
From (2) and (3)
2h² – 2√(a² + h²)·√(c² + h²)·cos B = -2ac cos v
Divide through by h² and use (1). This gives
1 – √(tan²A + 1)·√(tan²C + 1)·cos B = -tan A tan C cos v,
1 – 1/cos A·1/cos C·cos B = -tan A tan C cos v,
cos A·cos C – cos B = -sin A sin C cos v, and finally
cos v = (cos B – cos A·cos C)/(sin A sin C)
Example: A = B = C = 60⁰. Then v is the angle beween two neighbouring faces in a regular tetrahedreon (among other things).
cos v = (1/2 – 1/2·1/2)/(√3/2·√3/2) = 1/3, v = arccos(1/3).