# How can the dihedral angle (i.e. the angle between two planes) be defined sans a coordinate system?

Let $$P$$ and $$Q$$ two planes in the Euclidean $$3$$-dimensional space that intersect in a line $$L$$.

In the presence of a Cartesian coordinate system, the angle between $$P$$ and $$Q$$ can be defined as the angle between any two non-zero vectors $$v$$ and $$w$$ that are orthogonal to $$P$$ and to $$Q$$, respectively, and such that the angle between $$v$$ and $$w$$ is not obtuse. It can be shown with the tools of Analytic Geometry that this angle is the same regardless of the vectors $$v$$ and $$w$$ chosen from the many pairs of vectors that satisfy the requirements.

How can the angle between planes be defined in the absence of a Cartesian coordinate system, and without recourse to the tools of Analytic Geometry, but only to Hilbert's axioms? How can this definition be shown to be well-defined?

I expect the definition to go along the following lines.

Choose any point $$A$$ on $$L$$, and let $$S$$ and $$T$$ denote the lines lying in $$P$$ and $$Q$$, respectively, that are perpendicular to $$L$$ and pass through $$A$$. Then the non-obtuse angle between $$S$$ and $$T$$ is defined to be the angle between $$P$$ and $$Q$$.

The real question that I'm interested in is: how can it be shown that this definition is well-defined and is independent of the choice of the point $$A$$?

• Are Hilbert's axioms enough for a definition for an angle between two rays having the same vertex? [or at least one of the angles so formed] Jun 7, 2022 at 10:05
• @coffeemath I don't know. Are they not? Jun 7, 2022 at 10:07
• I don't know either, hence my question. If they were the dihedral might be something like the largest non-obtuse angle formed by two rays from a common point of the two planes, one ray lying in each plane. [Not sure about that, but this attempted definition would need a definition of angle between two rays. Jun 7, 2022 at 10:10
• @coffeemath The real question is: how can this definition be shown to be well-defined, i.e. independent of the choice of the common point? Jun 7, 2022 at 10:12
• One way might be to set up a correspondence between such angles from common point P and from common point Q which would preserve the measure of the angles. Jun 7, 2022 at 10:14

Let $$P$$ and $$Q$$ be planes which intersect along the line $$L$$ and let $$H,K$$ be chosen halfplanes of $$P$$ and $$Q$$ respectively (with the boundary being the line $$L$$).
Let $$o$$ be a point in $$L$$ and let $$A, B$$ be rays of origin $$o$$ perpendicular to $$L$$ contained in $$H, K$$ respectively. We will show that the angle $$AB$$ does not depend on the choice of $$o$$.
So let $$o'$$ be a point distinct from $$o$$ and let $$A',B'$$ be rays of origin $$o'$$ perpedicular to $$L$$ contained in $$H, K$$ respectively. There exist points $$a\in A, b\in B, a'\in A', b'\in B'$$ such that $$oa\equiv o'a', ob\equiv o'b'$$. Since also $$oa\parallel o'a', ob\parallel o'b'$$ we get that $$\square oaa'o'$$ and $$\square obb'o'$$ are parallelograms (in fact rectangles). Hence $$aa'\parallel oo'\parallel bb'$$ and $$aa'\equiv oo'\equiv bb'$$ so (by transitivity of parallelism and congruence) $$\square aa'b'b$$ is a parallelogram and $$ab\equiv a'b'$$. By SSS congruence criterion $$AB\equiv A'B'$$.