I'm trying to understand how equality of foreshortening in the axes of an isometric projection implies these axes form a 120° angle with one another.
As clarification, an axonometric projection is called isometric if all three axes are equally foreshortened (https://en.wikipedia.org/wiki/Isometric_projection). For instance, this is the isometric projection of a cube:
From intuition, is seems obvious that equality of foreshortening implies equality of angles. Still, while isometric projections are covered in many technical drawing textbooks (such as David Madsen's Engineering Drawing and Design), I could not find any proof of this implication therein.
I have formalized this problem through linear algebra. First, let A, B and C be vectors in $ℝ^3$ crossing one another in 90°. That is, let A, B and C be orthogonal vectors:
$ A \cdot B = B \cdot C = A \cdot C = 0$
Now, let P be the orthogonal projection matrix from $ℝ^2$ to $ℝ^3$ (as explained in: https://en.wikipedia.org/wiki/Projection_(linear_algebra)).
$P = \begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$
We have then, $P \cdot A$, $P \cdot B$ and $P \cdot C$ as projections of 3D vectors into the 2D plane (known in Art as the "picture plane").
The lengths of these projections are foreshortened with relation to their related vectors in $ℝ^3$. In isometric projection, this foreshortening is the same for all vectors, in other words:
$\frac{|P \cdot A|}{|A|} = \frac{|P \cdot B|}{|B|} = \frac{|P \cdot C|}{|C|}$
As stated, this equality of foreshortening must imply equality of angles. In other words, the dot products of all projections must be -1/2, the cosine of 120°:
$ (P \cdot A) \cdot (P \cdot B) = (P \cdot B) \cdot (P \cdot C) = (P \cdot A) \cdot (P \cdot C) = -\frac{1}{2}$
How could this be proven?
HINT:
Take direction cosines $ \cos^2 \alpha +\cos^2 \beta +\cos^2 \gamma =1$; if all the three angles are equal when viewed symmetric along a diagonal what do you get quantitatively by intuition ?