If $ABCD$ is a regular tetrahedron with $AB=l, $what is the measure of the angle between $AB$ and $ACD$ or mathematically $\angle(AB, (ACD))$ and I ask you, please, to explain the method. Thanks.
[Math] Regular Tetrahedron
geometry
Related Solutions
Here's an alternative approach. Let $r > 0$. The points $p_{0} = (r, r, r)$, $p_{1} = (r, -r, -r)$, $p_{2} = (-r, r, -r)$, and $p_{3} = (-r, -r, r)$ are vertices of a regular tetrahedron. The rest can be handled with coordinate geometry.
Barycenters of faces and radius of inscribed sphere:
Each lies at distance $\sqrt{3}r$ from the origin, and their mutual separation is $2\sqrt{2}r$. The barycenter of each face is a point of tangency for the inscribed sphere centered at the origin. Particularly, $\frac{1}{3}(p_{1} + p_{2} + p_{3}) = -\frac{1}{3}(r, r, r)$ lies on the inscribed sphere, which therefore has radius $R = \frac{1}{\sqrt{3}}r$.
Requested dimensions for unit inscribed sphere:
If $R = 1$, then $r = \sqrt{3}$, so each vertex is at distance $3$ from the origin, and the edge length is $2\sqrt{6}$.
This is not an answer. Just a visualization of the similar regular tetrahedron being rotated and remaining inscribed in the original tetrahedron, as the angle of rotation changes from $0$ to $\pi$.
To generate the view, I used a regular tetrahedron centered at the origin with one of its faces parallel to the $xy$ plane. A copy of that tetrahedron is generated, and its frame (the frame of reference attached to it) is rotated about an axis that passes through the origin and extending between the midpoint of one edge and the midpoint of the opposing edge as first suggested by Jaap Scherphius in his comment to my recent question.
The tetrahedron copy is then scaled uniformly by the scale factor
$ f = \dfrac{1}{2 |\sin \theta| + 1 } $
and this was also suggested by Jaap Scherphuis in his comment, where $\theta$ is the angle of rotation.
The curve in red is the trace of one of the vertices of the smaller tetrahedron as it rotates. After the small tetrahedron completes its rotation, the tetrahedrons (at that moment they are coincident) are rotated so that the plane containing the red curve faces the viewer.
The .GIF animation was created using my own MS-Excel software that uses VBA (Visual Basic for Applications) script. The generated Excel Chart frames were put together into a .GIF image online using this website
Best Answer
If the side length is 1, then the height of a face is $\sqrt3\over2$ by Pythagoras. Let $P$ be the projection of $B$ onto $ACD$. By symmetry it is the center of $ACD$, hence at a distance of $\frac23\cdot {\sqrt3\over2}={\sqrt3\over3}$ from $A$. Then ${\sqrt3\over3}$ is also the cosine of $\angle PAB$. Thus $\angle PAB = \arccos({\sqrt3\over3}) \approx 0.9553166181245 \approx 54.7356103172^o$.