So, each octahedron can be inscribed in a cube, so that the corner points of the octahedron are in the midpoints of the side areas of the cube, am I right?
From the octahedron $ABCDS_1S_2$, shown in the image, the vertices
$A =(13 | -5 | 3)$
, $B =(11 | 3 | 1)$
, $C =(5 | 3 | 7)$
$S_1 =(13 | 1 | 9)$
are given. This octahedron is inscribed in the illustrated cube with the corners $P_1$ to $P_8$.
$E_0: 2x_1 + x_2 + 2x_3 + 9 * (2a-5) = 0, a ∈ ℝ$
be a set of planes $aEa$; Let $h$ be the line passing through the points $S_1$ and $S_2 =(5 | -3 | 1).$
Now the task: For $0 <a ≤ 1$, the plane $E_a$ intersects a pyramid of the octahedron with the peak $S_1$:
I have to find the point of intersection $P_a$ of the plane $E_a$ with the line $h$ and then the volume $V_a$ of the truncated pyramid.
This is what I've done: I have determined
but how can I find the volume? If anyone needs the math for determining $P_a$, please say so and I'll add my calculations.