Question
Let the cube $ABCDA' B' C' D'$ be where the points $M$ and $P$ are the midpoints of the edges $(AB)$ and $(B B')$ and $P'$ and $N'$ the centers of the faces $A' B' C' D'$ respectively $CDD' C'$ . Calculate the distance between the lines $M P'$ , and $P N'$ , according to the edge of the cube $AB = a$ , where $a$ is strictly positive real number.
My idea
I know that this type of problems are solved expressing the volume in $2$ ways.
So using one of the volume formula we get that
$$
V(MP'N'P)=\frac{N'P\cdot MP'\cdot \operatorname{dis}(MP',N'P)\sin\angle (MP', N'P)}{6}
$$
We can easily express that $MP'=N'P=\frac{a \sqrt{5}}{2}$ and $ \angle (MP', N'P)= \frac{\sqrt{21}}{50}$ and to find the distance wanted i have to express the volume of the tetrahedron again.
I thought of showing what procent of its volume is from the volume of the cube which is equal to $a^3$, but i dont know how.
Hope one of you can help me! Thank you!
Best Answer
Here is a detailed computation of the volume of our tetrahedron.
The face $MN'P'$ of the tetrahedron $MN'P'P$ lies on a plane parallel to the cube face $AA'DD'$ and passing through the center of the cube. The area of triangle $MN'P'$ is equal to $$ \frac{3a^2}{8}. $$ The altitude from the vertex $P$ to the face $MN'P'$ is $a/2$. Therefore, the volume of the tetrahedron is $$ \frac{1}{3}\cdot \frac{3a^2}{8}\cdot\frac{a}{2}=\frac{a^3}{16}. $$
Adding.
In the figure, the plane passing through the center of the cube and parallel to $ADD'A'$ is green. The points $M$, $P'$, $N'$ lie on this plane. The altitude $PQ$ of the tetrahedron from vertex $P$ to face $MP'N'$ must be perpendicular to the green plane and $PQ=a/2$.