Question

If volume of regular tetrahedron of edge length $k$ is $V$ and shortest distance between any pair of opposite edges of same regular tetrahedron is $d,$ then the value of $\frac{d^3}{V}$ is

Answer 1

Length of edge of regular tetrahedron $=k$
Consider a tetrahedron $ABCD$
and let $A(0,0,0),$ $B(k,0,0),$ $C(\frac{k}{2},\frac{\sqrt{3}k}{2},0)$
Projection of point $D$ on plane containing points $A,B$ and $C$ is the centroid of equilateral triangle $ABC$
Projection of $D$ is $(\frac{k}{2},\frac{\sqrt{3}k}{6},0)$
$\therefore$ Let coordinates of $D$ be $(\frac{k}{2},\frac{\sqrt{3}k}{6},h)$

$\left|AD\right|=k$
$\Rightarrow \frac{k^2}{4}+\frac{k^2}{12}+h^2=k^2$
$\Rightarrow h^2=k^2-\frac{k^2}{3}=\frac{2k^2}{3}$
$\therefore h=\frac{\sqrt{2}}{\sqrt{3}}k$
$\therefore$ $D$ is $(\frac{k}{2},\frac{\sqrt{3}k}{6},\frac{\sqrt{2}k}{\sqrt{3}})$

Since, distance between any two skew lines of tetrahedron $ABCD=d,$
$\Rightarrow \overrightarrow{AC}\cdot \frac{\overrightarrow{CD}\times \overrightarrow{AB}}{|\overrightarrow{CD}\times \overrightarrow{AB}|}=d$
$\Rightarrow d=(\frac{k}{2}\hat{i}+\frac{\sqrt{3}k}{2}\hat{j})\cdot \frac{(-\frac{2\sqrt{3}k}{6}\hat{j}+\frac{\sqrt{2}k}{\sqrt{3}}\hat{k})\times k\hat{i}}{|(\frac{-2\sqrt{3}}{6}\hat{j}+\frac{\sqrt{2}k}{\sqrt{3}}\hat{k})\times k\hat{i}|}$
$\Rightarrow d=\frac{(\frac{k}{2}\hat{i}+\frac{\sqrt{3}k}{2}\hat{j})\cdot (\frac{2\sqrt{3}{k}^2}{6}\hat{k}+\frac{\sqrt{2}{k}^2}{\sqrt{3}}\hat{j})}{k^2\sqrt{{\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}^2}}$
$=\frac{\frac{\sqrt{3}k}{2}.\frac{\sqrt{2}{k}^2}{\sqrt{3}}}{k^2}$
$\Rightarrow d=\frac{1}{\sqrt{2}}k$

Area of equilateral $\Delta ABC=\frac{\sqrt{3}}{4}{k}^2$
Volume of tetrahedron, $V=\frac{1}{3}\times \left(\text{ar}\right(\Delta ABC\left)\right)\times h$
$=\frac{1}{3}\times \frac{\sqrt{3}}{4}{k}^2\times \frac{\sqrt{2}}{\sqrt{3}}k=\frac{\sqrt{2}{k}^3}{12}$

$\therefore \frac{d^3}{V}=\frac{{\left(\frac{k}{\sqrt{2}}\right)}^3}{\left(\frac{\sqrt{2}{k}^3}{12}\right)}$
$=\frac{k^3}{2\sqrt{2}}.\frac{12}{\sqrt{2}{k}^2}=\frac{12}{4}=3$

Alternate Solution :

We first draw a cube of side length $d$ and draw a tetrahedron in it.
By Pythagoras theorem,
$d^2+d^2=D{C}^2=k^2$
$\Rightarrow d=\frac{k}{\sqrt{2}}$

Volume of regular tetrahedron with edge length $k$ is $V=\frac{k^3}{6\sqrt{2}}$
$\therefore \frac{d^3}{V}=3$