Given a tetrahedron $ABCD$ with $D$ at the top and $AB=12,CD=6$.If the shortest distance between the skew lines $AB$ and $CD$ is $8$ and the angle between them is $\frac{\pi}{6}$,then find the volume of the tetrahedron.
Since shortest distance between the skew lines $AB$ and $CD=\frac{\vec{AB}\times\vec{CD}.\vec{AC}}{|\vec{AB}\times\vec{CD}|}=\frac{\vec{AB}\times\vec{CD}.\vec{BD}}{|\vec{AB}\times\vec{CD}|}=8$(given)
$|\vec{AB}\times\vec{CD}|=|\vec{AB}||\vec{CD}|\sin\frac{\pi}{6}=36$
Volume of tetrahedron $ABCD=\frac{1}{6}(\vec{AD}\times\vec{BD}).\vec{CD}$
I am stuck here and can not solve further.The answer given in my book is $48$ cubic units.
Please help me.
Best Answer
Take two planes $P_{AB},P_{CD}$ satisfying the followings :
The line $AB$ exists on $P_{AB}$.
The line $CD$ exists on $P_{CD}$.
$P_{AB}$ is parallel to $P_{CD}$.
The distance between $P_{AB}$ and $P_{CD}$ is $8$.
Take a point $A'$ on $P_{CD}$ such that $AB$ is parallel to $A'C$ and $|AB|=|A'C|$. Also, take a point $D'$ on $P_{AB}$ such that $CD$ is parallel to $BD'$ and $|CD|=|BD'|$.
Now $ABD'$-$A'CD$ is an oblique triangular prism whose volume is $$[\triangle{ABD'}]\times 8=\frac 12\times |AB|\times |CD|\times \sin\frac{\pi}{6}\times 8=144$$
Thus, the volume of the tetrahedron is $$\frac 13\times 144=\color{red}{48}.$$