The edges of a regular tetrahedron with vertices $A ,~ B,~ C$, and $D$ each have length one. 
Find the least possible distance between a pair of points $P$ and $Q$, where $P$ is on edge $AB$ and $Q$ is on edge $CD$.



$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{3}{4}\qquad \textbf{(C) }\frac{\sqrt{2}}{2}\qquad \textbf{(D) }\frac{\sqrt{3}}{2}\qquad \textbf{(E) }\frac{\sqrt{3}}{3}$
Note that the distance $PQ$ will be minimized when $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$.
To find this distance, consider triangle $\triangle PCQ$. $Q$ is the midpoint of $CD$, so $CQ=\frac{1}{2}$. Additionally, since $CP$ is the altitude of equilateral $\triangle ABC$, $CP=\frac{\sqrt{3}}{2}$.

Next, we need to find $\cos(\angle PCQ)$ in order to find $PQ$ by the Law of Cosines. To do so, drop down $D$ onto $\triangle ABC$ to get the point $D^\prime$.
$\angle PCD$ is congruent to $\angle D^\prime CD$, since $P$, $D^\prime$, and $C$ are collinear. Therefore, we can just find $\cos(\angle D^\prime CD)$.
Note that $\triangle CD^\prime D$ is a right triangle with $\angle CD^\prime D$ as a right angle. 



As given by the problem, $CD=1$.
Note that $D^\prime$ is the centroid of equilateral $\triangle ABC$. Additionally, since $\triangle ABC$ is equilateral, $D^\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\prime$ is $\frac{2}{3}$ of the altitude of $\triangle ABC$. Therefore, $CD^\prime=\frac{\sqrt{3}}{3}$. 
Since $\cos(\angle D^\prime CD)=\cos(\angle PCQ)=\frac{CD^\prime}{CD}$, $\cos(\angle PCQ)=\frac{\frac{\sqrt{3}}{3}}{1}=\frac{\sqrt{3}}{3}$
\[PQ^2=CP^2+CQ^2-2(CP)(CQ)\cos(\angle PCQ)\]
\[PQ^2=\frac{3}{4}+\frac{1}{4}-2\left(\frac{\sqrt{3}}{4}\right)\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{3}\right)\]
Simplifying, $PQ^2=\frac{1}{2}$.
Therefore, $PQ=\frac{\sqrt{2}}{2}\Rightarrow$ $\boxed{\textbf{C}}$
Solution by treetor10145