9. Let $O A B C$ be a regular tetrahedron with edge length 1, and let $E$ and $F$ be the midpoints of $A B$ and $O C$, respectively. Then the distance between the skew lines $\mathrm{OE}$ and $\mathrm{BF}$ is $\qquad$
Let $\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c}$, then $\overrightarrow{O E}=\frac{1}{2}(\vec{a}+\vec{b}), \overrightarrow{B F}=\frac{1}{2} \vec{c}-\vec{b}$,
Assume $\vec{n}=x \vec{a}+y \vec{b}+z \vec{c}$ is the common perpendicular vector of $\mathrm{OE}$ and $\mathrm{BF}$, then we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \vec { n } \cdot \vec { O E } = \frac { 1 } { 2 } ( x \vec { a } + y \vec { b } + z \vec { c } ) ( \vec { a } + \vec { b } ) = \frac { 1 } { 4 } ( 3 x + 3 y + 2 z ) = 0 } \\
{ \vec { n } \cdot \vec { B F } = ( x \vec { a } + y \vec { b } + z \vec { c } ) ( \frac { 1 } { 2 } \vec { c } - \vec { b } ) = - \frac { 1 } { 4 } ( x + 3 y ) = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3 x+3 y+2 z=0 \\
x+3 y=0
\end{array}\right.\right. \text {, take } \\
\vec{n}=(3,-1,-3)=3 \vec{a}-\vec{b}-3 \vec{c} \text {, then }|\vec{n}|=\sqrt{19-3-9+3}=\sqrt{10}, \vec{n} \cdot \vec{b}=(3 \vec{a}-\vec{b}-3 \vec{c}) \cdot \vec{b}=-1 \text {, }
\end{array}
$$

Therefore, the projection length $d=\frac{|\vec{n} \cdot \vec{b}|}{|\vec{n}|}=\frac{\sqrt{10}}{10}$ of vector $\overrightarrow{O B}=\vec{b}$ on $\vec{n}$ is the desired result.