Let $A$ and $B$ be two opposite corners of a unit cube, and let $C$ be another vertex of the cube other than $A$ and $B.$  Find the distance from $C$ to line $AB.$
We can take $A = (0,0,0),$ $B = (1,1,1),$ and $C = (0,0,1).$  Then line is $AB$ is parameterized by $(t,t,t).$

[asy]
import three;

size(180);
currentprojection = perspective(6,3,2);

triple I = (1,0,0), J = (0,1,0), K = (0,0,1), O = (0,0,0);
triple A = (0,0,0), B = (1,1,1), C = (0,0,1), P = interp(A,B,1/3);

draw((1,0,0)--(1,0,1)--(0,0,1)--(0,1,1)--(0,1,0)--(1,1,0)--cycle);
draw((0,0,0)--(1,0,0),dashed);
draw((0,0,0)--(0,1,0),dashed);
draw((0,0,0)--(0,0,1),dashed);
draw((0,1,1)--(1,1,1));
draw((1,0,1)--(1,1,1));
draw((1,1,0)--(1,1,1));
draw(A--B,dashed);
draw(C--P,dashed);

label("$A$", A, S);
label("$B$", B, N);
label("$C$", C, N);
label("$P$", P, SE);
[/asy]

Let $P = (t,t,t).$  Then lines $CP$ and $AB$ are perpendicular, so their respective vectors are orthgonal.  Hence,
\[\begin{pmatrix} t \\ t \\ t - 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0.\]Then $(t)(1) + (t)(1) + (t - 1)(1) = 0.$  Solving, we find $t = \frac{1}{3}.$

Then $P = \left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right),$ and so $CP = \boxed{\frac{\sqrt{6}}{3}}.$