The perpendicular distance of a corner of unit cube from a diagonal not passing through it is

Answer

`sqrt((2)/(3))`

Explanation

Let the edges OA, OB and OC of the unit cube be along OX, OY and OZ, respectively.
Since OA=OB=OC=1 unit,
`vec(OA)=hati,vec(OB)=hatjandvec(OC)=hatk`
Let CM be perpendicular from the corner C on the diagonal OP. The vector equation of OP is
`vecr=lamda(hati+hatj+hatk)`
OM=projection of `vec(OC)" on "vec(OP)=(vec(OC).vec(OP))/(|vec(OP)|)`
`=hatk.((hati+hatj+hatk))/(sqrt3)=(1)/(sqrt3)`
Now `OC^(2)=OM^(2)+CM^(2)`
or `CM^(2)=OC^(2)=OM^(2)=1-(1)/(3)=(2)/(3)`
or `CM=sqrt(2/(3))`