Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 11 / d / iii / Solution

The square $d(t{)}^2=(t+1{)}^2+1/2$ is minimized when its first term vanishes. ThereforeThis agrees with the distance between skew lines found in part (b). The corresponding parameter on the first line is its orthogonal projection valueThe closest pair is $(-1/2,-3/2,0)$ and $(-1,-1,0)$; their connecting vector $(-1/2,1/2,0)$ is perpendicular to both direction vectors.