Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 11

The direction vector is $b-a=(1,2,-2)$. The vector equation and symmetric Cartesian equation of the line areThe $(x,y)$-plane has $z=0$, which gives $t=2$. Thus its intersection with the line has position

The plane normals are $n_1=(2,-1,-1)$ and $n_2=(3,-1,-3)$. Their cross product is $(2,3,1)$, perpendicular to both normals and hence parallel to the intersection line. Setting $z=0$ in both equations gives the common point $(1,-1,0)$. ThereforeThe normals are not parallel, so their intersection is a unique line.

From the line point $r_0=(1,-1,0)$ to $P=(2,2,1)$, take $w=P-r_0=(1,3,1)$ and line direction $d=(2,3,1)$. Orthogonal projection gives the nearest parameter $t=(w\cdot d)/(d\cdot d)=12/14=6/7$. The perpendicular displacement has squared Euclidean normHence the distance from a point to a line in three dimensions isThe nearest point is $(19/7,11/7,6/7)$; subtracting it from $P$ gives a vector orthogonal to $d$, verifying the minimum.