Separate a line–plane distance from a point–plane distance

The plane normal $(2,-1,1)$ supplies the direction of a perpendicular line. Use A as the base point. This construction answers the initial unnumbered ask before (i)–(iii).

A point of the line has position $(3+2\lambda,1-\lambda,2+\lambda)$. At its intersection B,

$\begin{aligned}2(3+2\lambda)-(1-\lambda)+(2+\lambda)&=4,\\6+4\lambda-1+\lambda+2+\lambda&=4,\\\lambda&=-\frac12.\end{aligned}$

Hence $B=(3-1,1+1/2,2-1/2)=(2,3/2,3/2)$, the source point $(2,1.5,1.5)$. The line is perpendicular to the plane, so B is also A’s perpendicular foot.

B belongs to both sets. Therefore the shortest distance between the entire line and the plane is zero. This is the line–plane distance requested here; the point A itself need not be on the plane.

Method 1: use the perpendicular foot B.

$\begin{aligned}&\overrightarrow{AB}\\ & =B-A\\ & =(-1,\tfrac12,-\tfrac12).\end{aligned}$

Thus

$h=|\overrightarrow{AB}|=\sqrt{1+\tfrac14+\tfrac14}=\frac{\sqrt6}{2}.$

Method 2: project from a convenient point in the plane. Set $C=(c,0,0)$; the plane gives $2c=4$, so $C=(2,0,0)$. Then $\overrightarrow{CA}=A-C=(1,1,2)$ and the unit normal is $(2,-1,1)/\sqrt6$:

$\begin{aligned}h&=\left|\overrightarrow{CA}\cdot\widehat{\mathbf n}\right|\\&=\frac{|2-1+2|}{\sqrt6}\\&=\frac{\sqrt6}{2}.\end{aligned}$

Both printed methods agree. Unlike the full line in (ii), A is not in the plane: $A\cdot(2,-1,1)=7\ne4$, so its distance is positive.