Distances in the space

Point-Plane Distance

The distance from a point, P, to a line, π, is the smallest distance from the point to one of the infinite points on the line. The distance corresponds to the perpendicular line from the point to the plane.

The distance from a point, P(x₀,y₀,z₀), to a plane, $\pi\;\equiv\;Ax+By+Cz+D=0\;$ , is defined by:

$d(P,\pi)=\;\frac{|Ax_0+By_0+Cz_0+D|}{sqrt{A^2+B^2+C^2}}\;$

Calculate the distance from point P=(3,1,-2) to the plane

$\pi\;\equiv\;2x+y-z+1=0\;$

Solution:

$d(P,\pi)=\;\frac{|2\cdot3+1\cdot1+(-2)\cdot1+1|}{sqrt{2^2+1^2+(-1)^2}}=\frac{10}{sqrt{6}}\;$

Calculate the distance from point P=(3,1,-2) to the plane

$\pi\;\equiv\;2y-3=0\;$

Solution:

$d(P,\pi)=\;\frac{|2\cdot1-3|}{sqrt{0^2+2^2+0^2}}=\frac{1}{2}\;$

Calculate the distance from point Q=(5,5,3) to the plane

$\pi\;\equiv\;(x,y,z)=(0,0,4)+(2,2,-1)\lambda+(-3,2,0)\mu\;$

Solution:

x	2	-3
y	2	2
z+4	-1	0

$\Rightarrow\;$

$\pi\;\equiv\;2x+3y+10z+40=0\;$

$d(P,\pi)=\;\frac{|5\cdot2+3\cdot5+10\cdot3+40|}{sqrt{2^2+3^2+10^2}}=\frac{95}{sqrt{113}}\;$