Skill: Calculate distances involving points, lines, and planes

Distance means the shortest Euclidean length between two objects. For a point and a plane, the shortest segment is always perpendicular to the plane.

For a point $(P(x_0,y_0,z_0))$ and a plane

$$Ax+By+Cz+D=0,$$

the perpendicular distance is

$$d=\frac{ \vert Ax_0+By_0+Cz_0+D \vert }{\sqrt{A^2+B^2+C^2}}.$$

The vector

$$\mathbf n=\langle A,B,C\rangle$$

is a normal vector to the plane, so it gives the direction of the shortest path.

Worked example

Find the distance from

$$P(1,2,3)$$

to the plane

$$2x-y+2z-3=0.$$

Step 1: Identify the coefficients

Compare the plane with $(Ax+By+Cz+D=0)$:

$$A=2,\qquad B=-1,\qquad C=2,\qquad D=-3.$$

Step 2: Substitute the point’s coordinates

Calculate the numerator:

$$\vert Ax_0+By_0+Cz_0+D \vert = \vert 2(1)-1(2)+2(3)-3 \vert .$$ $$= \vert 2-2+6-3 \vert =3.$$

Step 3: Calculate the length of the normal vector

$$\sqrt{A^2+B^2+C^2} = \sqrt{2^2+(-1)^2+2^2} = \sqrt{9}=3.$$

Step 4: Divide

$$d=\frac{3}{3}=1.$$

Therefore, the distance from $P$ to the plane is

$$\boxed{1\text{ unit}}.$$

The shortest segment from $P$ to the plane follows the normal direction $\langle 2,-1,2\rangle$. If a point already lies on the plane, substituting its coordinates gives $0$, so its distance to the plane is $0$. Similarly, intersecting lines or planes have distance $0$; only separate parallel or skew objects have a positive distance.