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Distance from a point to a plane

The distance from a point to a plane is the length of the shortest perpendicular segment. You insert the coordinates of the point into the equation of the plane, take the absolute value, and divide by the length of the normal vector. For the origin the coordinates drop out, and the constant term is what remains. The normal vector is read from the coefficients of $x$ , $y$ and $z$ . The unit is the same as the coordinates.

\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}

distance to the plane

Symbols

$a,\ b,\ c$	the normal vector
$d$	the constant term when the equation equals 0

Example

The plane $x + 2y + 2z - 6 = 0$ and the origin: $\dfrac{|-6|}{\sqrt{1 + 4 + 4}} = \dfrac{6}{3} = 2$ .

Write the plane as

ax + by + cz + d = 0

before you use the formula. Move the constant to the left-hand side.

Practise lines and planes in space for free →

← Sum of squares and a geometric sum · Intersection of a line and a plane →

Part of Mathematics R2: Lines and planes in space.