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Distance from a Point to a Plane — Definition, Formula & Examples

Distance from a point to a plane is the shortest (perpendicular) length between a given point in 3D space and a flat plane. You compute it by plugging the point's coordinates into the plane's equation and dividing by the length of the normal vector.

Given a plane ax+by+cz+d=0ax + by + cz + d = 0 and a point P0=(x0,y0,z0)P_0 = (x_0, y_0, z_0), the distance from P0P_0 to the plane is the absolute value of the signed distance ax0+by0+cz0+da2+b2+c2\frac{ax_0 + by_0 + cz_0 + d}{\sqrt{a^2 + b^2 + c^2}}. This quantity equals the magnitude of the projection of any vector from the plane to P0P_0 onto the unit normal of the plane.

Key Formula

D=∣ax0+by0+cz0+d∣a2+b2+c2D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}
Where:
  • a,b,ca, b, c = Coefficients of the plane equation, forming the normal vector $\langle a, b, c \rangle$
  • dd = Constant term when the plane is written as $ax + by + cz + d = 0$
  • (x0,y0,z0)(x_0, y_0, z_0) = Coordinates of the given point
  • DD = Perpendicular distance from the point to the plane

How It Works

Start by writing the plane equation in standard form ax+by+cz+d=0ax + by + cz + d = 0. Then substitute the coordinates of the point into ax0+by0+cz0+dax_0 + by_0 + cz_0 + d to get the numerator. Compute a2+b2+c2\sqrt{a^2 + b^2 + c^2} from the normal vector coefficients for the denominator. Take the absolute value of the ratio to obtain the distance. The result is always non-negative, and it equals zero exactly when the point lies on the plane.

Worked Example

Problem: Find the distance from the point (1, 2, 3) to the plane 2x − 2y + z − 6 = 0.
Substitute the point: Plug the coordinates into the numerator expression.
2(1)+(−2)(2)+1(3)−6=2−4+3−6=−52(1) + (-2)(2) + 1(3) - 6 = 2 - 4 + 3 - 6 = -5
Compute the normal vector's magnitude: Use the coefficients a = 2, b = −2, c = 1.
22+(−2)2+12=4+4+1=3\sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3
Calculate the distance: Take the absolute value of the numerator and divide by the denominator.
D=∣−5∣3=53D = \frac{|-5|}{3} = \frac{5}{3}
Answer: The distance is 53\frac{5}{3} units.

Why It Matters

This formula appears in multivariable calculus, linear algebra, and computer graphics whenever you need to measure how far an object is from a surface. In physics and engineering, it is used to compute clearances, reflection distances, and error margins in 3D modeling.

Common Mistakes

Mistake: Forgetting the absolute value and reporting a negative distance.
Correction: The signed quantity ax0+by0+cz0+dax_0 + by_0 + cz_0 + d can be negative depending on which side of the plane the point lies. Always take the absolute value to get the geometric distance.