Distance of a Point from a Plane: Explained with Examples

The distance from a point P(x1, y1, z1) to the plane ax + by + cz + d = 0 is the length of the perpendicular dropped from the point to the plane. It is the shortest possible distance between the point and any point of the plane.

The formula is d = |ax1 + by1 + cz1 + d| / √(a² + b² + c²): plug the point into the left side of the plane equation, take the absolute value, and divide by the normal's length.

Key Concepts

• Numerator: |ax1 + by1 + cz1 + d| measures how far the point is from satisfying the plane equation; it is zero exactly when the point lies on the plane.
• Denominator: √(a² + b² + c²) is the length of the normal vector (a, b, c). Dividing by it converts the numerator into true length units.
• The absolute value is essential — distance is never negative.
• The sign of the raw numerator tells which side of the plane the point is on: points on opposite sides give opposite signs.
• If the plane is already in normal form lx + my + nz = p, the distance is simply |lx1 + my1 + nz1 − p|.

Formulas

• d = |ax1 + by1 + cz1 + d| / √(a² + b² + c²).
• Normal-form shortcut: d = |lx1 + my1 + nz1 − p|.
• Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0: d = |d1 − d2| / √(a² + b² + c²).

Worked Example

Find the distance of (2, 3, −1) from the plane 3x − 4y + 12z = 25.

1. Rewrite as 3x − 4y + 12z − 25 = 0, so a = 3, b = −4, c = 12, d = −25.
2. Numerator: |3×2 + (−4)×3 + 12×(−1) − 25| = |6 − 12 − 12 − 25| = |−43| = 43.
3. Denominator: √(9 + 16 + 144) = √169 = 13.
4. Distance = 43/13 ≈ 3.31 units.
5. Check the side: the raw value −43 is negative, meaning the point sits on the opposite side from the direction the normal (3, −4, 12) points. Fine.

Common Misconceptions

• Forget the absolute value. A negative "distance" is meaningless; the modulus guarantees a non-negative answer.
• Just plug in and stop. The numerator alone is not a distance — you must divide by √(a² + b² + c²).
• The constant d can be ignored. d positions the plane; dropping it measures distance from a different plane through the origin.

Real-World Applications

• Collision detection: how close a point, for example a drone, is to a surface.
• Machining: clearance between a tool tip and a work surface.
• Computer graphics: depth of a point relative to a clipping plane.
• Surveying: height of a point above a reference plane.

Plug the point into the plane equation, take the absolute value, divide by the normal's length. That is the whole formula — the only traps are forgetting the modulus and forgetting the division.