Maths & numbers
Point-to-plane distance
Find signed distance and the closest point on a 3D plane.
A clear answer. Here’s how.
Uses the plane equation nx·X+ny·Y+nz·Z+d=0. Signed distance is (n·P+d)/|n|; subtracting that distance along the unit normal gives the orthogonal foot. Reports absolute distance too. The normal must be nonzero; scaling all plane coefficients preserves the plane and flips signed distance only when the scale is negative. Coordinates share one Euclidean unit.
See a worked example
Worked example
A fixed demonstration, separate from the values you enter above.
- Point X
- 1
- Point Y
- 2
- Point Z
- 5
- Plane normal X
- 0
- Plane normal Y
- 0
- Plane normal Z
- 1
- Plane constant d
- 0
Example result
| Quantity | Value |
|---|---|
| Signed distance | 5 |
| Absolute distance | 5 |
| Closest point X | 1 |
| Closest point Y | 2 |
| Closest point Z | 0 |
Plane equation nx·X + ny·Y + nz·Z + d = 0. Signed distance follows the supplied normal direction.