Plane Equation by 3 Points

Plane through 3 points: ax + by + cz = d

Three points that don't sit on one line fix a single plane in 3D. To find it, take two edges of the little triangle they form and cross them: n = (P₂ − P₁) × (P₃ − P₁) = (a, b, c) gives the normal, then d = a·x₁ + b·y₁ + c·z₁ pins down the constant. Try it with P₁=(1,0,0), P₂=(0,1,0), P₃=(0,0,1). The edges are (−1,1,0) and (−1,0,1), their cross product is (1,1,1), and the plane works out to x + y + z = 1. Which way the normal points tells you the front face from the back, and that matters for back-face culling and lighting.

Applications

It shows up in computer graphics, where triangle rendering needs normal vectors for Phong shading and back-face culling. GIS work uses it to interpolate terrain height between sample points in Digital Elevation Models. You'll also find it in CAD and 3D modeling, in aviation for working out target slope and approach angle from waypoints, in structural engineering when fitting a reference plane to surveyed points, and in the collision detection of physics engines.

FAQ

What if the three points are collinear? Then the cross product collapses to the zero vector and there's no single answer. Any number of planes can pass through one line.

Does point order matter? Only for orientation. Swap two of the points and the normal flips to the other side (n → −n), but you're still describing the exact same plane.

How to normalize the equation? Divide a, b, c, d by ‖n‖ so the normal has length 1. That makes distance-to-plane calculations clean: dist(P) = |a·x + b·y + c·z − d|.