Plane Through Three Points

Find the equation of the plane ax+by+cz+d=0 that passes through three points in 3D space, with the normal vector and distance from the origin.

Input

Enter the coordinates of three points in 3D space to compute the equation of the plane ax+by+cz+d=0 that contains them.

Point P1

x coordinate

y coordinate

z coordinate

Point P2

x coordinate

y coordinate

z coordinate

Point P3

x coordinate

y coordinate

z coordinate

Result

Plane equation

1x + 1y + 1z - 1 = 0

Normal vector

(1, 1, 1)

Normal vector magnitude

1.732051

Distance from origin

0.57735

Equation coefficients

Coefficient a

Coefficient b

Coefficient c

Coefficient d

-1

The normal vector is computed from the cross product (P2 - P1) × (P3 - P1). The coefficients may differ from another equivalent plane equation by a constant scale factor.

How it works

Two edge vectors v1 = P2 - P1 and v2 = P3 - P1 lie in the plane, and their cross product v1 × v2 gives the normal vector (a, b, c).
The constant term d is found from the dot product of the normal vector and point P1 as d = -(a, b, c) · P1, which fixes the plane equation ax + by + cz + d = 0.
The distance from the origin to the plane equals |d| divided by the magnitude of the normal vector sqrt(a² + b² + c²).
When the three points are collinear, the magnitude of the cross product is near zero and no unique plane exists, so the tool reports that case.

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