How this calculation works
Two differences of the entered points lie in the plane. Their cross product is normal to it. Substitution of one point finds the constant term. The equation is normalized to primitive integer coefficients with the first nonzero coefficient positive. Collinear points do not define a unique plane.
n = (B−A) × (C−A); n · (x−A) = 0A worked example
The points (1,0,0), (0,1,0), (0,0,1) define x + y + z − 1 = 0, with normal (1,1,1).
Good to know
Flat Cartesian coordinates with matching units. Each component is between −1,000,000 and 1,000,000; decimal inputs allow up to 6 places and integer fractions have denominators up to 1,000,000. Expressions are not parsed. Exact intermediate fractions have a size limit; reduce unusually complex inputs if that limit is reached. Lengths and angles are decimal approximations.
Reference: Lamar University: equations of planes (opens in a new tab) ↗
Two directions establish the plane’s orientation
The vectors B − A and C − A both lie within the desired plane. Their cross product is perpendicular to both, giving a normal direction. One of the entered points then sets the plane’s position rather than only its orientation.
If the three points lie on one line, the two differences do not supply independent directions. Infinitely many planes can contain that line, so there is no unique plane to report. Repeating a point causes the same loss of information.
The output equation uses primitive integer coefficients with the first nonzero coefficient positive. Multiplying every coefficient by another nonzero constant describes the same plane, even though the written equation looks different.
Read the normal and the constant together
For x + y + z − 1 = 0, the vector (1,1,1) is normal to the plane. The constant −1 locates the plane relative to the origin. Omitting it would describe a parallel plane passing through the origin.
The reported normal is not necessarily a unit vector. Its direction is what establishes perpendicularity. The origin distance divides the magnitude of the constant by the normal’s length, so rescaling the complete equation leaves that distance unchanged.
Verify a result by substituting all three original points into the equation. Every substitution must give zero for the exact rational input. Measured points with uncertainty can still define an exact algebraic plane for the numbers typed, but that does not make the physical measurements exact.
A few questions, answered
Can a plane be vertical?
Yes. There is no requirement to write it as z = f(x,y). A plane such as x = 2 is represented directly.
What if the plane passes through the origin?
Its constant term is zero, and the reported origin distance is zero.
Will swapping the points change the result?
It may reverse the raw cross product. The normalized equation represents the same plane and uses a consistent sign convention.