Plane Equation Calculator From Three Points
Enter the plane data
Examples: 2, -0.75, or 1/3. Three-point inputs must be distinct and non-collinear.
| A | B | C | D |
|---|---|---|---|
| — | — | — | — |
Show personalized calculation steps
Enter valid values to see the full working.
Complete equation forms
Interactive 3D preview
The labeled blue patch is the plane; P1, P2, and P3 are plotted points; the teal arrow labeled n is its normal. Drag to rotate or use the buttons. The numerical result remains complete without this graphic.
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Formula for a plane through three points
For P1=(x1,y1,z1), P2=(x2,y2,z2), and P3=(x3,y3,z3), form two in-plane vectors:
u = P2 − P1, v = P3 − P1
The cross product n = u × v = (A,B,C) is perpendicular to both vectors, so it is a normal to the plane. Substitute any supplied point into:
A(x − x1) + B(y − y1) + C(z − z1) = 0
Expanding gives the general form Ax + By + Cz + D = 0, where D = −(Ax1 + By1 + Cz1).
Step-by-step method
- Subtract P1 from P2 and P3 to obtain two direction vectors in the plane.
- Take their cross product component by component.
- Use the cross product as the normal vector (A,B,C).
- Insert P1 into point-normal form, expand the brackets, and collect the constant term.
- Clear rational denominators, divide all integer coefficients by their greatest common divisor, and make the leading nonzero coefficient positive.
- Substitute all three original points to verify that each exact residual is zero.
Worked example
Let P1=(1,2,3), P2=(4,1,0), and P3=(2,3,5). Then:
u=(3,−1,−3), v=(1,1,2)
u × v = ((−1)(2)−(−3)(1), (−3)(1)−(3)(2), (3)(1)−(−1)(1)) = (1,−9,4)
Substituting P1 gives 1(x−1)−9(y−2)+4(z−3)=0, which expands to x − 9y + 4z + 5 = 0.
Verification: P1 gives 1−18+12+5=0; P2 gives 4−9+0+5=0; and P3 gives 2−27+20+5=0. Multiplying the whole equation by any nonzero constant produces a different-looking equation with exactly the same plane.
Equation forms
| Form | Pattern | Useful when |
|---|---|---|
| General | Ax+By+Cz+D=0 | Testing points, comparing coefficients, and most algebra problems. |
| Point-normal | n·(X−P)=0 | A point and perpendicular direction are already known. |
| Normalized | n̂·X+d=0 | Signed distances; the normal has length 1. |
| Vector / parametric | X=P+su+tv | Generating points, lines, meshes, and intersections. |
| Determinant | det[X−P1; P2−P1; P3−P1]=0 | Showing directly that the plane passes through three points. |
| Intercept | x/a+y/b+z/c=1 | Reading finite axis crossings. It is unavailable when any required intercept is undefined or zero. |
Collinear points
Three collinear points do not determine a unique plane: infinitely many planes can rotate around their shared line. Algebraically, P2−P1 and P3−P1 are parallel, so their cross product is the zero vector. Repeated points fail for the same reason because one direction vector becomes zero. The calculator distinguishes repeated points, a zero normal, and collinearity with inline messages.
Frequently asked questions
How do I find a plane from three points?
Subtract P1 from P2 and P3, take the cross product of those two vectors to get a normal (A, B, C), then substitute P1 into A(x−x1)+B(y−y1)+C(z−z1)=0 and simplify.
Why must the three points be non-collinear?
Collinear points lie on one line, and infinitely many planes contain that line. Their two direction vectors have a zero cross product, so there is no unique normal or plane.
What happens if two points are identical?
A repeated point creates a zero direction vector, so the three entries cannot define a unique plane. Enter three distinct, non-collinear points.
Why can different-looking equations describe the same plane?
Multiplying every coefficient A, B, C, and D by the same nonzero number does not change the solution set. This calculator divides out common factors and makes the leading nonzero coefficient positive.
How do I identify the normal vector?
In Ax+By+Cz+D=0, the vector (A, B, C) is normal to the plane. For three points it is calculated as (P2−P1) × (P3−P1).
Does the calculator support fractions?
Yes. Enter integers, decimals, or fractions such as 1/3 and −5/7. The calculator keeps rational inputs exact and simplifies the final integer coefficients.
How can I test whether a fourth point is coplanar?
Substitute the fourth point into Ax+By+Cz+D. It is coplanar when the exact residual is zero; for measured decimal data, also inspect the signed distance and stated tolerance.
What are the equations of the xy, xz, and yz planes?
The xy-plane is z=0, the xz-plane is y=0, and the yz-plane is x=0. Their normal vectors point along the z-, y-, and x-axes, respectively.