Equation of a Plane Through Three Points

Understanding the plane equation through three points — concept and principle

Three points that do not lie on one line determine exactly one plane. This calculator finds that plane in the standard form ax + by + cz = d, gives its normal vector (a, b, c), and as a bonus reports the area of the triangle formed by the three points. It is used in analytic geometry and linear algebra homework, in 3D modelling and game engines (every triangle of a mesh has a plane and a normal), in surveying and in CAD.

How it is calculated

1. Enter the points as "x y z; x y z; x y z" (P₁; P₂; P₃).
2. Build two edge vectors: u = P₂ − P₁ and v = P₃ − P₁.
3. The normal is the cross product n = u × v = (u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁).
4. With n = (a, b, c), the constant is d = a·x₁ + b·y₁ + c·z₁, using P₁.
5. Triangle area = |n| ÷ 2.

The principle

The cross product of two vectors is perpendicular to both of them. Since u and v lie in the plane, n = u × v is perpendicular to the whole plane. A point X is in the plane exactly when the vector X − P₁ is perpendicular to n, that is n · (X − P₁) = 0, which expands to ax + by + cz = d. The length of u × v equals |u||v| sin(angle between them), the area of the parallelogram spanned by u and v; the triangle is half of it.

Worked example

1. Default points: P₁ = (1, 0, 0), P₂ = (0, 1, 0), P₃ = (0, 0, 1).
2. u = P₂ − P₁ = (−1, 1, 0); v = P₃ − P₁ = (−1, 0, 1).
3. n = (1·1 − 0·0, 0·(−1) − (−1)·1, (−1)·0 − 1·(−1)) = (1, 1, 1).
4. d = 1·1 + 1·0 + 1·0 = 1, so the plane is 1x + 1y + 1z = 1.
5. |n| = √3 = 1.7320508, so the triangle area is √3 ÷ 2 = 0.8660254038.

How to read the result

The coefficients (a, b, c) are the normal vector; any non-zero multiple of the equation describes the same plane, so 2x + 2y + 2z = 2 is identical. If the normal comes out (0, 0, 0), the three points are collinear and no unique plane exists. The distance from any point (x₀, y₀, z₀) to the plane is |ax₀ + by₀ + cz₀ − d| ÷ √(a² + b² + c²); for the origin here that is 1 ÷ √3 ≈ 0.577.

Common mistakes

• Entering two identical or collinear points, which gives a zero normal.
• Computing the cross product components in the wrong order and getting a sign error.
• Forgetting that reversing the point order flips the normal direction; the plane is the same, but the "front side" in graphics changes.

Frequently asked questions

How do I check that a fourth point lies on the plane?

Substitute it into ax + by + cz and see whether you get d. For the example, (1/3, 1/3, 1/3) gives 1, so it lies on the plane.

What is the intercept form of the plane?

When a, b, c and d are all non-zero, x/(d/a) + y/(d/b) + z/(d/c) = 1. The example plane cuts each axis at 1.

How do I make the normal a unit vector?

Divide (a, b, c) by its length. For (1, 1, 1) that gives (0.577, 0.577, 0.577).

Sources: Gilbert Strang, Introduction to Linear Algebra, chapter on vectors, dot and cross products · James Stewart, Calculus: Early Transcendentals, section on equations of lines and planes