By Chirantan Basu | Updated Aug 30, 2022

The equation of a plane in three‑dimensional space can be expressed as ax + by + cz = d, where at least one of the constants a, b, or c is non‑zero. When three points are known, the plane can be derived using vector cross products, a reliable geometric technique that guarantees an exact solution.

Step 1 – Identify the Three Points

Label the points A, B, and C. For illustration, let A = (3, 1, 1), B = (1, 4, 2), and C = (1, 3, 4).

Step 2 – Form Two Vectors on the Plane

Choose any two vectors that lie on the plane. A convenient choice is AB and AC:

• AB = B – A = (1–3, 4–1, 2–1) = (–2, 3, 1)
• AC = C – A = (1–3, 3–1, 4–1) = (–2, 2, 3)

Step 3 – Compute the Normal Vector via Cross Product

The cross product of AB and AC yields a vector normal to the plane:

AB × AC = (a₂b₃ – a₃b₂, a₃b₁ – a₁b₃, a₁b₂ – a₂b₁)

Substituting the coordinates gives:

AB × AC = (3·3 – 1·2, 1·(–2) – (–2)·3, (–2)·2 – 3·(–2)) = (7, 4, 2)

Thus, the normal vector N is (7, 4, 2).

Step 4 – Write the Plane Equation

Using point C (or any known point) and the normal vector, the plane equation is:

7(x – 1) + 4(y – 3) + 2(z – 4) = 0

Expanding and simplifying yields the standard form:

7x + 4y + 2z = 27

Step 5 – Verify the Result

Substitute each of the original points into the equation to confirm they satisfy it. All three points satisfy 7x + 4y + 2z = 27, validating the calculation.

TL;DR

Use vector cross products to find a plane’s normal vector, then plug any point into the dot‑product form to obtain the plane’s equation.