Calculators / Calculus 3 / Plane Through Three Points (3D) Calculator

Plane Through Three Points (3D) Calculator

Published on: May 17, 2026

This Plane Through Three Points (3D) Calculator helps you find the equation of a plane in three-dimensional space that passes through three given points. First form two direction vectors from the points, then use their cross product to find a normal vector to the plane. Substitute that normal vector and one of the points into the point-normal form and simplify to get the equation. It is a simple way to check answers, understand the method clearly, and practise 3D coordinate geometry step by step.

Step-by-step method

  1. Identify the three points P₁, P₂, P₃.
  2. Find two direction vectors in the plane and compute the normal vector n = v × w.
  3. Use the point-normal plane equation and simplify.

Formula:

n = v × w
a(x − x₁) + b(y − y₁) + c(z − z₁) = 0
ax + by + cz = d, where d = ax₁ + by₁ + cz₁

Example 1: (1,2,3),(4,0,-1),(2,5,6)

Step 1 - Identify the three points.

In this problem: Use the three given points P₁, P₂, P₃.

P₁ = ( 1, 2, 3 )
P₂ = ( 4, 0, -1 )
P₃ = ( 2, 5, 6 )

Step 2a - Compute two direction vectors in the plane.

In this problem: Compute v = P₂ − P₁ and w = P₃ − P₁.

v = P₂ − P₁ = ( 3, -2, -4 )
w = P₃ − P₁ = ( 1, 3, 3 )

Step 2b - Write the cross product.

In this problem: The normal vector is n = v × w.

n = v × w

Step 2c - Compute the normal vector.

In this problem: Compute n from v × w.

n = ⟨6, -13, 11⟩

Step 3a - Write the point-normal plane equation.

In this problem: Use a(x − x₁) + b(y − y₁) + c(z − z₁) = 0.

a(x − x₁) + b(y − y₁) + c(z − z₁) = 0

Step 3b - Substitute values.

In this problem: Use P₁ as the point and n = ⟨a,b,c⟩ as the normal.

6(x − 1) + -13(y − 2) + 11(z − 3) = 0

Step 3c - Write standard form.

In this problem: Rearrange into ax + by + cz = d.

−13y + 6x + 11z = 13

Final answer: n=<6,-13,11>, plane: 6*x - 13*y + 11*z=13

Example 2: (1/2,0,-3),(3,2,-4),(2,1,1)

Step 1 - Identify the three points.

In this problem: Use the three given points P₁, P₂, P₃.

P₁ = (
1
2
, 0, -3 )
P₂ = ( 3, 2, -4 )
P₃ = ( 2, 1, 1 )

Step 2a - Compute two direction vectors in the plane.

In this problem: Compute v = P₂ − P₁ and w = P₃ − P₁.

v = P₂ − P₁ = (
5
2
, 2, -1 )
w = P₃ − P₁ = (
3
2
, 1, 4 )

Step 2b - Write the cross product.

In this problem: The normal vector is n = v × w.

n = v × w

Step 2c - Compute the normal vector.

In this problem: Compute n from v × w.

n = ⟨9,
-23
2
,
-1
2
⟩

Step 3a - Write the point-normal plane equation.

In this problem: Use a(x − x₁) + b(y − y₁) + c(z − z₁) = 0.

a(x − x₁) + b(y − y₁) + c(z − z₁) = 0

Step 3b - Substitute values.

In this problem: Use P₁ as the point and n = ⟨a,b,c⟩ as the normal.

9(x −
1
2
) +
-23
2
(y − 0) +
-1
2
(z − -3) = 0

Step 3c - Write standard form.

In this problem: Rearrange into ax + by + cz = d.

9x −
23
2
y −
1
2
z = 6

Final answer: n=<9,-23/2,-1/2>, plane: 9*x - 23*y/2 - z/2=6