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A plane in 3D space — a perfectly flat, infinitely-extending surface — is fully and uniquely determined either by three non-collinear points that lie on it, or by a single point on the plane plus a normal vector that points perpendicular (straight out) from it. Finding a plane's equation, Ax+By+Cz+D=0, is a routine step in 3D computer graphics (defining a wall, floor, or clipping surface), structural and mechanical engineering (describing a flat face of a part or a load-bearing surface), and multivariable calculus (finding a tangent plane to a surface at a point).
How it works
Choose your given information. If you provide three points, the calculator first builds two direction vectors lying within the plane (by subtracting one point from each of the other two), then takes their cross product — a vector operation guaranteed to produce a new vector perpendicular to both original vectors, and therefore perpendicular to the entire plane. That cross product becomes the plane's normal vector (A, B, C). If you instead provide a point plus a normal vector directly, that normal is used as-is. Either way, the constant D is then solved using any one known point (x₀, y₀, z₀) on the plane: D = −(Ax₀ + By₀ + Cz₀), which comes from requiring that point satisfy the plane's equation exactly.
- Enter given information.
- Enter point 1 — x (three points).
- Enter point 1 — y (three points).
- Enter point 1 — z (three points).
- Enter point 2 — x (three points).
- Enter point 2 — y (three points).
- Enter point 2 — z (three points).
- Enter point 3 — x (three points).
- Enter point 3 — y (three points).
- Enter point 3 — z (three points).
- Enter point — x (point + normal).
- Enter point — y (point + normal).
- Enter point — z (point + normal).
- Enter normal — x (point + normal).
- Enter normal — y (point + normal).
- Enter normal — z (point + normal).
- Click Calculate to see your results.
Examples
Points (1,0,0), (0,1,0), (0,0,1)
Two direction vectors are (-1,1,0) and (-1,0,1). Their cross product is (1,1,1), giving the plane 1(x-1)+1(y-0)+1(z-0)=0, which simplifies to x+y+z-1=0 — the familiar unit-intercept plane.
Points (2,0,0), (0,3,0), (0,0,4)
The direction vectors' cross product simplifies to (6,4,3), giving the plane 6(x-2)+4(y-0)+3(z-0)=0, which simplifies to 6x+4y+3z-12=0 — matching the intercept form x/2+y/3+z/4=1.
Point (1,2,3), normal vector (2,-1,4)
Using the point-normal form directly: 2(x-1)-1(y-2)+4(z-3)=0, which simplifies to 2x-y+4z-12=0.
Who should use it
- Finding a tangent plane to a surface in multivariable calculus.
- Defining a flat face or wall in 3D modeling or CAD.
- Solving a physics or engineering problem involving a flat load-bearing surface.
Industry applications
- 3D computer graphics and CAD
- Structural and mechanical engineering
- Multivariable calculus education
Advantages
- Handles both the three-point and point-plus-normal input methods.
- Directly usable for further 3D geometry (distance, intersection, tangent-plane problems).
Limitations
- Requires genuinely non-collinear points — collinear input produces no valid result.
Common mistakes to avoid
- Entering three points in a way that makes two of the direction vectors parallel (collinear points), which fails to define a unique plane.
- Forgetting that swapping the order of the cross product's two vectors flips the sign of the resulting normal vector.
- Mixing up a point-based and normal-based approach midway through, using coordinates meant for one method with the formula for the other.
Best practices
- Double check that your three points aren't collinear (all lying on one straight line) before computing, since that produces an invalid, non-unique plane.
- Verify your final equation by substituting one of the original points back in — it should make the equation equal exactly zero.
- Remember that scaling all four coefficients by the same constant doesn't change the plane at all — different-looking equations can be equivalent.
Tips
- Need to find where two planes intersect? Use the Line of Intersection of Two Planes Calculator.