The formula this page uses
Plane Ax + By + Cz = D → n = (A, B, C) From three points: n = PQ × PR Surface F = 0: n = ∇F
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Normal Vector Calculator
Solve normal vector problems with clear steps, notation, and a final check.
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Calculate without using AI.
Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.
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Vectors
Normal Vector Calculator explained
The short version
- A normal vector sticks out at a right angle to a surface, a plane, or a line.
- For a plane written Ax + By + Cz = D, the normal is simply (A, B, C). The D plays no part.
- Given three points, the cross product of two edge vectors gives a normal in one step.
What each part means
SymbolWhat it means
n — Normal vectorPerpendicular to the surface. Any non-zero multiple of it is also a normal.
A, B, C — Plane coefficientsThe numbers in front of x, y and z once the plane is in that form.
PQ, PR — Two edge vectorsBuilt by subtracting one corner from each of the others.
∇F — GradientFor a surface written F(x, y, z) = 0, the gradient points along the normal.
Show your work: a full example
- Three points on a planeP(1, 0, 2), Q(3, 1, 2), R(2, −1, 5)
- Build two edge vectors from PPQ = (2, 1, 0) and PR = (1, −1, 3)
- Cross product, i component(1)(3) − (0)(−1) = 3
- Cross product, j component (note the minus)−[(2)(3) − (0)(1)] = −6
- Cross product, k component(2)(−1) − (1)(1) = −3
- So the normal isn = (3, −6, −3), which simplifies to (1, −2, −1)
- Write the plane and checkx − 2y − z = −1, and Q gives 3 − 2 − 2 = −1 while R gives 2 + 2 − 5 = −1
A second, different case
- A different case: a curved surfacez = x² + y² at the point (1, 2, 5)
- Rewrite it so one side is zeroF(x, y, z) = x² + y² − z = 0
- Take the gradient∇F = (2x, 2y, −1)
- Evaluate at (1, 2, 5)n = (2, 4, −1)
- Make it a unit normal|n| = √(4 + 16 + 1) = √21 ≈ 4.583, so n̂ ≈ (0.4364, 0.8729, −0.2182)
- Write the tangent plane2(x − 1) + 4(y − 2) − (z − 5) = 0, which tidies to 2x + 4y − z = 5
- Check the point sits on it2(1) + 4(2) − 5 = 2 + 8 − 5 = 5 ✓
Copy-ready example
P(1, 0, 2), Q(3, 1, 2), R(2, −1, 5)
Three points on a plane
Questions about the Normal Vector Calculator
How many normal vectors does a plane have?
Infinitely many, but they all lie on one line through the origin. (3, −6, −3), (1, −2, −1) and (−1, 2, 1) all describe the same plane; only the length and the arrow direction differ.
Why does the cross product give a normal?
Because the cross product of two vectors is perpendicular to both by construction. PQ and PR both lie flat inside the plane, so anything perpendicular to both must stick straight out of it.
What if the three points are in a straight line?
Then PQ and PR are parallel, their cross product is the zero vector, and no plane is determined. Three collinear points sit on infinitely many planes, so the calculation correctly refuses to pick one.
How do I get the tangent plane from a normal?
Use n · (r − r₀) = 0 with r₀ the point of contact. For n = (2, 4, −1) at (1, 2, 5) that expands to 2(x − 1) + 4(y − 2) − (z − 5) = 0, or 2x + 4y − z = 5.