Home›Calculus 3›Normal lines and normal vectors to surfaces›Problem 10.611

Normal lines and normal vectors to surfaces

Problem 10.611 · medium

Find a normal vector and the normal line to the surface \( \displaystyle x y z = 0 \) at \( \displaystyle (0, 1, 2) \).

\[ \left. x y z \right|_{\substack{ x=0\\ y=1\\ z=2 }} = 0 \]
The point is on the surface.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial x} x y z \right|_{\substack{ x=0\\ y=1\\ z=2 }}\\\left. \frac{\partial}{\partial y} x y z \right|_{\substack{ x=0\\ y=1\\ z=2 }}\\\left. \frac{\partial}{\partial z} x y z \right|_{\substack{ x=0\\ y=1\\ z=2 }}\end{matrix}\right] = \left[\begin{matrix}2\\0\\0\end{matrix}\right] \]
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t \right|_{\substack{ t=0 }}\\\left. 1 \right|_{\substack{ t=0 }}\\\left. 2 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}0\\1\\2\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved

Answer \( \mathbf n = \left\langle 2, 0, 0 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t, 1, 2 \right\rangle \)

Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the claimed normal is parallel to a finite-difference gradient

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The surface xyz=0 is a union of coordinate planes and is not smooth at (0,1,2) (where x=0), so the gradient method for finding a unique normal vector is invalid. The gradient <2,0,0> is not a well-defined normal to the surface at that point in the standard sense.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-17 — [domain objection, downgraded to style] The surface xyz=0 is a union of coordinate planes and is not smooth at (0,1,2) (where x=0), so the gradient method for finding a unique normal vector is invalid. The gradient <2,0,0> is not a well-defined normal to the surface at that point in the standard sense.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The surface xyz = 0 is a union of coordinate planes, which is not smooth at (0, 1, 2) (it lies on the intersection of the plane yz=0? No, x=0. At x=0
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The point (0, 1, 2) is a singular point of the surface xyz=0 because the gradient is zero there. Therefore, the surface does not have a well-defined tangent plane or normal vector at this point, making the calculation of a normal line via the gradient invalid.
gpt-oss:20b: pass 2026-10-14

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/normal_line, checked 2026-10-17 with SymPy 1.14.0.