Home›Calculus 3›Normal lines and normal vectors to surfaces›Problem 10.698

Normal lines and normal vectors to surfaces

Problem 10.698 · medium

Find a normal vector and the normal line to the surface \( \displaystyle x^{2} + y^{2} - z = 1 \) at \( \displaystyle (1, 1, 1) \).

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

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

✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation. 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: pass — The solution correctly identifies the gradient as the normal vector and constructs the parametric equation for the normal line. All steps are logically sound and algebraically verified.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the gradient as the normal vector and constructs the parametric equation for the normal line. All steps are logically sound and algebraically verified.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the gradient as the normal vector and constructs the parametric equation for the normal line. All steps are logically sound and algebraically verified.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15
gpt-oss:20b: pass 2026-10-15

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 2026-10-18 with SymPy 1.14.0.