Home›Calculus 3›Normal lines and normal vectors to surfaces›Problem 10.792

Normal lines and normal vectors to surfaces

Problem 10.792 · medium

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

\[ \left. 2 x^{2} + 2 y^{2} - z \right|_{\substack{ x=1\\ y=0\\ z=1 }} = 1 \]
The point is on the surface.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial x} \left(2 x^{2} + 2 y^{2} - z\right) \right|_{\substack{ x=1\\ y=0\\ z=1 }}\\\left. \frac{\partial}{\partial y} \left(2 x^{2} + 2 y^{2} - z\right) \right|_{\substack{ x=1\\ y=0\\ z=1 }}\\\left. \frac{\partial}{\partial z} \left(2 x^{2} + 2 y^{2} - z\right) \right|_{\substack{ x=1\\ y=0\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}4\\0\\-1\end{matrix}\right] \]
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 4 t + 1 \right|_{\substack{ t=0 }}\\\left. 0 \right|_{\substack{ t=0 }}\\\left. 1 - t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\0\\1\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved

Answer \( \mathbf n = \left\langle 4, 0, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 4 t + 1, 0, 1 - t \right\rangle \)

✓ 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

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-11-03
gpt-oss:20b: pass 2026-11-03
qwen3.6:27b-mlx: pass 2026-11-02 — The solution correctly identifies the gradient as the normal vector and constructs the parametric equation for the normal line. The algebraic steps are verified and the logic is sound.
gpt-oss:20b: pass 2026-11-02
qwen3.6:27b-mlx: pass 2026-11-02
gpt-oss:20b: pass 2026-11-02

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 2026-11-03 with SymPy 1.14.0.