Normal lines and normal vectors to surfaces
Problem 10.696 · medium
Find a normal vector and the normal line to the surface \( \displaystyle x^{2} + y^{2} + z^{2} = 3 \) at \( \displaystyle (1, 1, 1) \).
- \[ \left. x^{2} + y^{2} + z^{2} \right|_{\substack{ x=1\\ y=1\\ z=1 }} = 3 \]The point is on the surface.✓ Proved
- \[ \left[\begin{matrix}\left. \frac{\partial}{\partial x} \left(x^{2} + y^{2} + z^{2}\right) \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial y} \left(x^{2} + y^{2} + z^{2}\right) \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial z} \left(x^{2} + y^{2} + z^{2}\right) \right|_{\substack{ x=1\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}2\\2\\2\end{matrix}\right] \]∇F at the point is normal to the level surface.✓ Proved
- \[ \left[\begin{matrix}\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. 2 t + 1 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\1\\1\end{matrix}\right] \]The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, 2, 2 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, 2 t + 1, 2 t + 1 \right\rangle \)
✓ 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: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the gradient as the normal vector and constructs the parametric equation of the normal line using the point and the gradient direction. All steps are logically sound and algebraically verified.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the gradient as the normal vector and constructs the parametric equation for the normal line. The algebraic checks confirm the point lies on the surface and the parameterization passes through the point.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.