Vectors in the plane
Problem 9.600 · easy
Find the angle \( \displaystyle \theta \in [0, 2\pi) \) that \( \displaystyle \mathbf u = \left\langle -1, - \sqrt{3} \right\rangle \) makes with the positive \( \displaystyle x \)-axis.
- \[ 2 \]‖u‖.✓ Proved
- \[ \left[\begin{matrix}-1\\- \sqrt{3}\end{matrix}\right] \]u = ‖u‖⟨cos θ, sin θ⟩ with θ = 4*pi/3, in the right quadrant.✓ Proved
Answer \( \theta = \frac{4 \pi}{3} \)
✓ 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.
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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | atan2 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the magnitude and verifies the angle using the standard polar form definition. The steps are logically sound and lead to the correct answer.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly identifies the magnitude and verifies the angle using the standard polar form definition. The steps are logically sound and lead to the correct answer.gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly identifies the magnitude and verifies that the vector components match the polar form for the given angle in the third quadrant.
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/vectors_plane, checked 2026-10-13 with SymPy 1.14.0.