Vectors in the plane
Problem 9.433 · easy
Find the angle \( \displaystyle \theta \in [0, 2\pi) \) that \( \displaystyle \mathbf u = \left\langle \sqrt{2}, - \sqrt{2} \right\rangle \) makes with the positive \( \displaystyle x \)-axis.
- \[ 2 \]‖u‖.✓ Proved
- \[ \left[\begin{matrix}\sqrt{2}\\- \sqrt{2}\end{matrix}\right] \]u = ‖u‖⟨cos θ, sin θ⟩ with θ = 7*pi/4, in the right quadrant.✓ Proved
Answer \( \theta = \frac{7 \pi}{4} \)
✓ 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 that the given angle corresponds to the vector components in the correct quadrant.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly identifies the magnitude and verifies that the given angle corresponds to the vector components in the correct quadrant.gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly identifies the magnitude and verifies the angle using the standard polar form, confirming the vector lies in the fourth quadrant with the correct reference angle.
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 2026-10-09 with SymPy 1.14.0.