Vectors in the plane
Problem 9.273 · easy
Find the component form of the vector \( \displaystyle \mathbf u \) with \( \displaystyle \|\mathbf u\| = 2 \) at angle \( \displaystyle \theta = \frac{11 \pi}{6} \) from the positive \( \displaystyle x \)-axis.
- \[ \left[\begin{matrix}\sqrt{3}\\-1\end{matrix}\right] \]u = ‖u‖⟨cos θ, sin θ⟩.✓ Proved
Answer \( \left\langle \sqrt{3}, -1 \right\rangle \)
✓ 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ 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 | floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the formula for converting polar coordinates to component form, yielding the correct vector.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly applies the formula for converting polar coordinates to component form, yielding the correct vector.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly applies the formula for converting polar coordinates to component form, yielding the correct vector.gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.