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Cross product

Problem 9.64 · easy

Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 0, -4, -2 \rangle,\ \mathbf{v} = \langle 2, -2, 1 \rangle \), and the area of the parallelogram they span.

Expand the determinant with rows i j k, u, v.
Reviewed
\[ \left[\begin{matrix}-8\\-4\\8\end{matrix}\right] \]
The components.Reviewed
\[ 12 \]
Area = |u × v|.Reviewed

Answer \( \mathbf{u}\times\mathbf{v} = \langle -8, -4, 8 \rangle,\quad \text{area} = 12 \)

Lines: 3 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
3	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	Lagrange's identity √(\|u\|²\|v\|² − (u·v)²) gives the same area

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly computes the cross product components and the magnitude. The steps are logically sound and the final answer is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly computes the cross product components and the magnitude. The steps are logically sound and the final answer is correct.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly computes the cross product components and the magnitude. The final answer 12 matches the calculated area.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly state the cross product vector \mathbf{u} \times \mathbf{v} = \langle -8, -4, 8 \rangle, which is the first part of the question. It jumps directly to the area calculation without presenting the required vector result.
gpt-oss:20b: pass 2026-09-26

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/cross_product_area, checked 2026-10-13 with SymPy 1.14.0.