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Cross product

Problem 9.147 · medium

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

Expand the determinant with rows i j k, u, v.
Reviewed
\[ \left[\begin{matrix}4\\-16\\6\end{matrix}\right] \]
The components.✓ Proved
\[ 2 \sqrt{77} \]
Area = |u × v|.✓ Proved

Answer \( \mathbf{u}\times\mathbf{v} = \langle 4, -16, 6 \rangle,\quad \text{area} = 2 \sqrt{77} \)

Lines: 2 proved, 1 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.

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
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	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 and its magnitude to find the area of the parallelogram. The steps are logically sound and the final answer is correct.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly computes the cross product and its magnitude to find the area of the parallelogram. The steps are logically sound and the final answer is correct.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The cross product calculation is incorrect. The correct cross product is <4, -16, 6>, but the magnitude is sqrt(4^2 + (-16)^2 + 6^2) = sqrt(16 + 256 + 36) = sqrt(308) = 2*sqrt(77). Wait, let me re-calculate. 4i - 16j + 6k. Magnitude squared is 16 + 256 + 36 = 308. 308 = 4 * 77. So 2*sqrt(77) is correct. Why did I think it was wrong? Let me re-read the solution. The solution says the components are <4, -16, 6>. Let's check the cross product manually. u = <4, 1, 0>, v = <-2, 1, 4>. i(1*4 - 0*1) - j(4*4 - 0*(-2)) + k(4*1 - 1*(-2)) = i(4) - j(16) + k(4+2) = <4, -16, 6>. The magnitude is sqrt(16 + 256 + 36) = sqrt(308) = 2*sqrt(77). The solution is actually correct. Why would I flag it? Ah, looking at the provided solution text: 'Matrix([...]) ... -- The components.' It lists the vector twice? No, it just shows the vector. Is there a subtle error? The problem asks for u x v AND the area. The solution provides the vector components and the area. The area calculation is correct. The vector calculation is correct. Is there a style issue? The sentence 'Expand the determinant...' is a bit vague but acceptable. Let me look closer. Is the vector <4, -16, 6> correct? Yes. Is the area 2*sqrt(77) correct? Yes. So why 'fail'? Maybe I'm missing something. Let's re-read the prompt. 'Find u x v ... and the area'. The solution provides the vector and the area. It seems correct. Wait, looking at the JSON output format, I need to be adversarial. Is there a sign error? u x v = <4, -16, 6>. v x u = <-4, 16, -6>. The problem asks for u x v. The solution gives <4, -16, 6>. This is correct. The area is the magnitude. This is correct. Perhaps the 'error' is that the solution doesn't explicitly state the final vector answer in a standard form, just shows a matrix? Or maybe the 'unchecked' equation line is confusing? No, the instructions say equations are checked by CAS. The sentences are what I check. Sentence 1: 'Expand the determinant with rows i j k, u, v.' This is a valid method. Sentence 2 is an equation. Sentence 3 is an equation. There are no other sentences. The solution seems sound. Why would it fail? Let me double check the arithmetic one more time. u = <4, 1, 0>, v = <-2, 1, 4>. i component: 1*4 - 0*1 = 4. j component: -(4*4 - 0*(-2)) = -16. k component: 4*1 - 1*(-2) = 6. Vector: <4, -16, 6>. Magnitude: sqrt(16 + 256 + 36) = sqrt(308) = sqrt(4*77) = 2*sqrt(77). Everything is correct. I will mark it as pass.
gpt-oss:20b: pass 2026-09-28

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 2026-09-28 with SymPy 1.14.0.