Cross product
Problem 9.104 · easy
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle -4, -3, 0 \rangle,\ \mathbf{v} = \langle -2, 2, -1 \rangle \), and the area of the parallelogram they span.
- Expand the determinant with rows i j k, u, v.Reviewed
- \[ \left[\begin{matrix}3\\-4\\-14\end{matrix}\right] \]The components.✓ Proved
- \[ \sqrt{221} \]Area = |u × v|.✓ Proved
Answer \( \mathbf{u}\times\mathbf{v} = \langle 3, -4, -14 \rangle,\quad \text{area} = \sqrt{221} \)
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: passqwen3.6:27b-mlx: pass — The solution correctly computes the cross product and its magnitude. The steps are logically sound and the final answer is correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly computes the cross product and its magnitude. The steps are logically sound and the final answer is correct.gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: fail (error) 2026-09-26 — The cross product calculation is incorrect. The correct cross product is <3, -4, 14>, not <3, -4, -14>. The z-component should be (-4)(2) - (-3)(-2) = -8 - 6 = -14? Wait. u=<-4,-3,0>, v=<-2,2,-1>. i(-3*-1 - 0*2) - j(-4*-1 - 0*-2) + k(-4*2 - -3*-2) = i(3) - j(4) + k(-8-6) = <3, -4, -14>. Actually, the solution's vector <3, -4, -14> is correct. Let me re-read the prompt's provided solution carefully. The solution says the components are <3, -4, -14>. My calculation: i( (-3)(-1) - (0)(2) ) = 3. j( (-4)(-1) - (0)(-2) ) = 4, so -j(4) = -4. k( (-4)(2) - (-3)(-2) ) = -8 - 6 = -14. So <3, -4, -14> is correct. The magnitude is sqrt(9 + 16 + 196) = sqrt(221). The solution is actually correct. Why did I think it was wrong? Let me double check the problem statement. u = <-4, -3, 0>, v = <-2, 2, -1>. Cross product: i(3) - j(4) + k(-14). Vector is <3, -4, -14>. Magnitude sqrt(9+16+196) = sqrt(221). The solution is correct. I will mark it as pass.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 2026-09-26 with SymPy 1.14.0.