Cross product
Problem 9.72 · medium
Find \( \displaystyle \mathbf{u} \times \mathbf{v} \) for \( \displaystyle \mathbf{u} = \langle 3, 1, 1 \rangle,\ \mathbf{v} = \langle -4, 4, 2 \rangle \), and the area of the parallelogram they span.
- Expand the determinant with rows i j k, u, v.
- \[ \left[\begin{matrix}-2\\-10\\16\end{matrix}\right] \]The components.✓ Proved
- \[ 6 \sqrt{10} \]Area = |u × v|.✓ Proved
Answer \( \mathbf{u}\times\mathbf{v} = \langle -2, -10, 16 \rangle,\quad \text{area} = 6 \sqrt{10} \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | 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 |
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.