∫Calc Practice

Vectors in three dimensions

Problem 9.281 · medium

Find the vector \( \displaystyle \mathbf u \) with \( \displaystyle \|\mathbf u\| = 10 \) pointing in the same direction as \( \displaystyle \mathbf v = \left\langle 3, 0, -4 \right\rangle \).
  1. \[ 5 \]
    ‖v‖.✓ Proved
  2. \[ \left[\begin{matrix}6\\0\\-8\end{matrix}\right] \]
    10·v/‖v‖.✓ Proved
Answer \( \left\langle 6, 0, -8 \right\rangle \)

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. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0length checked, and parallel by the equality case of Cauchy–Schwarz

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly calculates the magnitude of v as 5 and scales it by 10/5 to get the final vector. The steps are logically sound and the algebra is verified.
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly calculates the magnitude of v as 5 and scales it by 10/5 to get the final vector. The steps are logically sound and the algebra is verified.
  • gpt-oss:20b: pass 2026-10-05
  • qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to state the method (scaling the vector by the ratio of desired magnitude to current magnitude) and provides no justification for the intermediate steps, making it impossible to verify the logic or learn the procedure.
  • 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_space, checked 2026-10-05 with SymPy 1.14.0.