∫Calc Practice

Projections and orthogonality

Problem 9.350 · medium

Find \( \displaystyle \operatorname{proj}_{\mathbf v} \mathbf u \) for \( \displaystyle \mathbf u = \left\langle -2, 1 \right\rangle \), \( \displaystyle \mathbf v = \left\langle 0, 4 \right\rangle \).
  1. \[ 0 + 4 = 4 \]
    u·v.✓ Proved
  2. \[ 0 + 16 = 16 \]
    ‖v‖².✓ Proved
  3. \[ \left[\begin{matrix}0\\1\end{matrix}\right] \]
    proj = (u·v/‖v‖²) v.✓ Proved
Answer \( \left\langle 0, 1 \right\rangle \)

✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.

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
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the remainder is orthogonal to v and the projection is parallel to it

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly applies the vector projection formula and the arithmetic is verified.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-07
  • qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the vector projection formula and the arithmetic is verified.
  • gpt-oss:20b: pass 2026-10-07
  • qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the vector projection formula and computes the result accurately.

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/vector_projection, checked 2026-10-07 with SymPy 1.14.0.