Projections and orthogonality
Problem 9.585 · easy
Which pairs of \( \displaystyle \mathbf a = \left\langle -1, -2, -3 \right\rangle \), \( \displaystyle \mathbf b = \left\langle -14, 1, 4 \right\rangle \), \( \displaystyle \mathbf c = \left\langle -1, 2, 4 \right\rangle \) are orthogonal?
- \[ \left[\begin{matrix}0\\-15\\32\end{matrix}\right] \]a·b, a·c, b·c.✓ Proved
- Two nonzero vectors are orthogonal exactly when their dot product is 0.Reviewed
Answer \( \mathbf a \perp \mathbf b \)
Lines: 1 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | dot products recomputed |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly computes the dot products and applies the definition of orthogonality. Since only a·b is 0, the conclusion that a and b are orthogonal is correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly computes the dot products and applies the definition of orthogonality. Since only a·b is 0, the conclusion that a and b are orthogonal is correct.qwen3.6:27b-mlx: fail (error) 2026-10-13 — The solution calculates the dot products but fails to explicitly state which pairs are orthogonal based on the results. Specifically, it identifies a·b = 0, but ignores that a·c = -15 and b·c = 32 are non-zero, so it does not fully answer 'which pairs' (implying a check of all pairs) or explicitly conclude that only a and b are orthogonal. More critically, the final stated answer is just 'a perp b', but the solution text provided ends at the definition of orthogonality without connecting the calculated values to the conclusion. The sentence provided is a definition, not a conclusion derived from the calculation. It misses the logical step: 'Since a·b = 0, a and b are orthogonal. Since a·c != 0 and b·c != 0, the other pairs are not.'gpt-oss:20b: pass 2026-10-13
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 2026-10-13 with SymPy 1.14.0.