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Lines in space

Problem 9.461 · easy

Are the lines \( \displaystyle \mathbf r_1(t) = \left\langle 4 - 3 t, 2 t + 3, 2 t + 3 \right\rangle \) and \( \displaystyle \mathbf r_2(s) = \left\langle 3 - 3 s, 2 s + 2, 2 s + 6 \right\rangle \) parallel, intersecting, or skew? If they intersect, find the point.

\[ \left[\begin{matrix}0\\0\\0\end{matrix}\right] \]
d₁ × d₂: zero exactly when the directions are parallel.✓ Proved
The directions are parallel and P₂ is not on L₁, so the lines are parallel and distinct.

Answer \( \text{parallel} \)

Lines: 1 proved, 1 not checked. 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	Not checked	—	a sentence; read, not computed
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the closest approach of the two lines, found numerically

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution claims the lines are parallel and distinct, but the lines actually intersect at t=1, s=1 (point (1, 5, 5)). The cross product of direction vectors is zero, confirming they are parallel, but the check for distinctness (P2 not on L1) is incorrect; P2(3, 2, 6) lies on L1 when t=1/3? No, let's recheck. L1: <4-3t, 2t+3, 2t+3>. If t=1/3, x=3, y=3.66, z=3.66. P2 is (3,2,6). Wait. L1 direction <-3, 2, 2>. L2 direction <-3, 2, 2>. They are parallel. Do they intersect? 4-3t = 3-3s => 1 = 3(t-s). 2t+3 = 2s+2 => 2(t-s) = -1 => t-s = -1/2. 2t+3 = 2s+6 => 2(t-s) = 3 => t-s = 3/2. Contradiction. So they are parallel and distinct. Wait, let me re-read the vectors. r1 = <4-3t, 2t+3, 2t+3>. r2 = <3-3s, 2s+2, 2s+6>. d1 = <-3, 2, 2>. d2 = <-3, 2, 2>. Parallel. Check intersection: 4-3t = 3-3s => 3s - 3t = -1. 2t+3 = 2s+2 => 2t - 2s = -1 => 2(s-t) = 1 => s-t = 1/2. 2t+3 = 2s+6 => 2t - 2s = 3 => 2(s-t) = -3 => s-t = -3/2. 1/2 != -3/2. So they are parallel and distinct. The solution says 'parallel'. The stated answer is 'parallel'. The solution logic is: directions parallel, P2 not on L1. Is P2 on L1? P2 is a point on L2, e.g., s=0 -> (3,2,6). Is (3,2,6) on L1? 4-3t=3 -> t=1/3. 2(1/3)+3 = 11/3 != 2. So P2 is not on L1. Thus lines are distinct. The conclusion 'parallel' is correct. Why did I think they intersect? I misread the problem or my previous mental check was wrong. Let's re-verify. d1 = <-3, 2, 2>. d2 = <-3, 2, 2>. Parallel. P1(4,3,3). P2(3,2,6). Vector P1P2 = <-1, -1, 3>. Cross product d1 x P1P2 = <-3, 2, 2> x <-1, -1, 3> = <6 - (-2), -3 - (-2), 3 - (-2)> = <8, -1, 5> != 0. So they are not the same line. They are parallel and distinct. The solution is correct. Wait, the prompt asks me to find what is WRONG. If it's correct, I say pass. Let me double check the 'Stated answer'. Stated answer: parallel. Solution concludes parallel. The logic holds. Is there a subtle error? 'P2 is not on L1'. P2 is just a point on L2. If P2 is not on L1, and directions are parallel, then L1 and L2 are distinct parallel lines. This is correct. So why would this fail? Maybe the sentence is too brief? 'The directions are parallel and P2 is not on L1, so the lines are parallel and distinct.' This is a valid logical step. Verdict: pass.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-10
qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution claims the lines are parallel and distinct, but the lines actually intersect at t=1, s=1 (point (1, 5, 5)). The cross product of direction vectors is zero, confirming they are parallel, but the check for distinctness (P2 not on L1) is incorrect; P2(3, 2, 6) lies on L1 when t=1/3? No, let's recheck. L1: <4-3t, 2t+3, 2t+3>. If t=1/3, x=3, y=3.66, z=3.66. P2 is (3,2,6). Wait. L1 direction <-3, 2, 2>. L2 direction <-3, 2, 2>. They are parallel. Do they intersect? 4-3t = 3-3s => 1 = 3(t-s). 2t+3 = 2s+2 => 2(t-s) = -1 => t-s = -1/2. 2t+3 = 2s+6 => 2(t-s) = 3 => t-s = 3/2. Contradiction. So they are parallel and distinct. Wait, let me re-read the vectors. r1 = <4-3t, 2t+3, 2t+3>. r2 = <3-3s, 2s+2, 2s+6>. d1 = <-3, 2, 2>. d2 = <-3, 2, 2>. Parallel. Check intersection: 4-3t = 3-3s => 3s - 3t = -1. 2t+3 = 2s+2 => 2t - 2s = -1 => 2(s-t) = 1 => s-t = 1/2. 2t+3 = 2s+6 => 2t - 2s = 3 => 2(s-t) = -3 => s-t = -3/2. 1/2 != -3/2. So they are parallel and distinct. The solution says 'parallel'. The stated answer is 'parallel'. The solution logic is: directions parallel, P2 not on L1. Is P2 on L1? P2 is a point on L2, e.g., s=0 -> (3,2,6). Is (3,2,6) on L1? 4-3t=3 -> t=1/3. 2(1/3)+3 = 11/3 != 2. So P2 is not on L1. Thus lines are distinct. The conclusion 'parallel' is correct. Why did I think they intersect? I misread the problem or my previous mental check was wrong. Let's re-verify. d1 = <-3, 2, 2>. d2 = <-3, 2, 2>. Parallel. P1(4,3,3). P2(3,2,6). Vector P1P2 = <-1, -1, 3>. Cross product d1 x P1P2 = <-3, 2, 2> x <-1, -1, 3> = <6 - (-2), -3 - (-2), 3 - (-2)> = <8, -1, 5> != 0. So they are not the same line. They are parallel and distinct. The solution is correct. Wait, the prompt asks me to find what is WRONG. If it's correct, I say pass. Let me double check the 'Stated answer'. Stated answer: parallel. Solution concludes parallel. The logic holds. Is there a subtle error? 'P2 is not on L1'. P2 is just a point on L2. If P2 is not on L1, and directions are parallel, then L1 and L2 are distinct parallel lines. This is correct. So why would this fail? Maybe the sentence is too brief? 'The directions are parallel and P2 is not on L1, so the lines are parallel and distinct.' This is a valid logical step. Verdict: pass.
qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-10

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/lines_in_space, checked 2026-10-10 with SymPy 1.14.0.