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Lines in space

Problem 9.229 · easy

Are the lines \( \displaystyle \mathbf r_1(t) = \left\langle - t - 1, - t, - t - 2 \right\rangle \) and \( \displaystyle \mathbf r_2(s) = \left\langle - 3 s - 3, 1 - s, 2 s - 4 \right\rangle \) parallel, intersecting, or skew? If they intersect, find the point.

\[ \left[\begin{matrix}-3\\5\\-2\end{matrix}\right] \]
d₁ × d₂: zero exactly when the directions are parallel.✓ Proved
\[ 15 \]
(P₂ − P₁)·(d₁ × d₂): zero means coplanar, so the lines meet.✓ Proved

Answer \( \text{skew} \)

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

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	✓ 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	the closest approach of the two lines, found numerically

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution incorrectly concludes the lines are skew despite finding that the scalar triple product is non-zero, which actually implies the lines are skew. However, the comment on line 2 claims 'zero means coplanar, so the lines meet', but the calculated value is 15 (non-zero), meaning they are NOT coplanar and thus skew. The logic in the comments is contradictory to the result shown, or rather, the result 15 proves they are skew, but the text implies a check for intersection that was misinterpreted. Wait, if the cross product is non-zero (line 1), they are not parallel. If the scalar triple product is non-zero (line 2), they are not coplanar, hence skew. The stated answer is 'skew'. The calculation 15=15 is correct for non-coplanarity. The comment 'zero means coplanar, so the lines meet' is a general statement, but since the value is 15, they do NOT meet. The solution arrives at the correct classification 'skew' but the explanatory text is confusingly phrased or potentially misleading if it implies the calculation showed they meet. Actually, looking closer: Line 1 shows d1 x d2 is non-zero, so not parallel. Line 2 shows (P2-P1).(d1xd2) = 15 != 0, so not coplanar. Therefore, they are skew. The solution is mathematically correct in its derivation of 'skew'. The comment on line 2 is a conditional statement 'zero means...'. It does not claim the value IS zero. It explains the test. Since the value is 15, the condition for coplanarity is not met. Thus, the lines are skew. The solution is correct. Why did I think it was wrong? Let me re-read carefully. Line 1: Cross product is non-zero -> Not parallel. Line 2: Scalar triple product is 15 -> Not coplanar -> Skew. Conclusion: Skew. The comments are: 1. 'zero exactly when the directions are parallel.' -> Implies if non-zero, not parallel. Correct. 2. 'zero means coplanar, so the lines meet.' -> This is slightly imprecise (coplanar lines might be parallel, but we already know they aren't), but the implication is that if it were zero, they would intersect. Since it is 15, they don't. Is there an error? Let's check the algebra quickly to ensure the 'proved' lines are actually correct, as the prompt says 'do not re-check the algebra of a line marked proved'. I must trust the algebra. If the algebra is trusted: 1. d1 x d2 != 0 => Not parallel. 2. (P2-P1) . (d1 x d2) != 0 => Not coplanar => Skew. The solution concludes 'skew'. This is correct. Wait, look at the comment on line 2 again: 'zero means coplanar, so the lines meet.' If the lines were coplanar and not parallel, they would meet. Since we established they are not parallel, this implication holds. Is there a subtle error? Maybe the vectors? r1(t) = <-t-1, -t, -t-2>. d1 = <-1, -1, -1>. P1 = <-1, 0, -2> (at t=0). r2(s) = <-3s-3, 1-s, 2s-4>. d2 = <-3, -1, 2>. P2 = <-3, 1, -4> (at s=0). d1 x d2 = |i j k; -1 -1 -1; -3 -1 2| = i(-2-1) - j(-2-3) + k(1-3) = <-3, 5, -2>. Matches line 1. P2 - P1 = <-3 - (-1), 1 - 0, -4 - (-2)> = <-2, 1, -2>. (P2-P1) . (d1 x d2) = <-2, 1, -2> . <-3, 5, -2> = 6 + 5 + 4 = 15. Matches line 2. Since 15 != 0, they are skew. The solution is correct. The comments are standard explanations of the tests. Verdict: pass.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly concludes the lines are skew despite finding that the scalar triple product is non-zero, which actually implies the lines are skew. However, the comment on line 2 claims 'zero means coplanar, so the lines meet', but the calculated value is 15 (non-zero), meaning they are NOT coplanar and thus skew. The logic in the comments is contradictory to the result shown, or rather, the result 15 proves they are skew, but the text implies a check for intersection that was misinterpreted. Wait, if the cross product is non-zero (line 1), they are not parallel. If the scalar triple product is non-zero (line 2), they are not coplanar, hence skew. The stated answer is 'skew'. The calculation 15=15 is correct for non-coplanarity. The comment 'zero means coplanar, so the lines meet' is a general statement, but since the value is 15, they do NOT meet. The solution arrives at the correct classification 'skew' but the explanatory text is confusingly phrased or potentially misleading if it implies the calculation showed they meet. Actually, looking closer: Line 1 shows d1 x d2 is non-zero, so not parallel. Line 2 shows (P2-P1).(d1xd2) = 15 != 0, so not coplanar. Therefore, they are skew. The solution is mathematically correct in its derivation of 'skew'. The comment on line 2 is a conditional statement 'zero means...'. It does not claim the value IS zero. It explains the test. Since the value is 15, the condition for coplanarity is not met. Thus, the lines are skew. The solution is correct. Why did I think it was wrong? Let me re-read carefully. Line 1: Cross product is non-zero -> Not parallel. Line 2: Scalar triple product is 15 -> Not coplanar -> Skew. Conclusion: Skew. The comments are: 1. 'zero exactly when the directions are parallel.' -> Implies if non-zero, not parallel. Correct. 2. 'zero means coplanar, so the lines meet.' -> This is slightly imprecise (coplanar lines might be parallel, but we already know they aren't), but the implication is that if it were zero, they would intersect. Since it is 15, they don't. Is there an error? Let's check the algebra quickly to ensure the 'proved' lines are actually correct, as the prompt says 'do not re-check the algebra of a line marked proved'. I must trust the algebra. If the algebra is trusted: 1. d1 x d2 != 0 => Not parallel. 2. (P2-P1) . (d1 x d2) != 0 => Not coplanar => Skew. The solution concludes 'skew'. This is correct. Wait, look at the comment on line 2 again: 'zero means coplanar, so the lines meet.' If the lines were coplanar and not parallel, they would meet. Since we established they are not parallel, this implication holds. Is there a subtle error? Maybe the vectors? r1(t) = <-t-1, -t, -t-2>. d1 = <-1, -1, -1>. P1 = <-1, 0, -2> (at t=0). r2(s) = <-3s-3, 1-s, 2s-4>. d2 = <-3, -1, 2>. P2 = <-3, 1, -4> (at s=0). d1 x d2 = |i j k; -1 -1 -1; -3 -1 2| = i(-2-1) - j(-2-3) + k(1-3) = <-3, 5, -2>. Matches line 1. P2 - P1 = <-3 - (-1), 1 - 0, -4 - (-2)> = <-2, 1, -2>. (P2-P1) . (d1 x d2) = <-2, 1, -2> . <-3, 5, -2> = 6 + 5 + 4 = 15. Matches line 2. Since 15 != 0, they are skew. The solution is correct. The comments are standard explanations of the tests. Verdict: pass.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly concludes the lines are skew based on a non-zero cross product, but fails to check for intersection. The lines actually intersect at t=1, s=0 (point (-2, -1, -3)), so they are not skew. The comment on line 2 is also mathematically confused regarding coplanarity.
gpt-oss:20b: fail (misleading) 2026-10-04 — The solution misinterprets the condition for coplanarity: (P₂−P₁)·(d₁×d₂)=15≠0, so the lines are not coplanar and therefore skew. The conclusion is correct, but the reasoning is false.

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 2026-10-04 with SymPy 1.14.0.