Lines in space
Problem 9.224 · medium
Are the lines \( \displaystyle \mathbf r_1(t) = \left\langle 1 - 3 t, 2 t - 1, 3 - 2 t \right\rangle \) and \( \displaystyle \mathbf r_2(s) = \left\langle - s - 3, - s, 1 \right\rangle \) parallel, intersecting, or skew? If they intersect, find the point.
- \[ \left[\begin{matrix}-2\\2\\5\end{matrix}\right] \]d₁ × d₂: zero exactly when the directions are parallel.✓ Proved
- \[ 0 \](P₂ − P₁)·(d₁ × d₂): zero means coplanar, so the lines meet.✓ Proved
- \[ \left[\begin{matrix}-2\\1\\1\end{matrix}\right] \]At t = 1, s = -1 both lines pass through <-2, 1, 1>.✓ Proved
Answer \( \text{intersecting}\text{ at } (-2, 1, 1) \)
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 |
| 3 | ✓ 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: passqwen3.6:27b-mlx: fail (error) — The solution claims the lines are parallel because the cross product of direction vectors is zero, but the direction vectors < -3, 2, -2 > and < -1, -1, 0 > are not scalar multiples. Additionally, it claims they intersect at (-2, 1, 1), but this point does not lie on the second line (z-coordinate is 1, but y-coordinate for s=-1 is 1, x is -2; wait, let's check: r2(-1) = < -(-1)-3, -(-1), 1 > = < -2, 1, 1 >. So the point is on both lines. However, the first claim that they are parallel is false. The cross product is < 2, -2, 5 >, not zero. The solution's equation 1 is marked 'proved' but is mathematically false. The logic 'zero exactly when directions are parallel' is correct, but the premise that the cross product is zero is false. Therefore, the lines are not parallel. They are intersecting. The final answer 'intersecting' is correct, but the reasoning contains a major error claiming they are parallel first or that the cross product is zero. Actually, looking at the structure: Line 1 says d1 x d2 = < -2, 2, 5 >. Then it says 'zero exactly when...'. It doesn't explicitly say 'it is zero'. But Line 2 says '(P2-P1) . (d1 x d2) = 0'. If d1 x d2 is not zero, this just means they are coplanar. Coplanar lines are either parallel or intersecting. Since they are not parallel (cross product non-zero), they must intersect. The solution's text for line 1 is confusing/wrong if it implies the cross product is zero. But more importantly, line 1 is marked 'proved' with a false equality if it implies the cross product is the zero vector? No, it shows the vector < -2, 2, 5 >. The comment says 'zero exactly when...'. This is a conditional statement. It doesn't assert it is zero. However, usually, one checks if cross product is zero to test for parallel. Here it is not zero, so not parallel. Then one checks for intersection. The solution jumps to checking coplanarity. If coplanar and not parallel, they intersect. The solution finds a point. The logic is actually sound if interpreted as: 1. Calculate cross product (not zero, so not parallel). 2. Check coplanarity (dot product is zero, so coplanar). 3. Since coplanar and not parallel, they intersect. Find point. The only issue is the comment on line 1 might be misleading if it implies the result is zero, but it just states the condition for parallelism. Wait, the prompt says 'find what is wrong'. Is there an error? The lines are indeed intersecting at (-2, 1, 1). The cross product is < 2, -2, 5 >. The solution writes < -2, 2, 5 >. Let's recompute d1 x d2. d1 = <-3, 2, -2>, d2 = <-1, -1, 0>. i(0 - 2) - j(0 - 2) + k(3 - (-2)) = <-2, 2, 5>. The solution has < -2, 2, 5 >. This is correct. The comment 'zero exactly when...' is a general statement. It doesn't say 'this is zero'. So the logic holds: Not parallel (cross != 0). Coplanar (dot with normal = 0). Therefore intersecting. Find point. This seems correct. Why would it fail? Maybe the 'proved' tag on line 1 is suspicious if the system checked the vector arithmetic? The vector arithmetic is correct. The comment is a bit loose but not false. Let's look closer. 'd1 x d2: zero exactly when the directions are parallel.' This is a true statement. It doesn't claim d1 x d2 is zero. It just defines the test. Then line 2 checks coplanarity. If coplanar and not parallel, they intersect. The solution doesn't explicitly state 'since cross product is not zero, they are not parallel'. It just presents the cross product. A student might miss the implication. But is it an error? It's arguably 'misleading' or 'style' if it doesn't explicitly conclude 'not parallel'. However, the final answer is correct. Let's look for a harder error. Did it check the point correctly? r1(1) = < -2, 1, 1 >. r2(-1) = < -2, 1, 1 >. Yes. Is the cross product calculation correct? Yes. Is the coplanarity check correct? P2 - P1 = <-3 - 1, 0 - 2, 1 - 3> = <-4, -2, -2>. Dot with <-2, 2, 5> = 8 - 4 - 10 = -6. Wait. Line 2 says 0 = 0. Let's recompute (P2 - P1) . (d1 x d2). P1 = <1, -1, 3>. P2 = <-3, 0, 1>. P2 - P1 = <-4, 1, -2>. d1 x d2 = <-2, 2, 5>. Dot product: (-4)(-2) + (1)(2) + (-2)(5) = 8 + 2 - 10 = 0. Okay, my previous mental math for P2-P1 was wrong (I used y=0 for P2, but r2(s) = <-s-3, -s, 1>, so P2 is at s=0, which is <-3, 0, 1>. Correct. P1 is at t=0, <1, -1, 3>. Correct. Difference <-4, 1, -2>. Correct. Dot product is 0. So they are coplanar. The solution is mathematically correct. Is there a style issue? The comment on line 1 is a bit ambiguous. It doesn't explicitly say 'Since this is not zero, they are not parallel'. It just states the condition. A strict reviewer might say it fails to explicitly conclude 'not parallel'. But is that an error? It's a gap in exposition. Let's look at the options. 'pass', 'fail'. If it's correct, pass. If it's misleading, fail. The comment 'zero exactly when...' is a definition. It doesn't assert the value is zero. It's fine. I will pass it.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims the lines are parallel because the cross product of direction vectors is zero, but the direction vectors < -3, 2, -2 > and < -1, -1, 0 > are not scalar multiples. Additionally, it claims they intersect at (-2, 1, 1), but this point does not lie on the second line (z-coordinate is 1, but y-coordinate for s=-1 is 1, x is -2; wait, let's check: r2(-1) = < -(-1)-3, -(-1), 1 > = < -2, 1, 1 >. So the point is on both lines. However, the first claim that they are parallel is false. The cross product is < 2, -2, 5 >, not zero. The solution's equation 1 is marked 'proved' but is mathematically false. The logic 'zero exactly when directions are parallel' is correct, but the premise that the cross product is zero is false. Therefore, the lines are not parallel. They are intersecting. The final answer 'intersecting' is correct, but the reasoning contains a major error claiming they are parallel first or that the cross product is zero. Actually, looking at the structure: Line 1 says d1 x d2 = < -2, 2, 5 >. Then it says 'zero exactly when...'. It doesn't explicitly say 'it is zero'. But Line 2 says '(P2-P1) . (d1 x d2) = 0'. If d1 x d2 is not zero, this just means they are coplanar. Coplanar lines are either parallel or intersecting. Since they are not parallel (cross product non-zero), they must intersect. The solution's text for line 1 is confusing/wrong if it implies the cross product is zero. But more importantly, line 1 is marked 'proved' with a false equality if it implies the cross product is the zero vector? No, it shows the vector < -2, 2, 5 >. The comment says 'zero exactly when...'. This is a conditional statement. It doesn't assert it is zero. However, usually, one checks if cross product is zero to test for parallel. Here it is not zero, so not parallel. Then one checks for intersection. The solution jumps to checking coplanarity. If coplanar and not parallel, they intersect. The solution finds a point. The logic is actually sound if interpreted as: 1. Calculate cross product (not zero, so not parallel). 2. Check coplanarity (dot product is zero, so coplanar). 3. Since coplanar and not parallel, they intersect. Find point. The only issue is the comment on line 1 might be misleading if it implies the result is zero, but it just states the condition for parallelism. Wait, the prompt says 'find what is wrong'. Is there an error? The lines are indeed intersecting at (-2, 1, 1). The cross product is < 2, -2, 5 >. The solution writes < -2, 2, 5 >. Let's recompute d1 x d2. d1 = <-3, 2, -2>, d2 = <-1, -1, 0>. i(0 - 2) - j(0 - 2) + k(3 - (-2)) = <-2, 2, 5>. The solution has < -2, 2, 5 >. This is correct. The comment 'zero exactly when...' is a general statement. It doesn't say 'this is zero'. So the logic holds: Not parallel (cross != 0). Coplanar (dot with normal = 0). Therefore intersecting. Find point. This seems correct. Why would it fail? Maybe the 'proved' tag on line 1 is suspicious if the system checked the vector arithmetic? The vector arithmetic is correct. The comment is a bit loose but not false. Let's look closer. 'd1 x d2: zero exactly when the directions are parallel.' This is a true statement. It doesn't claim d1 x d2 is zero. It just defines the test. Then line 2 checks coplanarity. If coplanar and not parallel, they intersect. The solution doesn't explicitly state 'since cross product is not zero, they are not parallel'. It just presents the cross product. A student might miss the implication. But is it an error? It's arguably 'misleading' or 'style' if it doesn't explicitly conclude 'not parallel'. However, the final answer is correct. Let's look for a harder error. Did it check the point correctly? r1(1) = < -2, 1, 1 >. r2(-1) = < -2, 1, 1 >. Yes. Is the cross product calculation correct? Yes. Is the coplanarity check correct? P2 - P1 = <-3 - 1, 0 - 2, 1 - 3> = <-4, -2, -2>. Dot with <-2, 2, 5> = 8 - 4 - 10 = -6. Wait. Line 2 says 0 = 0. Let's recompute (P2 - P1) . (d1 x d2). P1 = <1, -1, 3>. P2 = <-3, 0, 1>. P2 - P1 = <-4, 1, -2>. d1 x d2 = <-2, 2, 5>. Dot product: (-4)(-2) + (1)(2) + (-2)(5) = 8 + 2 - 10 = 0. Okay, my previous mental math for P2-P1 was wrong (I used y=0 for P2, but r2(s) = <-s-3, -s, 1>, so P2 is at s=0, which is <-3, 0, 1>. Correct. P1 is at t=0, <1, -1, 3>. Correct. Difference <-4, 1, -2>. Correct. Dot product is 0. So they are coplanar. The solution is mathematically correct. Is there a style issue? The comment on line 1 is a bit ambiguous. It doesn't explicitly say 'Since this is not zero, they are not parallel'. It just states the condition. A strict reviewer might say it fails to explicitly conclude 'not parallel'. But is that an error? It's a gap in exposition. Let's look at the options. 'pass', 'fail'. If it's correct, pass. If it's misleading, fail. The comment 'zero exactly when...' is a definition. It doesn't assert the value is zero. It's fine. I will pass it.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims the lines are parallel because the cross product of direction vectors is zero, but the direction vectors < -3, 2, -2 > and < -1, -1, 0 > are not scalar multiples, so their cross product is not zero. Additionally, the logic that coplanarity implies intersection is flawed for parallel lines (which could be distinct and parallel), though here the lines are actually skew.gpt-oss:20b: pass 2026-10-04
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.