Lines in space
Problem 9.222 · medium
Are the lines \( \displaystyle \mathbf r_1(t) = \left\langle 2 - 3 t, 4 - 3 t, 3 t - 3 \right\rangle \) and \( \displaystyle \mathbf r_2(s) = \left\langle 5, 8 - s, 2 s - 8 \right\rangle \) parallel, intersecting, or skew? If they intersect, find the point.
- \[ \left[\begin{matrix}-3\\6\\3\end{matrix}\right] \]d₁ × d₂: zero exactly when the directions are parallel.✓ Proved
- \[ 0 \](P₂ − P₁)·(d₁ × d₂): zero means coplanar, so the lines meet.✓ Proved
- \[ \left[\begin{matrix}5\\7\\-6\end{matrix}\right] \]At t = -1, s = 1 both lines pass through <5, 7, -6>.✓ Proved
Answer \( \text{intersecting}\text{ at } (5, 7, -6) \)
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 incorrectly concludes the lines intersect based on them being coplanar (cross product of directions is zero and they share a plane), but parallel coplanar lines do not necessarily intersect. The lines are actually parallel and distinct, as the point found in step 3 does not satisfy the first line's equation for t=-1 (r1(-1) = <5, 7, -6> is correct, wait. Let's recheck. r1(-1) = <2-3(-1), 4-3(-1), 3(-1)-3> = <5, 7, -6>. r2(1) = <5, 8-1, 2(1)-8> = <5, 7, -6>. So they DO intersect. Wait, let me re-read the vectors. d1 = <-3, -3, 3>. d2 = <0, -1, 2>. d1 x d2 = <-3*2 - 3*(-1), 3*0 - (-3)*2, (-3)*(-1) - (-3)*0> = <-6+3, 0+6, 3+0> = <-3, 6, 3>. This is NOT zero. So they are NOT parallel. The first equation claims d1 x d2 is zero? No, the equation shows a non-zero vector equal to itself. The comment says 'zero exactly when the directions are parallel'. The vector is NOT zero. So they are not parallel. Step 2 checks if they are coplanar. (P2-P1) dot (d1 x d2). P1=<2,4,-3>, P2=<5,8,-8>. P2-P1 = <3, 4, -5>. Dot product with <-3, 6, 3> is -9 + 24 - 15 = 0. So they are coplanar. Since they are not parallel and are coplanar, they intersect. Step 3 finds the intersection point. The logic holds, but the comment in step 1 is misleading/wrong because it implies the cross product IS zero, or that the test for parallelism passed. Actually, the equation is just an identity A=A. It doesn't prove they are parallel. It just calculates the cross product. The comment 'zero exactly when...' is a general statement, but the vector shown is not zero. A student might think the calculation resulted in zero. However, the final conclusion is correct. Is it 'error' or 'misleading'? The comment suggests the condition for parallelism was met or is being checked. Since the vector is non-zero, they are not parallel. The solution proceeds to check coplanarity. If they are coplanar and not parallel, they intersect. The logic is sound, but the presentation of step 1 is confusing. It shows a non-zero vector and comments on when it is zero. It doesn't explicitly state 'Since this is not zero, they are not parallel'. It just shows the vector. Then step 2 shows the scalar triple product is 0. This implies coplanar. Non-parallel + Coplanar = Intersecting. This is correct. However, step 1's comment is potentially misleading if interpreted as 'the result is zero'. But the result is clearly not zero. Let's look closer. 'd1 x d2: zero exactly when the directions are parallel.' This is a true statement about the cross product. It doesn't claim the current result is zero. It's a definition. So the logic is: 1. Calculate cross product (non-zero, so not parallel). 2. Check coplanarity (zero, so coplanar). 3. Find intersection. This is a valid method. Why did I think it was wrong? Ah, I misread the vector in step 1 as being zero or the comment as claiming it was zero. The comment is just a reminder of the property. The solution is actually correct. Wait, let me double check the arithmetic. d1 = <-3, -3, 3>. d2 = <0, -1, 2>. Cross: i( -3*2 - 3*(-1) ) = -6 + 3 = -3. j( 3*0 - (-3)*2 ) = 6. k( (-3)*(-1) - (-3)*0 ) = 3. Vector <-3, 6, 3>. Correct. P2-P1 = <3, 4, -5>. Dot: -9 + 24 - 15 = 0. Correct. Intersection: 2-3t = 5 => -3t=3 => t=-1. 4-3t = 8-s => 4+3 = 8-s => 7=8-s => s=1. 3t-3 = 2s-8 => -3-3 = 2-8 => -6 = -6. Correct. The solution is correct. The comments are slightly terse but not false. Step 1 comment explains what the cross product tests for. Step 2 comment explains what the scalar triple product tests for. Step 3 verifies the point. It is sound.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly concludes the lines intersect based on them being coplanar (cross product of directions is zero and they share a plane), but parallel coplanar lines do not necessarily intersect. The lines are actually parallel and distinct, as the point found in step 3 does not satisfy the first line's equation for t=-1 (r1(-1) = <5, 7, -6> is correct, wait. Let's recheck. r1(-1) = <2-3(-1), 4-3(-1), 3(-1)-3> = <5, 7, -6>. r2(1) = <5, 8-1, 2(1)-8> = <5, 7, -6>. So they DO intersect. Wait, let me re-read the vectors. d1 = <-3, -3, 3>. d2 = <0, -1, 2>. d1 x d2 = <-3*2 - 3*(-1), 3*0 - (-3)*2, (-3)*(-1) - (-3)*0> = <-6+3, 0+6, 3+0> = <-3, 6, 3>. This is NOT zero. So they are NOT parallel. The first equation claims d1 x d2 is zero? No, the equation shows a non-zero vector equal to itself. The comment says 'zero exactly when the directions are parallel'. The vector is NOT zero. So they are not parallel. Step 2 checks if they are coplanar. (P2-P1) dot (d1 x d2). P1=<2,4,-3>, P2=<5,8,-8>. P2-P1 = <3, 4, -5>. Dot product with <-3, 6, 3> is -9 + 24 - 15 = 0. So they are coplanar. Since they are not parallel and are coplanar, they intersect. Step 3 finds the intersection point. The logic holds, but the comment in step 1 is misleading/wrong because it implies the cross product IS zero, or that the test for parallelism passed. Actually, the equation is just an identity A=A. It doesn't prove they are parallel. It just calculates the cross product. The comment 'zero exactly when...' is a general statement, but the vector shown is not zero. A student might think the calculation resulted in zero. However, the final conclusion is correct. Is it 'error' or 'misleading'? The comment suggests the condition for parallelism was met or is being checked. Since the vector is non-zero, they are not parallel. The solution proceeds to check coplanarity. If they are coplanar and not parallel, they intersect. The logic is sound, but the presentation of step 1 is confusing. It shows a non-zero vector and comments on when it is zero. It doesn't explicitly state 'Since this is not zero, they are not parallel'. It just shows the vector. Then step 2 shows the scalar triple product is 0. This implies coplanar. Non-parallel + Coplanar = Intersecting. This is correct. However, step 1's comment is potentially misleading if interpreted as 'the result is zero'. But the result is clearly not zero. Let's look closer. 'd1 x d2: zero exactly when the directions are parallel.' This is a true statement about the cross product. It doesn't claim the current result is zero. It's a definition. So the logic is: 1. Calculate cross product (non-zero, so not parallel). 2. Check coplanarity (zero, so coplanar). 3. Find intersection. This is a valid method. Why did I think it was wrong? Ah, I misread the vector in step 1 as being zero or the comment as claiming it was zero. The comment is just a reminder of the property. The solution is actually correct. Wait, let me double check the arithmetic. d1 = <-3, -3, 3>. d2 = <0, -1, 2>. Cross: i( -3*2 - 3*(-1) ) = -6 + 3 = -3. j( 3*0 - (-3)*2 ) = 6. k( (-3)*(-1) - (-3)*0 ) = 3. Vector <-3, 6, 3>. Correct. P2-P1 = <3, 4, -5>. Dot: -9 + 24 - 15 = 0. Correct. Intersection: 2-3t = 5 => -3t=3 => t=-1. 4-3t = 8-s => 4+3 = 8-s => 7=8-s => s=1. 3t-3 = 2s-8 => -3-3 = 2-8 => -6 = -6. Correct. The solution is correct. The comments are slightly terse but not false. Step 1 comment explains what the cross product tests for. Step 2 comment explains what the scalar triple product tests for. Step 3 verifies the point. It is sound.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly concludes the lines intersect because they are coplanar. Coplanarity is a necessary but not sufficient condition for intersection; the solution fails to verify that the intersection point is consistent for both lines (i.e., it does not show that the same t and s values produce the same point, although the final line claims this, the logical step from line 2 to the conclusion is invalid without solving the system). Furthermore, the cross product in line 1 is not zero (d1=<-3,-3,3>, d2=<0,-1,2>, d1xd2 = <-3, 6, 3> != 0), so the lines are not parallel, but the comment implies the zero result of line 2 proves intersection directly, which is false logic.gpt-oss:20b: fail (error) 2026-10-04 — The dot product (P₂−P₁)·(d₁×d₂) is 24, not 0, so the lines are not coplanar and thus are skew, not intersecting. The conclusion about the intersection point is incorrect.
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.