In 3D, two lines have THREE possibilities: On flat paper two lines are either parallel or they cross. In 3D there is a third option, because there's room for lines to pass by each other without touching:
Parallel — same direction (one direction vector is a scalar multiple of the other). They never meet (unless they're actually the same line).
Intersecting — different directions, and they meet at exactly one point.
Skew — different directions, but they still never meet. Think of one road on a bridge crossing over another road below: not parallel, yet they never touch.
The decision procedure: Step 1 — directions. Is d₂ a scalar multiple of d₁?
• Yes → the lines are parallel (check one point to see if they're the same line or genuinely parallel).
• No → go to Step 2.
Step 2 — try to intersect. Set the two position expressions equal, giving three component equations in s and t. Solve two of them for s and t, then substitute into the third.
• Third equation holds → the lines intersect (put s back in to get the point).
• Third equation fails → the lines are skew.
How to tell parallel from skew: Both parallel and skew lines never meet, so 'no intersection' alone is not enough to call them skew. The difference is the directions: parallel = directions are multiples; skew = directions are NOT multiples but the lines still don't meet. Always check the directions first.
IB-style question — are these lines parallel?
Lines L₁ = (1, 0, 2) + s(2, 1, 3) and L₂ = (0, 4, 1) + t(4, 2, 6).
Determine whether L₁ and L₂ are parallel.
Step by step
- Compare the directions: is (4, 2, 6) a scalar multiple of (2, 1, 3)?
- Yes — every component is exactly doubled, so the directions are parallel.
- Check they aren't the same line: is L₂'s point (0, 4, 1) on L₁? Need 1 + 2s = 0 ⇒ s = −½, but then y = 0 + (−½) = −½ ≠ 4.
- The point isn't on L₁, so the lines are distinct.
Final answer
The directions are multiples (d₂ = 2d₁) and the lines are distinct, so L₁ and L₂ are parallel (and never meet).
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Solve two equations, then TEST the third: When the directions differ, set the position vectors equal. Matching components gives three equations in just two unknowns (s and t). Two equations are enough to pin down s and t — the third is the consistency test:
• It checks out → the lines really do meet. Substitute s back to read off the point.
• It contradicts → no values of s, t satisfy all three, so the lines never meet → since the directions differ, they are skew.
IB-style question — find the point of intersection
Lines L₁ = (1, 2, 1) + s(1, 1, 2) and L₂ = (7, 2, 5) + t(−2, 1, 0).
Show that the lines intersect and find the point of intersection.
Step by step
- Set the position vectors equal, component by component.
- The z-equation has only s — solve it first.
- Put s = 2 into the y-equation to get t.
- TEST in the x-equation (the one not yet used).
- It holds, so they intersect. Put s = 2 into L₁ to find the point.
Final answer
All three equations are satisfied, so the lines intersect at the point (3, 4, 5).
IB-style question — show two lines are skew
Lines L₁ = (1, 2, 0) + s(1, 1, 1) and L₂ = (2, 1, 5) + t(1, −1, 0).
Show that the lines are skew.
Step by step
- Directions first: is (1, −1, 0) a multiple of (1, 1, 1)? No — the third component would need 0 = k and 1 = k. So NOT parallel.
- Try to intersect: equate components.
- Solve the x- and y-equations. Subtracting: (1+s)−(2+s) = (2+t)−(1−t) ⇒ −1 = 1 + 2t ⇒ t = −1, then s = 0.
- TEST in the z-equation. Left side s = 0; right side is 5.
- The third equation fails, so there is no common point. Combined with non-parallel directions, the lines are skew.
Final answer
The directions aren't parallel and the equations are inconsistent (0 ≠ 5), so the lines are skew.