IB Maths AI HL Vector Equations of Lines Paper 1 & 2 Skew lines ~10 min read

Shortest Distance Between Two Lines

Two skew lines (not parallel, not intersecting) have one unique line segment perpendicular to both — that’s the shortest connection. Two methods: scalar-product (set F₁F₂ · b₁ = 0 and F₁F₂ · b₂ = 0) or vector-product (use b₁ × b₂, which is perpendicular to both).

📘 What you need to know

Skew lines: not parallel, never meet. Live in 3D, not in 2D (where lines must either meet or be parallel).
Shortest distance = length of the common perpendicular between the two lines.
Cross product trick: d = b₁ × b₂ is automatically perpendicular to both directions. Use it to align the connecting segment.
Two methods:
· Scalar product: 2 perpendicularity equations in λ, μ → solve → |F₁F₂|
· Vector product: distance = |(a₂ − a₁) · (b₁ × b₂)||b₁ × b₂|
Parallel lines: pick any point on one line, treat as point-to-line distance (use the previous sub-topic).
Intersecting lines: shortest distance = 0.
How to tell parallel/skew/intersecting:
· b₁ = k b₂ ⇒ parallel (or same line)
· otherwise try to solve a₁ + λb₁ = a₂ + μb₂: solution exists ⇒ intersect; no solution ⇒ skew

Scalar-product method — two perpendicular conditions

Let F₁ on l₁ and F₂ on l₂ be the two closest points. The segment F₁F₂ must be perpendicular to both lines. So you get two scalar-product equations:

Scalar product method F₁F₂ · b₁ = 0 and F₁F₂ · b₂ = 0 two equations in λ, μ → solve → substitute → distance = |F₁F₂|

The teal segment F₁F₂ is the only line segment perpendicular to both skew lines. Its direction is parallel to b₁ × b₂ — the cross product of the direction vectors.

Vector-product shortcut — one formula

Since the common perpendicular direction is b₁ × b₂, project the gap vector a₂ − a₁ (between the two lines’ base points) onto this direction. The magnitude of that projection is the shortest distance.

Vector-product shortcut d = |(a₂ − a₁) · (b₁ × b₂)||b₁ × b₂| numerator = absolute value of the scalar triple product; denominator = magnitude of the cross product

Shortcut warning: this formula only works when the lines are skew (not parallel). If b₁ × b₂ = 0, the lines are parallel — use the point-to-line method instead. Also: if the lines actually intersect, the numerator comes out to 0 (distance = 0).

🧭 Recipe — shortest distance between two skew lines

Check first: are they parallel (b₁ ∥ b₂)? Are they intersecting (set the equations equal and solve)?
If parallel: pick a point on one, find its distance to the other (point-to-line).
If intersecting: d = 0.
If skew — pick a method:
· Vector product: d = |(a₂ − a₁) · (b₁ × b₂)| / |b₁ × b₂|. Three computations, one answer.
· Scalar product: write F₁ = a₁ + λb₁, F₂ = a₂ + μb₂, form F₁F₂, set F₁F₂ · b₁ = 0 and F₁F₂ · b₂ = 0, solve for λ and μ, then d = |F₁F₂|.
Use the GDC for cross products and simultaneous equations.

Worked examples

WE 1

Vector-product method (from the PDF)

l₁: r = (3−1−2) + λ(1−20), l₂: r = (99−9) + μ(−141). Find the shortest distance between them.

b₁ × b₂ = (1−20) × (−141) = (−2·1 − 0·4, 0·(−1) − 1·1, 1·4 − (−2)(−1)) = (−2, −1, 2) a₂ − a₁ = (9−3, 9−(−1), −9−(−2)) = (6, 10, −7) scalar triple product (a₂−a₁) · (b₁ × b₂) = 6(−2) + 10(−1) + (−7)(2) = −12 − 10 − 14 = −36 magnitudes |b₁ × b₂| = √(4 + 1 + 4) = 3 distance d = |−36| / 3 = 12 shortest distance = 12 units

WE 2

Scalar-product method (same lines as WE 1)

Solve WE 1 using the scalar-product method: find λ, μ giving the closest points, then compute |F₁F₂|.

F₁ on l₁, F₂ on l₂ F₁ = (3+λ, −1−2λ, −2) F₂ = (9−μ, 9+4μ, −9+μ) F₁F₂ = F₂ − F₁ F₁F₂ = (6−μ−λ, 10+4μ+2λ, −7+μ) F₁F₂ · b₁ = 0 (6−μ−λ)(1) + (10+4μ+2λ)(−2) + (−7+μ)(0) = 0 −5λ − 9μ = 14 …(i) F₁F₂ · b₂ = 0 (6−μ−λ)(−1) + (10+4μ+2λ)(4) + (−7+μ)(1) = 0 9λ + 18μ = −27 …(ii) solve (i) & (ii) simultaneously λ = −1, μ = −1 substitute back F₁F₂ = (6+1+1, 10−4−2, −7−1) = (8, 4, −8) |F₁F₂| = √(64 + 16 + 64) = √144 = 12 d = 12 units ✓ matches WE 1 two methods, same answer — choose whichever is faster on the day.

WE 3

Distance between parallel lines

l₁: r = (000) + λ(100) and l₂: r = (034) + μ(200). Find the shortest distance.

b₂ = 2·b₁ → directions parallel pick point A = (0, 3, 4) on l₂ point-to-line distance from A to l₁ AP = OA − origin = (0, 3, 4) AP × b₁ = (0, 3, 4) × (1, 0, 0) = (3·0−4·0, 4·1−0·0, 0·0−3·1) = (0, 4, −3) |AP × b₁| = √(0 + 16 + 9) = 5 |b₁| = 1 d = 5 / 1 = 5 units never use the cross-product 2-line formula on parallel lines — b₁ × b₂ = 0 causes division by zero.

WE 4

Classify the relationship

Determine whether the lines l₁: r = (100) + λ(120) and l₂: r = (001) + μ(211) are parallel, intersecting, or skew.

directions parallel? (1, 2, 0) vs (2, 1, 1) — not scalar multiples not parallel do they intersect? equate x: 1 + λ = 2μ y: 2λ = μ z: 0 = 1 + μ → μ = −1 from y: λ = −1/2; check x x: 1 − 1/2 = 1/2; 2(−1) = −2 ✗ no consistent (λ, μ) → don’t intersect lines are skew 3 equations, 2 unknowns — solve any 2, verify the third.

WE 5

Vector-product formula in practice

Find the shortest distance between the skew lines from WE 4.

b₁ × b₂ = (1, 2, 0) × (2, 1, 1) = (2·1 − 0·1, 0·2 − 1·1, 1·1 − 2·2) = (2, −1, −3) |b₁ × b₂| = √(4 + 1 + 9) = √14 a₂ − a₁ = (−1, 0, 1) (a₂ − a₁) · (b₁ × b₂) = (−1)(2) + 0(−1) + 1(−3) = −5 distance d = |−5| / √14 = 5/√14 d = 5/√14 ≈ 1.34 units

WE 6

Drone and bird (from the PDF)

A drone starts at (−5, 4, −8) and moves in direction (−121). At the same time a bird starts at (6, −4, 3) and moves in direction (2−34). Find the minimum distance between them.

b₁ × b₂ = (−1, 2, 1) × (2, −3, 4) = (2·4 − 1·(−3), 1·2 − (−1)(4), (−1)(−3) − 2·2) = (11, 6, −1) |b₁ × b₂| = √(121 + 36 + 1) = √158 a₂ − a₁ = (6−(−5), −4−4, 3−(−8)) = (11, −8, 11) scalar triple product (11, −8, 11) · (11, 6, −1) = 121 − 48 − 11 = 62 distance d = |62| / √158 = 62/√158 d ≈ 4.93 units (3 sf) PDF’s exact answer — vector-product formula is much faster than the scalar method on this one.

💡 Top tips

Classify first — check parallel (scalar multiple of directions), then intersecting (solve simultaneously), then skew.
Vector-product formula is faster when lines are skew — fewer steps and the GDC handles the cross product.
Scalar method gives F₁ and F₂ too — if the question asks for the closest points themselves, use this method.
Use different parameters: λ for l₁, μ for l₂. They’re independent.
Cross product = 0 means parallel — drop to the point-to-line method.

⚠ Common mistakes

Using one perpendicularity equation instead of two in the scalar method — you need F₁F₂ perpendicular to both directions, giving two equations.
Applying the skew-line formula to parallel lines — b₁ × b₂ = 0 gives division by zero. Use point-to-line instead.
Using a₁ − a₂ instead of a₂ − a₁ — magnitudes are the same so the distance still works, but be consistent.
Forgetting absolute value in the numerator — the triple product can be negative; the distance can’t.
Treating the cross product like a number — b₁ × b₂ is a vector; the formula uses its magnitude on the bottom and a dot product with it on top.

Chapter complete — you now have all five Vector Equations of Lines sub-topics: vector form (r = a + λb), parametric form, angle between two lines (scalar product), shortest distance from a point to a line (perpendicular), and shortest distance between two lines (skew, parallel, intersecting). Together they cover every Paper 1 & 2 vector-line question on the AI HL syllabus.

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