IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read HL only

Shortest Distance Between Two Lines

For two skew lines, the shortest distance lies along the common perpendicular — the unique direction perpendicular to both. The vector product b₁ × b₂ gives that direction directly, so a clean one-step formula handles most exam questions.

📘 What you need to know

Skew-line formula: d = |(a₂ − a₁) · (b₁ × b₂)||b₁ × b₂| (not in formula booklet — memorise).
Geometric meaning: project the displacement a₂ − a₁ onto the common perpendicular direction b₁ × b₂.
Scalar product method (alternative): find feet F₁, F₂ with b₁ · F₁F₂ = 0 AND b₂ · F₁F₂ = 0; then |F₁F₂|.
Parallel lines (b₁ × b₂ = 0): pick any point on one line, find the point-to-line distance to the other.
Intersecting lines: distance = 0 (they meet).
Coincident lines: distance = 0 (they’re the same line).
Use different parameters (λ, μ) for the two lines.

The shortest-distance formula

Shortest distance between skew lines d = |(a₂ − a₁) · (b₁ × b₂)||b₁ × b₂|

The vector b₁ × b₂ is perpendicular to both lines — exactly the direction of the common perpendicular. Projecting the displacement a₂ − a₁ onto this direction (and taking magnitude) gives the shortest distance.

Two methods

Vector product (faster)

d = |(a₂ − a₁) · n||n|
where n = b₁ × b₂

no parameters — three vector operations

Scalar product (also gives feet)

b₁ · F₁F₂ = 0
b₂ · F₁F₂ = 0

solve for λ, μ; gives feet F₁, F₂ too

When to use which: vector product if you only need the distance — it’s faster. Scalar product if you also need the feet of the perpendicular F₁ and F₂ on each line.

Special cases

Lines are…	Approach	Distance
Skew	vector product OR scalar product method	positive value from formula
Parallel (not coincident)	pick a point on one line, use point-to-line distance	positive constant gap
Intersecting	—	0
Coincident	—	0

🧭 Recipe — shortest distance between two skew lines

Compute b₁ × b₂ — the common perpendicular direction. If it’s 0, the lines are parallel — switch method.
Compute a₂ − a₁ — the displacement between anchors.
Take the dot product (a₂ − a₁) · (b₁ × b₂).
Take absolute value and divide by |b₁ × b₂|.
Sanity check: if your answer is 0, the lines must intersect or be coincident — verify.

Worked examples

WE 1

Shortest distance between two skew lines

Find the shortest distance between the lines r₁ = (1, 2, 3) + λ(1, 1, 0) and r₂ = (4, 1, 4) + μ(1, 0, 1).

Step 1: b₁ × b₂ i: (1)(1) − (0)(0) = 1 j: −[(1)(1) − (0)(1)] = −1 k: (1)(0) − (1)(1) = −1 b₁ × b₂ = (1, −1, −1); |b₁ × b₂| = √3 Step 2: a₂ − a₁ and dot product a₂ − a₁ = (3, −1, 1) (a₂−a₁)·(b₁×b₂) = (3)(1) + (−1)(−1) + (1)(−1) = 3 Distance = |3|/√3 = √3 positive value → lines are skew, not intersecting

WE 2

Shortest distance with integer answer

Find the shortest distance between the lines r₁ = (3, 1, 2) + λ(1, 0, −1) and r₂ = (5, 1, 3) + μ(1, −2, 0).

Step 1: b₁ × b₂ i: (0)(0) − (−1)(−2) = −2 j: −[(1)(0) − (−1)(1)] = −1 k: (1)(−2) − (0)(1) = −2 b₁ × b₂ = (−2, −1, −2); |…| = √(4+1+4) = 3 Step 2: a₂ − a₁ = (2, 0, 1) (a₂−a₁)·(b₁×b₂) = (2)(−2) + (0)(−1) + (1)(−2) = −6 Distance = 6/3 = 2 always take the absolute value of the dot product before dividing

WE 3

Shortest distance using the scalar product method

Find the shortest distance between the lines r₁ = (1, 2, 3) + λ(1, −1, 0) and r₂ = (3, 4, 5) + μ(0, 1, −1) using the scalar product method.

Step 1: General points and F₁F₂ F₁ = (1+λ, 2−λ, 3); F₂ = (3, 4+μ, 5−μ) F₁F₂ = (2−λ, 2+μ+λ, 2−μ) Step 2: Two perpendicularity equations b₁·F₁F₂ = (2−λ) − (2+μ+λ) = −2λ − μ = 0 … (i) b₂·F₁F₂ = (2+μ+λ) − (2−μ) = λ + 2μ = 0 … (ii) Step 3: Solve From (i): μ = −2λ; sub (ii): λ − 4λ = 0 → λ = 0, μ = 0 Step 4: Substitute, find magnitude F₁F₂ = (2, 2, 2); |F₁F₂| = √12 = 2√3 Distance = 2√3 two equations, two unknowns — solve simultaneously

WE 4

Show two lines are skew, then find their shortest distance

(a) Show that the lines r₁ = (1, 1, 0) + λ(1, 0, 1) and r₂ = (2, 3, 1) + μ(0, 1, 1) are skew. (b) Find the shortest distance between them.

Part (a): Show skew x: 1 + λ = 2 → λ = 1 y: 1 = 3 + μ → μ = −2 z: λ = 1 + μ → 1 = 1 + (−2) = −1 ✗ Inconsistent → skew (and directions aren’t parallel) Part (b): Vector product method b₁ × b₂ = (−1, −1, 1); |…| = √3 a₂ − a₁ = (1, 2, 1) (a₂−a₁)·(b₁×b₂) = −1 − 2 + 1 = −2 Distance = 2/√3 = 2√3/3 always rationalise the denominator unless the question says otherwise

WE 5

Distance between two parallel lines

Show that the lines r₁ = (3, 2, 1) + λ(1, 2, 2) and r₂ = (5, 1, 1) + μ(2, 4, 4) are parallel, and find the shortest distance between them.

Step 1: Parallel check b₂ = (2, 4, 4) = 2(1, 2, 2) = 2b₁ ✓ → parallel Step 2: Pick point P = (5, 1, 1) on l₂; find distance from P to l₁ A = (3, 2, 1); AP = P − A = (2, −1, 0) Step 3: AP × b₁ (using b₁ for direction of l₁) i: (−1)(2) − (0)(2) = −2 j: −[(2)(2) − (0)(1)] = −4 k: (2)(2) − (−1)(1) = 5 AP × b₁ = (−2, −4, 5); |…| = √(4+16+25) = √45 = 3√5 Step 4: Apply point-to-line formula |b₁| = 3; distance = 3√5/3 Distance = √5 for parallel lines, the b₁ × b₂ formula fails (cross product is 0) — switch to point-to-line

WE 6

Application — minimum separation between flight paths

Two aircraft fly in straight lines. Aircraft A’s path is given by r = (4, 2, −1) + t(2, 1, −2) and aircraft B’s path by r = (6, 2, 1) + s(1, 2, 2), where coordinates are in km. Find the minimum distance between the two flight paths.

Step 1: Cross product of directions b₁ × b₂: i: (1)(2)−(−2)(2) = 6 j: −[(2)(2)−(−2)(1)] = −6 k: (2)(2)−(1)(1) = 3 b₁ × b₂ = (6, −6, 3); |…| = √81 = 9 Step 2: a₂ − a₁ = (2, 0, 2) (a₂−a₁)·(b₁×b₂) = 12 + 0 + 6 = 18 Minimum distance = 18/9 = 2 km paths treated as full lines — minimum separation along their geometric paths, regardless of timing

💡 Top tips

Check parallel first: if b₁ × b₂ = 0, switch to the point-to-line method.
Take absolute value of the dot product — distance is always non-negative.
Always rationalise a single √ in the denominator (e.g., 2/√3 → 2√3/3) unless told otherwise.
If the answer is 0, lines intersect or are coincident — go back and check the classification.
For parallel lines, any point on one will give the same distance to the other — pick the simplest.

⚠ Common mistakes

Forgetting absolute value in the numerator — distance must be non-negative.
Sign errors in the j-component of the cross product (the determinant flips sign).
Using the formula on parallel lines — the cross product is zero, formula breaks. Switch to point-to-line.
Confusing “shortest distance between paths” with “closest approach in time” for moving objects — the latter requires same time on both paths.
Computing |a₂ − a₁| as the answer — that’s the gap between anchors, not the perpendicular distance.

That closes Vector Equations of Lines. Up next: Vector Planes. A plane in 3D needs three pieces of information — a point, plus two non-parallel direction vectors (or equivalently, a normal vector). The same scalar/vector product toolkit you’ve built here transfers across, with a few new equations: vector form, parametric form, Cartesian form, and the all-important normal form n · (r − a) = 0.

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Shortest Distance Between Two Lines

📘 What you need to know

The shortest-distance formula

Two methods

Special cases

🧭 Recipe — shortest distance between two skew lines

Worked examples

💡 Top tips

⚠ Common mistakes

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