IB Maths AI HL Vector Equations of Lines Paper 1 & 2 Perpendicular distance ~9 min read

Shortest Distance Between a Point and a Line

From point P to line l, the shortest path is the perpendicular from P to l. Drop the perpendicular onto l; its foot is point F. The distance is |FP|, and the defining condition is FP · b = 0.

📘 What you need to know

Shortest = perpendicular: any other line from P to l is longer (Pythagoras).
Foot of the perpendicular F: the point on l closest to P. FP ⊥ direction b.
Scalar-product method (the main one):
· write F as a + λb (general point on line)
· form FP = p − F
· set FP · b = 0, solve for λ
· distance = |FP|
Vector-product shortcut (not in booklet): distance = |AP × b||b|, where A is any point on the line.
Magnitude: |v| = √(v₁² + v₂² + v₃²).
If P is already on the line, distance = 0 (and λ exists making FP = 0).

The scalar-product method — step by step

Set up F as a general point on l: F = a + λb. Build FP = p − F (the vector from F to P). Demand that FP is perpendicular to the direction of l: FP · b = 0. Solve that one equation for λ. Plug λ back in, compute |FP|.

The perpendicular path FP is shorter than any other connection from P to the line. The defining condition is FP · b = 0 — one equation, one unknown (λ).

Shortest distance from point P to line l Scalar method: FP · b = 0 ⇒ solve for λ ⇒ d = |FP| Vector-product shortcut: d = |AP × b||b| A is any point on l; b is the direction. Vector-product shortcut is not in formula booklet.

The vector-product shortcut

One formula, one answer. Pick any point A on the line (use the position vector a), form AP = p − a, take the cross product AP × b, divide its magnitude by |b|. No equation to solve — the answer pops out directly.

When to use which: scalar-product method always works and gives you F as well (useful if the question asks for the foot of the perpendicular). Vector-product shortcut is faster when only the distance is needed and your GDC computes cross products easily.

🧭 Recipe — shortest distance from a point to a line

Identify the line r = a + λb and the external point P with position vector p.
Write F as a general point on the line: F = a + λb.
Form FP = p − F (the vector from F to P), in terms of λ.
Set FP · b = 0 (perpendicularity condition).
Solve the resulting linear equation for λ.
Substitute λ back into FP, compute |FP| → that’s the shortest distance.
Or use the shortcut: d = |AP × b| / |b| with any point A on the line.

Worked examples

WE 1

Scalar-product method (from the PDF)

Point A(1, 2, 0). Line l: r = (206) + λ(012). Point B lies on l so that AB ⊥ l. Find the shortest distance from A to l.

general point B on l OB = (2λ6+2λ) vector AB = OB − OA AB = (1λ − 26 + 2λ) AB ⊥ b ⇒ AB · b = 0 0(1) + 1(λ−2) + 2(6+2λ) = 0 λ − 2 + 12 + 4λ = 0 5λ + 10 = 0 → λ = −2 substitute λ = −2 into AB AB = (1−42) distance = |AB| |AB| = √(1 + 16 + 4) = √21 shortest distance = √21 units

WE 2

Find the foot AND the distance

P(10, 5, −10). Line l: r = (3−1−2) + λ(1−20). Find the foot of the perpendicular F and the shortest distance from P to l.

F = a + λb OF = (3+λ−1−2λ−2) FP = OP − OF FP = (7−λ6+2λ−8) FP · b = 0 (7−λ)(1) + (6+2λ)(−2) + (−8)(0) = 0 7 − λ − 12 − 4λ = 0 −5λ − 5 = 0 → λ = −1 substitute λ = −1 F = (3−1, −1+2, −2) = (2, 1, −2) FP = (8, 4, −8) |FP| = √(64 + 16 + 64) = √144 = 12 F = (2, 1, −2); d = 12 units

WE 3

Vector-product shortcut

Find the shortest distance from P(3, 0, 1) to the line r = (100) + λ(010) using the vector-product formula.

take A = (1, 0, 0) AP = OP − OA = (2, 0, 1) cross product AP × b AP × b = (201) × (010) = ((0)(0)−(1)(1), (1)(0)−(2)(0), (2)(1)−(0)(0)) = (−1, 0, 2) magnitudes |AP × b| = √(1 + 0 + 4) = √5 |b| = 1 d = √5 / 1 = √5 units geometric check: P is √(2² + 1²) = √5 from the y-axis (which is what the line is). ✓

WE 4

Point already on the line

Find the shortest distance from C(2, 0, −1) to the line r = (40−5) + λ(−102).

F = a + λb OF = (4 − λ, 0, −5 + 2λ) FC = OC − OF FC = (λ − 2, 0, 4 − 2λ) FC · b = 0 (λ−2)(−1) + 0(0) + (4−2λ)(2) = 0 −λ + 2 + 8 − 4λ = 0 −5λ + 10 = 0 → λ = 2 substitute λ = 2 FC = (0, 0, 0) d = 0 → C is on the line if FP comes out as (0, 0, 0), the point lies on the line.

WE 5

Real-world wording — helicopter and tower

A helicopter flies in a straight line. Its position at time t (in seconds) is given by r = (201050) + t(2−10) (metres). A radio tower stands at T(40, 20, 50). Find the closest the helicopter gets to the tower, in metres.

general position F on flight path F = (20+2t, 10−t, 50) FT = OT − OF FT = (20−2t, 10+t, 0) FT · b = 0 (20−2t)(2) + (10+t)(−1) + 0(0) = 0 40 − 4t − 10 − t = 0 30 − 5t = 0 → t = 6 substitute t = 6 FT = (8, 16, 0) |FT| = √(64 + 256) = √320 closest distance = √320 ≈ 17.9 m t plays the same role as λ — just a parameter sliding along the line.

WE 6

Distance using the cross-product shortcut

Find the distance from P(1, 2, 3) to the line r = (000) + λ(111) using the cross-product formula.

A = (0, 0, 0) AP = (1, 2, 3) AP × b = (123) × (111) = (2·1 − 3·1, 3·1 − 1·1, 1·1 − 2·1) = (−1, 2, −1) magnitudes |AP × b| = √(1 + 4 + 1) = √6 |b| = √3 d = |AP × b| / |b| d = √6 / √3 = √2 d = √2 ≈ 1.41 units cross-product shortcut: 3 steps, no equation to solve.

💡 Top tips

Choose your method: scalar-product method gives both F and d; cross-product shortcut gives just d but is faster.
The condition is FP · b = 0, not OP · b = 0 — the perpendicular is from F to P, not from O.
Use any point A on the line for the cross-product shortcut — the simplest choice is the position vector a.
GDC computes cross products — learn how on your model, especially for Paper 2.
Sanity check: |FP| should be smaller than |AP| for any other A on the line (perpendicular is shortest).

⚠ Common mistakes

Computing |OP| or |AP| as the answer — neither equals the perpendicular distance unless P happens to be at F.
Setting FP · a = 0 instead of FP · b = 0 — perpendicularity is to the direction, not the position.
Cross product mistake: AP × b ≠ b × AP (the result flips sign), but magnitudes are equal — so for distance it doesn’t actually matter, but be careful for direction-sensitive questions.
Forgetting to divide by |b| in the shortcut — without it the units are wrong.
Not substituting λ back — solving for λ is step 4 of 5; you still need to plug it into FP and take the magnitude.

Next up — Shortest Distance Between Two Lines. Two non-parallel non-intersecting (skew) lines have a unique common perpendicular. Same idea as before, but now you need two parameters (λ and μ) and either the cross product of both directions or two perpendicularity conditions.

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