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

Shortest Distance Between a Point and a Line

The shortest distance from a point to a line is always the perpendicular distance. Two equally good methods get you there: scalar product (parameterise the foot of the perpendicular and minimise) or vector product (one-shot formula).

📘 What you need to know

Shortest distance = perpendicular distance: drop a perpendicular from the point P to the line.
Foot of the perpendicular F is the point on the line closest to P; FP ⊥ direction b.
Scalar product method: parameterise F as a + λb, set b · FP = 0, solve for λ, then |FP|.
Vector product method: distance = |AP × b||b|, where A is any point on the line.
Choose any anchor A on the line — the formula gives the same answer.
Formula not in booklet: memorise both methods.
If P lies on the line: AP × b = 0 → distance = 0.

Two methods, same answer

Scalar product method

b · FP = 0

find λ at the foot, then |FP|

Vector product method

d = |AP × b||b|

one calculation, no parameter

When to use which: scalar product if you also need the foot of the perpendicular (the closest point on the line). Vector product if you only need the distance — it’s faster.

The vector product formula

Shortest distance from P to line r = a + λb d = |AP × b||b|

Here A is any point on the line (typically the anchor) and P is the external point. Geometrically, |AP × b| is the area of the parallelogram on AP and b; dividing by |b| (the base) gives the height — which is the perpendicular distance.

🧭 Recipe — shortest distance using scalar product method

Write F as a function of λ: F = a + λb.
Compute FP = P − F in terms of λ.
Set the scalar product = 0: b · FP = 0 (perpendicularity).
Solve for λ; substitute back to find F (foot of perpendicular).
Compute |FP| for the shortest distance.

Worked examples

WE 1

Shortest distance using the scalar product method

Find the shortest distance from the point P(4, 3, 2) to the line r = (1, 0, 2) + λ(1, 2, 2).

Step 1: General point F on line F = (1+λ, 2λ, 2+2λ) Step 2: FP = P − F FP = (3 − λ, 3 − 2λ, −2λ) Step 3: Set b · FP = 0 (1)(3−λ) + (2)(3−2λ) + (2)(−2λ) = 0 9 − 9λ = 0 → λ = 1 Step 4: FP at λ = 1, then magnitude FP = (2, 1, −2); |FP| = √(4+1+4) = √9 Shortest distance = 3 F = (2, 2, 4) is the foot of perpendicular

WE 2

Shortest distance using the vector product method

Find the shortest distance from the point P(7, 2, −2) to the line r = (2, 1, −1) + λ(1, 2, −2).

Step 1: A = (2, 1, −1) is on the line; AP = P − A AP = (5, 1, −1) Step 2: AP × b AP × b = (1·(−2) − (−1)·2, −(5·(−2) − (−1)·1), 5·2 − 1·1) = (0, 9, 9) Step 3: Magnitudes |AP × b| = √(0+81+81) = √162 = 9√2 |b| = √(1+4+4) = 3 Distance = 9√2 / 3 = 3√2 vector product is faster when you only need the distance

WE 3

Find the foot of the perpendicular and the distance

Find the foot of the perpendicular F from the point P(4, 1, 0) to the line r = (4, 1, −3) + λ(1, −1, 2), and hence the shortest distance from P to the line.

Step 1: F in terms of λ, then FP F = (4+λ, 1−λ, −3+2λ) FP = P − F = (−λ, λ, 3 − 2λ) Step 2: b · FP = 0 (1)(−λ) + (−1)(λ) + (2)(3−2λ) = 0 6 − 6λ = 0 → λ = 1 Step 3: Substitute λ = 1 F = (5, 0, −1); FP = (−1, 1, 1) |FP| = √3 F = (5, 0, −1), distance = √3 when the question asks for the foot, scalar product is the natural method

WE 4

Shortest distance from the origin to a line

Find the shortest distance from the origin O to the line r = (3, −1, 4) + λ(2, 1, −2).

Step 1: A = (3, −1, 4); AO = O − A = (−3, 1, −4) Step 2: AO × b AO × b: i-comp = (1)(−2) − (−4)(1) = 2 j-comp = −[(−3)(−2) − (−4)(2)] = −14 k-comp = (−3)(1) − (1)(2) = −5 AO × b = (2, −14, −5) Step 3: Magnitudes |AO × b| = √(4 + 196 + 25) = √225 = 15 |b| = √(4+1+4) = 3 Distance = 15 / 3 = 5 treat the origin like any other point — same formula applies

WE 5

Distance from a point to a line through two given points

Find the shortest distance from the point P(0, 5, 5) to the line passing through A(1, 0, 2) and B(3, 4, 4).

Step 1: Direction AB and simplify AB = B − A = (2, 4, 2) → simplify to b = (1, 2, 1) Step 2: AP = P − A AP = (−1, 5, 3) Step 3: AP × b i: (5)(1) − (3)(2) = −1 j: −[(−1)(1) − (3)(1)] = 4 k: (−1)(2) − (5)(1) = −7 AP × b = (−1, 4, −7) Step 4: Magnitudes |AP × b| = √(1+16+49) = √66 |b| = √(1+4+1) = √6 Distance = √66 / √6 = √11 simplifying the direction first keeps the numbers manageable

WE 6

Closest approach of a moving particle to a fixed point

A particle moves with position vector r = (2, −1, 4) + t(1, 2, −1), where t is in seconds. A camera is fixed at C(6, 1, 6). Find the time at which the particle is closest to the camera, and find the minimum distance.

Step 1: Particle at time t and FC F(t) = (2+t, −1+2t, 4−t) FC = C − F = (4−t, 2−2t, 2+t) Step 2: Set b · FC = 0 (perpendicular at minimum) (1)(4−t) + (2)(2−2t) + (−1)(2+t) = 0 6 − 6t = 0 → t = 1 Step 3: Compute FC at t = 1 FC = (3, 0, 3); |FC| = √(9+0+9) = √18 = 3√2 t = 1 second, min distance = 3√2 shortest distance from the path = perpendicular distance, occurring at one instant

💡 Top tips

Vector product is faster when only the distance is needed — no parameter to solve for.
Scalar product is essential if the question asks for the foot of the perpendicular.
Any point A on the line works in the vector product formula — pick the simplest.
Simplify the direction by removing common factors before computing — fewer arithmetic mistakes.
Sanity check: b · FP should equal 0 with your final answer plugged in.

⚠ Common mistakes

Setting b · AP = 0 instead of b · FP = 0 — only FP is perpendicular to the line.
Sign errors in the vector product — the j-component flips sign in the determinant expansion.
Computing |AP| as the answer — that’s the distance from A to P, not the perpendicular distance.
Forgetting to divide by |b| in the vector product method.
Confusing P − F with F − P — for the magnitude it doesn’t matter, but for the perpendicularity check, write FP = P − F.

Final note in this section: Shortest Distance Between Two Lines. For two skew lines, the shortest distance is along the common perpendicular. Two methods again — vector product of the directions plus a connecting displacement, or two scalar product equations to find the feet of perpendicular on each line.

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