IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~8 min read HL only

Shortest Distances with Planes

The shortest distance from a point to a plane is always the perpendicular distance — measured along a line in the direction of the plane’s normal. Same idea applies for parallel planes and parallel-to-plane lines: pick any point on one and find its perpendicular distance to the other.

📘 What you need to know

Shortest distance from a point to a plane = perpendicular distance.
Method (line method): build a line through the point in the direction of the plane’s normal, find its intersection with the plane, then compute |λn|.
Direct formula: distance = |n · p − d| / |n|, where p is the point and the plane is n · r = d.
For a parallel-to-plane line: pick any point on the line, find its distance to the plane.
For two parallel planes: pick any point on one, find its distance to the other.
Shortcut for parallel planes with the same normal n: distance = |d₁ − d₂| / |n|.
Always check parallel first when the problem involves a line and a plane (or two planes) — otherwise the perpendicular distance idea doesn’t apply.

Two methods

Line method

d = |λn|

build a line through P in direction n, find λ at intersection

Direct formula

d = |n · p − d||n|

substitute the point into the LHS, divide by |n|

Both give the same answer. The line method is what most mark schemes show. The direct formula is faster — useful for quick calculations and finding unknowns.

Special cases

Setup	Approach
Point and plane	line method or direct formula
Line parallel to plane	pick any point on line, treat as point-to-plane
Two parallel planes	pick any point on one, treat as point-to-plane
Line crosses plane	distance = 0 (they intersect)

🧭 Recipe — shortest distance from a point to a plane (line method)

Identify the normal n from the plane.
Write a line through P perpendicular to the plane: r = p + λn.
Substitute the parametric form into the plane equation.
Solve for λ.
Distance = |λ| × |n|.

Worked examples

WE 1

Shortest distance from a point to a plane

Find the shortest distance from the point P(5, 0, 3) to the plane 2x − y + 2z = 7.

Step 1: Normal n = (2, −1, 2); |n| = 3 Step 2: Line through P perpendicular to plane r = (5, 0, 3) + λ(2, −1, 2) Step 3: Substitute into plane equation 2(5+2λ) − (0−λ) + 2(3+2λ) = 7 10 + 4λ + λ + 6 + 4λ = 7 16 + 9λ = 7 → λ = −1 Step 4: Distance = |λ| × |n| = 1 × 3 Distance = 3 verify with direct formula: |2(5) − 0 + 2(3) − 7|/3 = |9|/3 = 3 ✓

WE 2

Shortest distance from the origin to a plane

Find the shortest distance from the origin O to the plane 6x + 2y − 3z = 14.

Step 1: Normal n = (6, 2, −3); |n| = √(36+4+9) = √49 = 7 Step 2: Line through O perpendicular to plane r = λ(6, 2, −3) Step 3: Substitute 6(6λ) + 2(2λ) − 3(−3λ) = 14 36λ + 4λ + 9λ = 49λ = 14 → λ = 2/7 Step 4: Distance = (2/7) × 7 Distance = 2 direct formula: |0 − 14|/7 = 2 — same answer, faster

WE 3

Shortest distance from a line parallel to a plane

Show that the line r = (3, 4, 2) + s(2, 1, −2) is parallel to the plane x + 2y + 2z = 6, and find the shortest distance between them.

Step 1: Check parallel: b · n = (2)(1) + (1)(2) + (−2)(2) = 0 ✓ Step 2: Pick anchor P = (3, 4, 2) on line Step 3: Direct formula for distance from P to plane |n · P − d| = |3 + 8 + 4 − 6| = 9 |n| = √(1+4+4) = 3 Distance = 9/3 = 3 since the line is parallel, every point on it is the same distance from the plane

WE 4

Shortest distance between two parallel planes

Find the shortest distance between the parallel planes Π₁: 2x − y + 2z = 9 and Π₂: 2x − y + 2z = −3.

Step 1: Same normal (2, −1, 2) → planes parallel ✓ Step 2: Use parallel-plane shortcut distance = |d₁ − d₂| / |n| |n| = √(4+1+4) = 3 |9 − (−3)| / 3 = 12/3 Distance = 4 long way: pick (5, 1, 0) on Π₁; line method gives λ = −4/3, distance = 4 ✓

WE 5

Find a coordinate so a point is at a given distance from a plane

Find the values of k for which the point P(k, 0, 1) is at a distance of 3 from the plane 2x + y − 2z = 5.

Step 1: Direct formula sets up the equation |2k + 0 − 2(1) − 5| / 3 = 3 |2k − 7| = 9 Step 2: Two cases from absolute value 2k − 7 = 9 → k = 8 2k − 7 = −9 → k = −1 k = 8 or k = −1 two solutions because P can be 3 units on either side of the plane

WE 6

Shortest distance from a point to a plane in vector form

Find the shortest distance from the point P(2, 3, −1) to the plane Π: r = (1, 0, 0) + λ(1, 1, 0) + μ(0, 1, 1).

Step 1: Find normal n = b × c i: (1)(1) − (0)(1) = 1 j: −[(1)(1) − (0)(0)] = −1 k: (1)(1) − (1)(0) = 1 n = (1, −1, 1); |n| = √3 Step 2: Find d = n · a d = (1)(1) + (−1)(0) + (1)(0) = 1 Cartesian: x − y + z = 1 Step 3: Direct formula |n · P − d| = |2 − 3 + (−1) − 1| = |−3| = 3 Distance = 3/√3 = √3 converting to Cartesian form first makes the distance calculation faster

💡 Top tips

Direct formula is faster: distance = |n · p − d| / |n| — use it for quick calculations.
Parallel planes shortcut: distance = |d₁ − d₂| / |n|, but only if the LHS coefficients match exactly.
For vector-form planes, convert to Cartesian first (find n = b × c) before applying the distance formula.
Always take absolute value of the numerator — distances are non-negative.
Sanity check: a distance of 0 means the point is ON the plane.

⚠ Common mistakes

Forgetting absolute value in the numerator — gives a signed distance, which can be negative.
Using the position vector a as the normal for vector-form planes — must take b × c.
Forgetting to divide by |n| — the formula gives distance, not just the dot product.
Applying the parallel-plane shortcut to non-parallel planes — coefficients must match (or be the same scalar multiple, with constants scaled accordingly).
Mixing up the line method’s λ with distance — distance = |λ| × |n|, not just |λ|.

That closes Vector Planes — and Topic 3 of the AA HL syllabus. Up next is Topic 4: Statistics & Probability. The geometry intuition you’ve built (vectors, projections, perpendicularity) will reappear in unexpected places — correlation as a kind of “angle” between data vectors, regression as a projection. Different language, same structural ideas.

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