IB Maths AI SL Topic 3 — Geometry of 3D Shapes Paper 1 & 2 Add a z-coordinate ~6 min read

3D Coordinate Geometry

Move from (x, y) on the plane to (x, y, z) in space. The midpoint and distance formulae work the same way — you just include the third coordinate. No new ideas; same Pythagoras, with one extra term inside the square root.

📘 What you need to know

3D point: (x, y, z). Three axes mutually perpendicular — x & y on the floor, z going up.
Midpoint of A(x₁, y₁, z₁) and B(x₂, y₂, z₂) = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2). Average each coordinate.
Distance: d = √((x₁−x₂)² + (y₁−y₂)² + (z₁−z₂)²). Pythagoras with one extra term.
Both formulae are in the formula booklet — geometry & trigonometry section.
No gradient in 3D: a single number can’t describe a slope in space (you’d need a direction). At AI SL you won’t be asked for one.
Space diagonal of a cuboid with sides a, b, c: d = √(a² + b² + c²). Just the distance formula with one corner at the origin.

Pythagoras twice — that’s the whole idea

The distance formula in 3D is “Pythagoras applied twice”. Drop a perpendicular from the high point to the floor, find the floor distance with 2D Pythagoras (the legs dx, dy), then use that floor distance plus the rise dz as the legs of a second Pythagoras to get the full 3D distance.

From B to P: the three orange dashed legs (dx = 3, dy = 4, dz = 12) sit along the box edges. The red dashed line on the floor is the 2D Pythagoras result. The green solid line is the 3D distance d = 13.

In the formula booklet midpoint = x₁+x₂2, y₁+y₂2, z₁+z₂2

d = √((x₁−x₂)² + (y₁−y₂)² + (z₁−z₂)²)

🧭 Recipe — any 3D coordinate problem

Label the points: write (x₁, y₁, z₁) and (x₂, y₂, z₂) directly above the numbers.
Find each difference: dx, dy, dz. Watch the signs — negatives often flip when subtracted.
For distance: square each, add the three, then square-root. For midpoint: average each pair.
For “working backwards” (missing endpoint given midpoint): use x₂ = 2M_x − x₁, and same for y and z.
State the answer with units: “10 m” for distance, “(a, b, c)” for midpoint.

3D Pythagorean quadruples: (1, 2, 2)→3 · (2, 3, 6)→7 · (3, 4, 12)→13 · (1, 4, 8)→9 · (4, 4, 7)→9. Recognising these makes the distance instant — no calculator needed.

Worked examples

WE 1

Midpoint between two warehouses

A drone-delivery system has two warehouses at P(1, 4, 7) and Q(5, −2, 3) (coordinates in km, including altitude). A relay station is to be placed at the midpoint of [PQ]. Find its coordinates.

Average each coordinate Mₓ = (1 + 5)/2 = 3 Mᵢ = (4 + (−2))/2 = 1 Mẑ = (7 + 3)/2 = 5 M = (3, 1, 5) exactly the same idea as 2D — just three averages instead of two. The negative y stays as a negative inside the sum: 4 + (−2) = 2.

WE 2

Distance — drone flight to a mountain peak

A drone flies in a straight line from base camp B(2, 3, 5) to peak P(5, 7, 17), coordinates in km (third entry = altitude). Find the straight-line distance flown.

Step 1 — find the three differences dx = 5 − 2 = 3 dy = 7 − 3 = 4 dz = 17 − 5 = 12 Step 2 — Pythagoras d = √(3² + 4² + 12²) = √(9 + 16 + 144) = √169 = 13 d = 13 km (3, 4, 12, 13) is a Pythagorean quadruple — the answer comes out as a whole number. Spotting it saves you reaching for the calculator.

WE 3

Midpoint of two satellites

Two satellites are tracked at positions S₁(−3, 2, −1) and S₂(5, −6, 7) (coordinates in 100 km from Earth’s centre). Find the midpoint of [S₁S₂].

Average each coordinate, careful with signs Mₓ = (−3 + 5)/2 = 2/2 = 1 Mᵢ = (2 + (−6))/2 = −4/2 = −2 Mẑ = (−1 + 7)/2 = 6/2 = 3 M = (1, −2, 3) three negative-positive pairs to average. Brackets help: (2 + (−6))/2 = −4/2 = −2 is much clearer than 2 + −6 / 2.

WE 4

Both formulae — helicopter pair

Two helicopters are at H₁(2, 1, 3) and H₂(6, 4, 15) (coordinates in km, altitude is the third entry). Find (a) the midpoint of [H₁H₂] and (b) the distance between them.

(a) midpoint Mₓ = (2 + 6)/2 = 4 Mᵢ = (1 + 4)/2 = 2.5 Mẑ = (3 + 15)/2 = 9 M = (4, 2.5, 9) (b) distance dx = 4, dy = 3, dz = 12 d = √(16 + 9 + 144) = √169 = 13 (a) M = (4, 2.5, 9) · (b) d = 13 km midpoint coordinates may come out as decimals or fractions — that’s fine. The (3, 4, 12, 13) quadruple appears again — common in IB problems.

WE 5

Working backwards — find missing endpoint

The midpoint of [PQ] is M(2, 5, −1) and one endpoint is P(7, 3, 4). Find the other endpoint Q.

Use Q = 2M − P for each coordinate Qₓ = 2(2) − 7 = 4 − 7 = −3 Qᵢ = 2(5) − 3 = 10 − 3 = 7 Qẑ = 2(−1) − 4 = −2 − 4 = −6 Check: midpoint of P and Q ((7 + (−3))/2, (3 + 7)/2, (4 + (−6))/2) = (2, 5, −1) ✓ Q = (−3, 7, −6) the shortcut Q = 2M − P works coordinate-by-coordinate. Always verify by averaging P and Q — should give M back.

WE 6

Space diagonal of a warehouse

A rectangular warehouse has internal dimensions 12 m × 8 m × 9 m (length × width × height). A wire is run from one floor corner to the opposite top corner. Find the length of the wire.

Set up: place one corner at the origin corner A = (0, 0, 0) opposite corner B = (12, 8, 9) Apply the 3D distance formula d = √(12² + 8² + 9²) = √(144 + 64 + 81) = √289 = 17 m d = 17 m the space diagonal of a cuboid = √(length² + width² + height²). Placing one corner at the origin turns it into a 3D distance question instantly.

💡 Top tips

Treat 3D like 2D with one extra term: same Pythagoras structure, plus a (z₁−z₂)² on the inside.
Label (x₁, y₁, z₁) above each point: especially helpful with negatives.
Spot Pythagorean quadruples: (3, 4, 12, 13) and (2, 3, 6, 7) are common in exam questions — answer is a clean integer.
For “space diagonal of a box”: place one corner at the origin, the opposite at (length, width, height), and use the 3D distance formula.
Sketch a rough 3D box if it helps: draw the three legs dx, dy, dz as box edges and the diagonal as the line through the box.

⚠ Common mistakes

Forgetting the z term: an easy slip when copying the 2D formula. Always check you have three squared differences inside the root.
Squaring then taking the root of just one piece: √(dx²) + √(dy²) + √(dz²) is NOT the same as √(dx² + dy² + dz²). Add first, then root.
Sign mistakes with negatives: (−1 + 7)/2 = 3, not 4. Use brackets to keep the signs visible.
Midpoint coordinates left unsimplified: 5/2 is fine but write it as 2.5 if the question wants decimals. Mixing forms looks careless.
Confusing dimensions and coordinates for the space-diagonal question. A box with sides 12, 8, 9 has corners at (0,0,0) and (12, 8, 9), not at (12, 8, 9) and (24, 16, 18).

Up next: Volume & Surface Area — the formulae for prisms, cylinders, pyramids, cones and spheres. Most are in the formula booklet; the trick is recognising the shape, identifying r and h, and combining shapes (cone + sphere, prism − cylinder, etc.) for richer problems.

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