IB Maths AI HL Geometry of 3D Shapes Paper 1 & 2 Midpoint, distance in 3D ~8 min read

3D Coordinate Geometry

The 2D midpoint and distance formulas extend naturally to 3D: just include a third coordinate. The distance formula is Pythagoras applied twice — first across the floor of a box, then up its vertical edge — giving √((x₁−x₂)²+(y₁−y₂)²+(z₁−z₂)²).

📘 What you need to know

Midpoint in 3D of (x₁, y₁, z₁) and (x₂, y₂, z₂): M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) — average each coordinate.
Distance in 3D: d = √((x₁−x₂)²+(y₁−y₂)²+(z₁−z₂)²) — 3D Pythagoras.
Both formulas are in the formula booklet (geometry & trigonometry section).
Why the distance formula works: the distance is the space diagonal of a rectangular box with side lengths |Δx|, |Δy|, |Δz|. First find the floor diagonal √(Δx²+Δy²), then use Pythagoras again with the vertical edge Δz.
3D Pythagorean triples: (1, 2, 2, 3), (2, 3, 6, 7), (1, 4, 8, 9), (4, 4, 7, 9), (2, 6, 9, 11), (3, 4, 12, 13) — recognising them gives clean integer distances.
Notation: line segment between A and B is [AB]; its length is AB or |AB|.

Why the 3D distance formula works

Imagine the line segment from A(x₁, y₁, z₁) to B(x₂, y₂, z₂) as the long diagonal of a rectangular box with edges parallel to the axes. The box has length |Δx|, depth |Δy|, height |Δz|. First apply 2D Pythagoras to the box’s floor: the floor diagonal has length √(Δx²+Δy²). That floor diagonal and the vertical edge Δz form a right-angled triangle (the vertical edge is perpendicular to the floor), so Pythagoras applies again, giving the full space diagonal √(floor²+Δz²) = √(Δx²+Δy²+Δz²). The 3D formula is just Pythagoras applied twice.

The space diagonal from O to C (bold teal) is built from the floor diagonal (red dashed, length 5) and the vertical edge (red dashed, length 12) using Pythagoras — giving √(5²+12²) = 13.

3D coordinate formulas midpoint: ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) distance: √((x₁−x₂)²+(y₁−y₂)²+(z₁−z₂)²)

Working with 3D coordinates

The procedure for 3D problems is identical to 2D — just include the z-coordinate every time. Label the coordinates (x₁, y₁, z₁) and (x₂, y₂, z₂) on the points before substituting. For reverse problems — finding a missing coordinate given a midpoint or a distance — set up an equation and solve. Quick mental check: if you get a 3D distance that’s smaller than any of the individual gaps |Δx|, |Δy|, |Δz|, you’ve made an arithmetic error — the diagonal of a box can’t be shorter than any edge.

Spot 3D Pythagorean quadruples: the patterns (1,2,2,3), (2,3,6,7), (1,4,8,9), (4,4,7,9), (2,6,9,11), and (3,4,12,13) all give clean integer distances. They turn up surprisingly often in exam problems.

🧭 Recipe — 3D coordinate geometry

Label the points: write (x₁, y₁, z₁) and (x₂, y₂, z₂) under the coordinates to avoid sign errors.
Pick the formula: midpoint (average) or distance (3D Pythagoras).
Substitute and simplify; watch the signs on differences involving negatives.
For reverse problems, write the formula with the unknown, then solve.
Sanity check: distance must be ≥ the largest single-coordinate gap.

Worked examples

WE 1

Basic distance and midpoint

Points P(1, 2, 1) and Q(3, 5, 7). (a) Find the distance PQ. (b) Find the midpoint of [PQ].

(a) compute the gaps Δx = 1−3 = −2; Δy = 2−5 = −3; Δz = 1−7 = −6 d² = 4 + 9 + 36 = 49 d = 7 units (b) average each coordinate M = ((1+3)/2, (2+5)/2, (1+7)/2) M = (2, 3.5, 4) (2, 3, 6, 7) Pythagorean quadruple — clean answer ✓

WE 2

Exact-form distance

Points A(3, −2, 1) and B(6, 2, 5). (a) Find AB in exact form. (b) Find the midpoint of [AB].

(a) distance via 3D Pythagoras d² = (3−6)² + (−2−2)² + (1−5)² = 9 + 16 + 16 = 41 d = √41 ≈ 6.40 units (b) midpoint M = ((3+6)/2, (−2+2)/2, (1+5)/2) M = (4.5, 0, 3)

WE 3

Reverse: find an endpoint from the midpoint

The midpoint of [AB] is M(2, −1, 4). Point A has coordinates (−1, 3, 2). Find the coordinates of B.

use B = 2M − A coordinate-wise B_x = 2(2) − (−1) = 5 B_y = 2(−1) − 3 = −5 B_z = 2(4) − 2 = 6 B = (5, −5, 6) check: midpoint of (−1, 3, 2) and (5, −5, 6) = (2, −1, 4) ✓

WE 4

Median of a 3D triangle

A triangle in 3D space has vertices A(2, 0, −1), B(0, 4, 3), and C(−2, 2, 5). (a) Find the midpoint M of side [AC]. (b) Find the length of the median from B to M.

(a) midpoint of [AC] M = ((2+(−2))/2, (0+2)/2, (−1+5)/2) M = (0, 1, 2) (b) distance B to M BM² = (0−0)² + (4−1)² + (3−2)² = 0 + 9 + 1 = 10 BM = √10 ≈ 3.16 units medians of a triangle connect each vertex to the midpoint of the opposite side.

WE 5

Applied: drone flight path

A drone at position A(10, 20, 30) (coordinates in metres) flies in a straight line to position B(30, 50, 90). Find: (a) the total distance flown; (b) the drone’s position when exactly halfway through the flight.

(a) compute the gaps Δ = (20, 30, 60) d² = 400 + 900 + 3600 = 4900 d = 70 m (b) halfway = midpoint M = ((10+30)/2, (20+50)/2, (30+90)/2) M = (20, 35, 60) (2, 3, 6, 7) triple scaled by 10 → clean answer ✓

WE 6

Space diagonal and centre of a cuboid

A rectangular cuboid has one vertex at the origin O(0, 0, 0) and the opposite vertex at C(4, 3, 12). Find: (a) the length of the space diagonal OC; (b) the midpoint of OC (the centre of the cuboid); (c) show that the distance from this centre to each of the 8 vertices of the cuboid is the same.

(a) 3D distance OC² = 16 + 9 + 144 = 169 OC = 13 units (b) midpoint of OC M = (2, 1.5, 6) centre = (2, 1.5, 6) (c) check distances from M to all 8 vertices to (0,0,0): √(4 + 2.25 + 36) = √42.25 = 6.5 to (4,3,12): √(4 + 2.25 + 36) = 6.5 to (4,0,0), (0,3,0), (0,0,12),…all give 6.5 all 8 vertices at distance 6.5 = OC/2 ✓ all 8 vertices lie on a sphere centred at M with radius OC/2 — this is the cuboid’s circumscribed sphere.

💡 Top tips

Label coordinates before substituting: write (x₁, y₁, z₁) and (x₂, y₂, z₂) above the points.
Spot 3D Pythagorean quadruples: (1,2,2,3), (2,3,6,7), (1,4,8,9), (3,4,12,13) for clean integer answers.
Reverse midpoint shortcut: B = 2M − A in each coordinate.
For distances on coordinate planes: a point on the xy-plane has z = 0; on the x-axis has both y = z = 0. Use these constraints when locating unknown points.
Sphere centred at a point: the locus of points at fixed distance from a given centre is a sphere — comes up in cuboid and circumscribed-sphere problems.

⚠ Common mistakes

Forgetting the z-coordinate in midpoint or distance — gives a 2D answer for a 3D problem.
Sign errors with negatives: 5 − (−3) = 8, not 5 − 3 = 2.
Forgetting the square root at the end of the distance formula — you computed d², not d.
Using sum instead of average for midpoint — remember to divide by 2.
Adding coordinates instead of squaring differences in the distance formula.

Next up: Volume & Surface Area — the formulas for the basic 3D shapes (cuboids, cylinders, pyramids, cones, spheres), most of which are in the formula booklet.

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