IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read

3D Coordinate Geometry

Adding a third axis turns a flat plane into 3D space — but the geometry barely changes. The midpoint of two points is still the average of their coordinates, and the distance between two points is still Pythagoras. The only difference is one extra term for the z-coordinate. Once you’ve spotted the pattern, every 2D coordinate-geometry skill from Section 3.1 transfers directly.

📘 What you need to know

3D point notation: a point in space is written (x, y, z) — three coordinates instead of two.
Midpoint formula: average each coordinate independently — three averages instead of two.
Distance formula: 3D Pythagoras — sum the squared differences in x, y, and z, then take the positive square root.
Both formulas are in the formula booklet in the Geometry & Trigonometry section — you don’t need to memorise them.
Order doesn’t matter for distance: differences are squared, so the sign cancels.
Pattern: the 3D formulas are just the 2D formulas with a “+ z-term” added — same structure, one extra dimension.
No gradient in 3D: in 3D, direction is described by a vector with three components, not a single number — that comes in the Vectors topic.
Reverse problems: any unknown coordinate or endpoint can be solved using these two formulas combined with given conditions.

Setting up three-dimensional space

In 2D, a point is fixed by two coordinates: how far along the x-axis and how far up the y-axis. In 3D, we add a third axis — typically called the z-axis — that points “out of the page” perpendicular to the other two.

A point in 3D space is then written as a triple (x, y, z). The origin is (0, 0, 0). The three axes meet at right angles to each other, and together they coordinate-label every point in space uniquely.

Mental model: imagine a corner of a room. The two walls meeting the floor at the corner give you two perpendicular axes (think x and y); the vertical edge running up the wall gives you the third (z). Every point in the room has three measurements — distance along each wall and height off the floor.

Midpoint of two points in 3D

Same as in 2D — average each coordinate. There are now three averages to compute instead of two.

3D midpoint formula M = ( x₁ + x₂2 , y₁ + y₂2 , z₁ + z₂2 )

Order makes no difference, since addition is commutative. If you know the midpoint and one endpoint, you can solve for the other endpoint just like in 2D — apply Q_i = 2M_i − P_i to each coordinate independently.

Distance between two points in 3D

The distance between two points in 3D is the length of the line segment joining them — which is the straight-line “as the bird flies” distance through space.

3D distance formula d = √( (x₁ − x₂)² + (y₁ − y₂)² + (z₁ − z₂)² )

This is just Pythagoras applied twice. Drop a vertical from one point to the horizontal plane through the other; the horizontal leg is the 2D distance √((Δx)² + (Δy)²), and the vertical leg is |Δz|. The 3D segment is the hypotenuse — square the two legs and take the root.

If you forget the formula, picture a rectangular box. The diagonal from one corner to the opposite corner has length √(l² + w² + h²) — exactly the 3D distance formula with sides equal to the differences in coordinates.

2D vs 3D — same idea, one extra term

Quantity	2D version	3D version
Midpoint	( (x₁+x₂)/2 , (y₁+y₂)/2 )	( (x₁+x₂)/2 , (y₁+y₂)/2 , (z₁+z₂)/2 )
Distance	√( (Δx)² + (Δy)² )	√( (Δx)² + (Δy)² + (Δz)² )
Gradient/slope	(y₂−y₁)/(x₂−x₁)	not a single number — use a direction vector (Topic 4)

The pattern is consistent: every 2D formula gains a “z-component” term in 3D. The only thing that doesn’t transfer is the gradient — direction in 3D needs more than one number, which is why vectors take over from gradient when we move into space.

🧭 Recipe — 3D coordinate geometry workflow

Label the two points clearly: write (x₁, y₁, z₁) above the first and (x₂, y₂, z₂) above the second. Three coordinates each — easy to drop one.
Pick the right formula: midpoint for “halfway”, distance for “length” or “how far apart”.
Substitute carefully. Bracket all negatives: (x₂) − (x₁), not x₂ − x₁ with floating signs.
Simplify each term separately, then combine. For distance, work out each squared difference one at a time before summing.
Take the positive square root for distance. Distance is non-negative — discard the negative root.
Sanity check: does the midpoint sit visibly between the two endpoints? Is the distance positive and reasonable for the size of the differences?

Worked examples

WE 1

Distance between two 3D points

Find the distance between P(2, 1, 4) and Q(5, 5, 16).

Step 1: Label coordinates (x₁, y₁, z₁) = (2, 1, 4); (x₂, y₂, z₂) = (5, 5, 16) Step 2: Substitute into d = √((x₁−x₂)² + (y₁−y₂)² + (z₁−z₂)²) d = √((2−5)² + (1−5)² + (4−16)²) Step 3: Compute each squared difference d = √((−3)² + (−4)² + (−12)²) d = √(9 + 16 + 144) = √169 d = 13 units a 3D Pythagorean triple — (3, 4, 12, 13) — clean exact answer

WE 2

Midpoint of a 3D line segment

Find the midpoint of the line segment joining P(7, −2, 3) and Q(−3, 6, −5).

Step 1: Apply midpoint formula M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) Step 2: Substitute and simplify each coordinate separately M_x = (7 + (−3))/2 = 4/2 = 2 M_y = (−2 + 6)/2 = 4/2 = 2 M_z = (3 + (−5))/2 = −2/2 = −1 M = (2, 2, −1) three independent averages — work them out one at a time to avoid mixing coordinates

WE 3

Find an unknown coordinate from a known distance

The points A(1, 0, 2) and B(7, k, 5) are such that the distance between them is √61. Find all possible values of k.

Step 1: Apply the distance formula and square both sides 61 = (1−7)² + (0−k)² + (2−5)² Step 2: Compute the known squared differences 61 = 36 + k² + 9 61 = 45 + k² Step 3: Solve for k k² = 16 k = 4 or k = −4 squaring loses the sign, so both values are valid — the question gives no other condition to discriminate

WE 4

Find an unknown endpoint from the midpoint

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

Step 1: Set up midpoint equations for each coordinate ((−2 + Q_x)/2, (5 + Q_y)/2, (−1 + Q_z)/2) = (3, −1, 4) Step 2: Solve for Q_x (−2 + Q_x)/2 = 3 → Q_x = 8 Step 3: Solve for Q_y (5 + Q_y)/2 = −1 → Q_y = −7 Step 4: Solve for Q_z (−1 + Q_z)/2 = 4 → Q_z = 9 Q = (8, −7, 9) check: midpoint of (−2, 5, −1) and (8, −7, 9) is (3, −1, 4) ✓

WE 5

Distance and midpoint together

For the points A(2, −1, 5) and B(6, 7, −3), find (a) the length of [AB] and (b) the midpoint of [AB].

(a) Distance d = √((2−6)² + (−1−7)² + (5−(−3))²) d = √((−4)² + (−8)² + 8²) d = √(16 + 64 + 64) = √144 |AB| = 12 units (b) Midpoint M = ((2+6)/2, (−1+7)/2, (5+(−3))/2) M = (8/2, 6/2, 2/2) M = (4, 3, 1) two formulas, one set of points — no need to relabel between parts

WE 6

Show three points form a right-angled triangle

The points A(1, 2, 3), B(4, 6, 3), and C(1, 2, 9) are vertices of a triangle. Show that the triangle is right-angled at A, and find its area.

Step 1: Find the three side lengths AB² = (1−4)² + (2−6)² + (3−3)² = 9 + 16 + 0 = 25 → AB = 5 AC² = (1−1)² + (2−2)² + (3−9)² = 0 + 0 + 36 = 36 → AC = 6 BC² = (4−1)² + (6−2)² + (3−9)² = 9 + 16 + 36 = 61 Step 2: Check Pythagoras at A AB² + AC² = 25 + 36 = 61 = BC² ✓ so the right angle is at A, with legs AB and AC Step 3: Area = ½ × leg × leg A = ½ × 5 × 6 right-angled at A; area = 15 square units classic exam pattern — find all three sides, use Pythagoras to identify the right angle, then use ½ × base × height for area

💡 Top tips

Label coordinates above the points — three subscripted labels (x₁, y₁, z₁) prevent the most common error: dropping a coordinate.
Bracket every negative. The formula has so many subtractions that one missed bracket can flip the answer.
Compute squared differences one at a time. Stack them vertically: 9, 16, 144 — then sum. Cleaner than trying to do all three in one line.
Recognise 3D Pythagorean triples: (1, 2, 2, 3), (2, 3, 6, 7), (3, 4, 12, 13), (4, 4, 7, 9). These produce clean integer distances and are favoured by exam-setters.
Leave answers as exact surds when no nice integer answer appears — √89 is more precise than 9.43 and easier to verify.
Use the formula booklet in the exam — both formulas are listed under Geometry & Trigonometry. No need to memorise.
For 3D triangle problems: find all three side lengths first, then test Pythagoras to spot any right angle. The longest side is always the hypotenuse if a right angle exists.

⚠ Common mistakes

Forgetting the z-term entirely. Using the 2D distance formula in a 3D problem misses the third squared difference and gives a smaller (wrong) answer.
Mixing up which coordinate is which. (3, 5, 7) and (3, 7, 5) are different points — order matters within each triple.
Sign errors on negative coordinates. (−2) − (−5) = 3, not −7 or −3 or 7. Bracket negatives carefully.
Forgetting the square root for distance. d² = 49 gives d = 7, not 49.
Including a ± sign for distance. Distance is non-negative — only the positive square root matters.
Trying to use a 2D gradient formula in 3D. There’s no single-number gradient in 3D — direction needs a vector (Topic 4).
Halving only some of the coordinates for the midpoint. All three coordinates need to be averaged.

With 3D distance and midpoint settled, the next note — Volume & Surface Area — applies the same Pythagorean reasoning to actual 3D objects: cuboids, pyramids, cones, cylinders, and spheres. The internal diagonals of these solids often hide right-angled triangles, and the surface-area calculations rely on splitting solids into recognisable 2D faces. The toolkit you’ve built so far — Pythagoras, areas of 2D shapes, and now 3D coordinates — handles almost all of it.

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