IB Maths AA SLTopic 3 — Geometry & TrigPaper 1 & 2~7 min read
3D Coordinate Geometry
If you’ve already done 2D coordinate geometry, this note is essentially the same — just with an extra z-coordinate tagged on. Both formulas (midpoint and distance) are in the IB formula booklet, so most of the work is just plugging in numbers carefully.
📘 What you need to know
A 3D point is written as (x, y, z) — three numbers instead of two.
Midpoint: average each coordinate separately. Add an x-pair, average. Same for y. Same for z.
Distance: Pythagoras with three terms instead of two. Add (Δz)2 under the square root.
Both formulas are in the IB formula booklet (geometry & trigonometry section).
Always label your six coordinates (x1, y1, z1) and (x2, y2, z2) before substituting — saves you from sign errors.
What does a 3D coordinate look like?
In 2D you’ve got two axes: x (across) and y (up). In 3D, you add a third axis — z — that points out of the page, perpendicular to both. A point in 3D space needs three numbers to pin it down: how far across, how far up, and how far in or out.
A point P(x, y, z) in 3D space
🏠 Real-world analogy
Think of a room. The x-coordinate is how far across the floor you’ve walked. The y-coordinate is how far back into the room you’ve gone. The z-coordinate is how high you’ve climbed. Three numbers, three directions, one unique location.
The two formulas
If you remember the 2D versions, the 3D ones are just the obvious extensions — same structure, with an extra z term added.
📍
3D Midpoint
(x1 + x22, y1 + y22, z1 + z22)
✓ in formula booklet
📏
3D Distance
d = √((x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2)
✓ in formula booklet
No 3D gradient formula at SL. The “slope” idea doesn’t translate cleanly into 3D — that’s a job for vectors, which appear in HL only. So at SL you only need these two.
2D vs 3D — same idea, one extra term
Notice the pattern. Adding a dimension just adds another term inside the formula:
Formula
2D version
3D version
Midpoint
(x1+x22, y1+y22)
(x1+x22, y1+y22, z1+z22)
Distance
√((Δx)2 + (Δy)2)
√((Δx)2 + (Δy)2 + (Δz)2)
🤔 Why does Pythagoras still work in 3D?
Imagine a cuboid (box) with corners at your two points. The space diagonal of that box is the distance you want. By Pythagoras on the floor (the xy-plane), the floor diagonal squared is (Δx)2 + (Δy)2. Then by Pythagoras a second time — using that floor diagonal and the vertical (Δz) — the full diagonal squared is (Δx)2 + (Δy)2 + (Δz)2. Two applications of Pythagoras → the 3D formula.
Finding the midpoint
3D midpoint
M = (x1 + x22, y1 + y22, z1 + z22)
✓ in formula booklet
Quick example: Midpoint of (2, 4, 6) and (0, −2, 4) is (2+02, 4+(−2)2, 6+42) = (1, 1, 5).
Finding the distance
3D distanced = √( (x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 )
✓ in formula booklet
Order doesn’t matter: just like in 2D, the differences are squared so (x1 − x2)2 = (x2 − x1)2. Subtract either way round.
In IB notation, the line segment between A and B is written [AB] and its length is just AB. Use this notation in your written answers.
Worked examples
WE 1
Find the midpoint in 3D
Find the midpoint of P(3, −1, 7) and Q(−5, 5, 1).
Step 1: Label the coordinatesP: x1=3, y1=−1, z1=7Q: x2=−5, y2=5, z2=1Step 2: Average each coordinatex: 3+(−5)2 = −1y: −1+52 = 2z: 7+12 = 4Midpoint = (−1, 2, 4)
WE 2
Find the distance in 3D
Find the distance between C(0, 0, 0) and D(3, 4, 12).
Step 1: Substitute into the formulad = √( (0−3)2 + (0−4)2 + (0−12)2 )Step 2: Square each term= √( 9 + 16 + 144 ) = √169d = 13 unitsclassic 3D Pythagorean triple: 3, 4, 12 → 13
WE 3
Multi-part: distance and midpoint (SME-style)
The points A and B have coordinates (−2, 1, 5) and (4, −3, 2) respectively.
(i) Calculate the distance AB. (ii) Find the midpoint of [AB].
The midpoint of [PQ] is M(2, −1, 3). Given P(−4, 5, 0), find the coordinates of Q.
Step 1: Set up three midpoint equations−4 + x2 = 2 | 5 + y2 = −1 | 0 + z2 = 3Step 2: Multiply each by 2 and solve−4 + x = 4 → x = 85 + y = −2 → y = −70 + z = 6 → z = 6Q = (8, −7, 6)handle each axis independently — same as 2D, just one more!
WE 5
Application — space diagonal of a cuboid
A rectangular box has its corners at O(0, 0, 0) and F(6, 8, 24). Find the length of the space diagonal OF, giving your answer to 3 s.f.
Label all six coordinates first. Write them down with subscripts before substituting — it eliminates 90% of sign errors.
Use brackets for negatives. Write 1 − (−3) instead of 1 − −3 to avoid arithmetic slips.
Both formulas are in the formula booklet — but knowing them by heart saves time mid-paper.
Distance formulas always give a positive result. If you get a negative number, you’ve made a sign error.
Leave answers exact when possible. √61 is a better answer than 7.810…, unless a decimal is asked for.
Use proper IB notation: [AB] is the line segment, AB is its length.
For application problems (boxes, rooms, flight paths), drawing a quick 3D sketch with the points labelled often makes the question obvious.
⚠ Common mistakes
Forgetting the z-term entirely. Going from 2D to 3D, students sometimes forget to add the (Δz)2 in distance or the z-average in midpoint.
Sign errors with negatives. 1 − (−3) is +4, not −4. Use brackets to be safe.
Forgetting to square inside the root. Only the differences are squared, then summed, then square-rooted at the end.
Forgetting the square root. Computing (Δx)2 + (Δy)2 + (Δz)2 alone gives d2, not d.
Adding instead of averaging the midpoint. Midpoint is the average — divide by 2 each time.
Missing brackets in the answer. A point in 3D is written (1, −1, 3.5), with three numbers and parentheses — not 1, −1, 3.5 on their own.
Mixing up midpoint and distance formulas. Both involve six coordinates — read the question carefully to know which one is needed.
These two formulas underpin everything in 3D geometry — from finding centres of cuboids, to checking distances between points in space, to anything you’ll see in vectors. Master them now and the rest of the topic gets easier.
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