3D Geometry (AA SL)
Three-dimensional geometry extends coordinate geometry and trigonometry from the flat plane into space. This topic covers finding the distance between two points in three dimensions and their midpoint, calculating the volume and surface area of common 3D solids, and using right-angled trigonometry to find angles and lengths hidden inside a solid.
What the syllabus says
This topic maps onto two points in the official IB Analysis & Approaches syllabus, shared with Applications & Interpretation.
| Code | Syllabus content |
|---|---|
| SL3.1 | The distance between two points in three-dimensional space, and their midpoint. In SL examinations, only right-angled trigonometry questions will be set in reference to three-dimensional shapes. |
| SL3.1 | Volume and surface area of three-dimensional solids including the right-pyramid, right cone, sphere, hemisphere, and combinations of these solids. The size of an angle between two intersecting lines, or between a line and a plane. |
Students should be able to identify relevant right-angled triangles inside three-dimensional objects and use them to find unknown lengths and angles.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is the distance between two points in 3D?
It's the straight-line length connecting two points in space, found by extending Pythagoras' theorem to three coordinates: \(x\), \(y\), and \(z\). It's the diagonal of the rectangular box formed by the coordinate differences.
e.g. distance from the origin to \((2,3,6)\) is \(\sqrt{2^2+3^2+6^2}=\sqrt{49}=7\).
What is a right-pyramid?
A right-pyramid has a base (often square or triangular) and an apex positioned directly above the centre of the base, so its height is perpendicular to the base. This makes many of its internal triangles right-angled.
e.g. a square-based pyramid with base side \(6\) and height \(4\) has volume \(\dfrac13\times36\times4=48\).
What is the angle between a line and a plane?
It's the angle between a line and its own projection (shadow) onto the plane - found by dropping a perpendicular from a point on the line to the plane and using the right-angled triangle formed.
e.g. if a line's vertical rise is \(3\) and its horizontal projection is \(4\), the angle to the plane is \(\arctan\left(\dfrac34\right)\approx36.9^\circ\).
What is a net of a solid?
A net is a 2D shape that can be folded up to form a 3D solid - unfolding a solid's faces onto a flat plane. Nets are a useful way to see and calculate total surface area, since each face becomes a flat shape you can measure directly.
e.g. the net of a cube with edge \(3\) is six \(3\times3\) squares, giving surface area \(6\times9=54\).
What is a compound solid?
A compound solid is made by joining two or more basic solids together, such as a cone sitting on top of a cylinder. Its volume and surface area are found by combining the volumes/areas of the individual pieces - being careful not to double-count any joined faces.
e.g. a hemisphere of radius \(3\) on a cylinder of radius \(3\), height \(5\): total volume \(=\pi(3)^2(5)+\tfrac23\pi(3)^3=45\pi+18\pi=63\pi\).
Key formulas
These formulas cover every question on this topic. The two tables below summarise all of them at a glance - the explanations underneath go into more depth on each one.
Formula reference
Every volume and surface area formula below is on the official formula booklet; only the 3D distance and midpoint formulas are assumed prior knowledge, extended directly from their 2D versions.
| Formula | Used for | Booklet? |
|---|---|---|
| \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\) | Distance between two points in 3D | Not in booklet - prior knowledge |
| \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right)\) | Midpoint in 3D | Not in booklet - prior knowledge |
| \(V=\dfrac13\times\text{base area}\times\text{height}\) | Volume of a right-pyramid | ✓ Yes |
| \(V=\dfrac13\pi r^2 h\) | Volume of a right cone | ✓ Yes |
| \(V=\dfrac43\pi r^3\) | Volume of a sphere | ✓ Yes |
| \(A=4\pi r^2\) | Surface area of a sphere | ✓ Yes |
2D vs 3D distance and midpoint
Every 3D formula on this topic is the natural extension of its 2D equivalent - just add a \(z\)-term.
| Feature | 2D | 3D |
|---|---|---|
| Distance formula | \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) | \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\) |
| Midpoint formula | \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\) | \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right)\) |
| Example points | \((0,0)\) to \((3,4)\) | \((0,0,0)\) to \((2,3,6)\) |
| Example distance | \(\sqrt{9+16}=5\) | \(\sqrt{4+9+36}=7\) |
Distance, midpoint and right-angled trigonometry
Most 3D geometry questions come down to identifying a right-angled triangle hidden inside the solid, then applying one of these three tools to it.
3D distance formula
\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\]
Pythagoras' theorem extended to three coordinates - the straight-line distance between two points in space.
Not in the formula booklet - prior knowledgeRight-angled trigonometry in 3D
\[\sin\theta=\dfrac{\text{opp}}{\text{hyp}},\ \cos\theta=\dfrac{\text{adj}}{\text{hyp}},\ \tan\theta=\dfrac{\text{opp}}{\text{adj}}\]
Find the right-angled triangle inside the solid first - the ratios then work exactly as in 2D.
Not in the formula booklet - prior knowledgeAngle between a line and a plane
\[\theta=\arctan\!\left(\dfrac{\text{vertical rise}}{\text{horizontal projection}}\right)\]
Drop a perpendicular from the line to the plane, then use the right-angled triangle formed.
Not in the formula booklet - prior knowledgeVolume and surface area of solids
All three formulas below are given in the formula booklet - the skill being tested is identifying which solid (or combination of solids) is described, not memorising the formula.
Right-pyramid / cone
\[V=\dfrac13\times\text{base area}\times\text{height}\]
For a cone, base area is \(\pi r^2\), giving \(V=\dfrac13\pi r^2h\).
✓ In the formula bookletSphere / hemisphere
\[V=\dfrac43\pi r^3,\quad A=4\pi r^2\]
A hemisphere is exactly half these values for volume and curved surface area.
✓ In the formula bookletCompound solids
\[V_{\text{total}}=V_1+V_2\]
Add the volumes of each piece; for surface area, exclude any face where the solids are joined.
✓ In the formula bookletWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Point \(P\) has coordinates \((2, 3, 6)\) and point \(O\) is the origin.
Find the distance \(OP.\)
Worked solution
\(OP = \sqrt{2^2+3^2+6^2} = \sqrt{49}\) M1
\(OP = 7.\) A1
Points \(P(0,0,0),\) \(Q(3,4,0)\) and \(R(3,4,12)\) form a triangle.
(a) Find the lengths \(PQ,\) \(QR\) and \(PR.\)
(b) Hence show that triangle \(PQR\) has a right angle, stating at which vertex it occurs.
Worked solution
(a) \(PQ = \sqrt{3^2+4^2} = 5,\) \(QR = \sqrt{0^2+0^2+12^2} = 12,\) \(PR = \sqrt{3^2+4^2+12^2}\) M1
\(= 13.\) A1 A1
(b) \(PQ^2+QR^2 = 25+144 = 169 = PR^2,\) so by the converse of Pythagoras' theorem the right angle is at \(Q.\) R1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting the \(z\)-term in the distance or midpoint formula. Treating a 3D problem as 2D by dropping the third coordinate is the single most common error - always check all three coordinates are used.
- Using diameter instead of radius in volume/surface area formulas. Every booklet formula for a sphere, cone, or hemisphere uses radius \(r\) - halve a given diameter before substituting.
- Double-counting or missing joined faces in compound solids. When two solids are glued together, the surface where they join is internal and should not be included in the total surface area.
- Picking the wrong right-angled triangle inside a 3D shape. Not every triangle in a 3D diagram is right-angled - identify the perpendicular height or edge first, and check the angle really is \(90^\circ\) before applying trigonometry.
Using your GDC
On calculator papers, your GDC's square-root and trig functions handle the arithmetic for 3D distances, volumes and surface areas quickly and accurately once you've substituted the right values in - and its memory/answer-recall functions help avoid rounding errors across multi-step problems. The harder part of this topic is usually visualising the solid and identifying the correct right-angled triangle or formula to use, which is done by hand regardless of calculator model. See the full GDC guide for more general calculator techniques.
Ready to practise properly?
3D geometry questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
How is the 3D distance formula different from the 2D one?
It's the same idea with one extra term. The 2D distance between \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). In 3D you add a \(z\)-term: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\).
Do I need trigonometry for 3D geometry questions?
Often, yes - but only right-angled trigonometry at SL. A typical question asks you to spot a right-angled triangle hidden inside a 3D solid (like the diagonal of a cuboid) and apply Pythagoras or basic trig ratios to it.
Are the volume and surface area formulas given in the exam?
Yes. Every volume and surface area formula for pyramids, cones, spheres, hemispheres and their combinations is provided in the formula booklet, so you don't need to memorise them - just know which one to apply.
Can I use my GDC for 3D geometry questions?
Yes, on calculator papers your GDC handles the arithmetic (square roots, trig ratios, evaluating volume formulas) quickly and accurately. The harder part is usually setting up the right expression from the 3D diagram, which is done by hand regardless. See the GDC guide for more.
Sub-topics
3D Geometry broken down into its individual skills, each with its own focused page.
Related topics
More Geometry & Trigonometry topics from the same AA SL syllabus unit, in case you want to keep going.