IB Math AA SL Topic 3.1: 3D geometry | Notes & Flashcards | Aimnova

IB Math AA SL — 3D geometry

Topic 3.1 of IB Mathematics: Analysis and Approaches covers 3D geometry, which is part of Unit 3: Geometry & Trigonometry. Students explore key concepts including Distance & midpoint (3D), Volume & surface area, Angles in 3D. A strong understanding of 3d geometry is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in 3D geometry

Key Idea: 3D geometry turns a solid into something you can measure: the distance between corners, the volume and surface area of a shape, and the angle a line makes with a face. It runs across both papers — by hand on Paper 1, with the GDC doing the arithmetic on Paper 2.

📏 Distance & midpoint in 3D

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

$d$: distance between two 3D points (x, y, z)

Distance = subtract each coordinate, square, add, root (point order doesn't matter — the squaring kills the sign). Midpoint = add each coordinate pair and halve — $M = \left( \tfrac{x₁+ₓ₂}{2}, \tfrac{y₁+y₂}{2}, \tfrac{z₁+z₂}{2} \right)$. Given the midpoint and one end, the missing endpoint is B = 2M − A.

🧊 Volume, surface area & angles

Composite (e.g. a cylinder topped by a hemisphere) → add the part volumes, but for surface area count only exposed faces — never the shared join. Backwards (given a volume, find r or h) → set the formula equal and solve; cancel the π on both sides first to keep the numbers friendly.

Tip: Almost every 3D angle becomes a 2D right-angled triangle — find it, redraw it flat, use SOH-CAH-TOA. The angle between a line and a plane is measured to the line's projection (shadow) on the plane, and you usually need a face or base diagonal (Pythagoras) first — so it's a two-step problem: length, then angle.

✏️ IB-style worked examples

IB-style question — distance & midpoint in 3D

Points are P(1, 3, 2) and Q(7, 11, 26). Find (a) the distance PQ and (b) the midpoint of PQ.

Step by step:

(a) Coordinate gaps are 6, 8 and 24.
$d = \sqrt{6^2 + 8^2 + 24^2}$
Square, add, root.
$= \sqrt{36 + 64 + 576} = \sqrt{676} = 26$
(b) Midpoint = average each coordinate.
$M = \left( \tfrac{1+7}{2}, \tfrac{3+11}{2}, \tfrac{2+26}{2} \right) = (4, 7, 14)$

Final answer:

PQ = 26 and the midpoint is (4, 7, 14).

IB-style question — volume, surface area and a composite solid

A solid is a cone of base radius 5 and height 12, sitting on top of a cylinder of the same radius and height 8. Find (a) the cone's curved (slant) surface area and (b) the total volume of the solid. Leave answers in terms of π.

Step by step:

(a) Slant height first, then curved area = πrl.
$l = \sqrt{5^2 + 12^2} = 13, \quad \pi r l = \pi(5)(13) = 65\pi$
(b) Add the two volumes — cone is ⅓πr²h.
$V_{\text{cone}} = \tfrac{1}{3}\pi(5)^2(12) = 100\pi$
Cylinder is πr²h; total = sum.
$V_{\text{cyl}} = \pi(5)^2(8) = 200\pi \;\Rightarrow\; V = 300\pi$

Final answer:

Curved area = 65π; total volume = 300π ≈ 942.

IB-style question — angle between a line and a plane (Paper 2)

A cuboid has a square base of side 6 and height 8. Find the angle the space diagonal (corner to opposite corner) makes with the base.

Step by step:

Length first — the base diagonal by Pythagoras.
$\text{base diagonal} = \sqrt{6^2 + 6^2} = 6\sqrt{2}$
Right triangle: opposite = height 8, adjacent = base diagonal.
$\tan\theta = \frac{8}{6\sqrt{2}}$
Inverse tan on the GDC.
$\theta = \tan^{-1}\!\left(\tfrac{8}{6\sqrt{2}}\right) \approx 43.3^\circ$

Final answer:

The diagonal makes about 43.3° with the base.

Important: Two traps dominate this topic. Geometry: in $V = \tfrac{1}{3}\pi r² h$ only the radius is squared — never square h. And for an angle, find the right diagonal first (the adjacent side is often a base diagonal, not an edge). Calculator: on Paper 2 an inverse-trig answer is only right if the GDC is in DEGREE mode.

Tap each card to reveal the answer.

Exam Tips

Distance: subtract, square, add, root. Midpoint: add and halve. Missing end: B = 2M − A.
Volume/area formulas are in the booklet — pick the right one and read the correct r and h.
Composite solids: add the parts (subtract holes); count only exposed faces for surface area.
Angle problems are two-step — find the diagonal with Pythagoras first, then SOH-CAH-TOA.
On Paper 2, check the GDC is in DEGREE mode before any inverse-trig angle.

IB Math AA SL — 3D geometry

Key concepts in 3D geometry

Key Idea: 3D geometry turns a solid into something you can measure: the distance between corners, the volume and surface area of a shape, and the angle a line makes with a face. It runs across both papers — by hand on Paper 1, with the GDC doing the arithmetic on Paper 2.

📏 Distance & midpoint in 3D

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

$d$: distance between two 3D points (x, y, z)

Distance = subtract each coordinate, square, add, root (point order doesn't matter — the squaring kills the sign). Midpoint = add each coordinate pair and halve — $M = \left( \tfrac{x₁+ₓ₂}{2}, \tfrac{y₁+y₂}{2}, \tfrac{z₁+z₂}{2} \right)$. Given the midpoint and one end, the missing endpoint is B = 2M − A.

🧊 Volume, surface area & angles

Composite (e.g. a cylinder topped by a hemisphere) → add the part volumes, but for surface area count only exposed faces — never the shared join. Backwards (given a volume, find r or h) → set the formula equal and solve; cancel the π on both sides first to keep the numbers friendly.

Tip: Almost every 3D angle becomes a 2D right-angled triangle — find it, redraw it flat, use SOH-CAH-TOA. The angle between a line and a plane is measured to the line's projection (shadow) on the plane, and you usually need a face or base diagonal (Pythagoras) first — so it's a two-step problem: length, then angle.

✏️ IB-style worked examples

IB-style question — distance & midpoint in 3D

Points are P(1, 3, 2) and Q(7, 11, 26). Find (a) the distance PQ and (b) the midpoint of PQ.

Step by step:

(a) Coordinate gaps are 6, 8 and 24.
$d = \sqrt{6^2 + 8^2 + 24^2}$
Square, add, root.
$= \sqrt{36 + 64 + 576} = \sqrt{676} = 26$
(b) Midpoint = average each coordinate.
$M = \left( \tfrac{1+7}{2}, \tfrac{3+11}{2}, \tfrac{2+26}{2} \right) = (4, 7, 14)$

Final answer:

PQ = 26 and the midpoint is (4, 7, 14).

IB-style question — volume, surface area and a composite solid

Step by step:

(a) Slant height first, then curved area = πrl.
$l = \sqrt{5^2 + 12^2} = 13, \quad \pi r l = \pi(5)(13) = 65\pi$
(b) Add the two volumes — cone is ⅓πr²h.
$V_{\text{cone}} = \tfrac{1}{3}\pi(5)^2(12) = 100\pi$
Cylinder is πr²h; total = sum.
$V_{\text{cyl}} = \pi(5)^2(8) = 200\pi \;\Rightarrow\; V = 300\pi$

Final answer:

Curved area = 65π; total volume = 300π ≈ 942.

IB-style question — angle between a line and a plane (Paper 2)

A cuboid has a square base of side 6 and height 8. Find the angle the space diagonal (corner to opposite corner) makes with the base.

Step by step:

Length first — the base diagonal by Pythagoras.
$\text{base diagonal} = \sqrt{6^2 + 6^2} = 6\sqrt{2}$
Right triangle: opposite = height 8, adjacent = base diagonal.
$\tan\theta = \frac{8}{6\sqrt{2}}$
Inverse tan on the GDC.
$\theta = \tan^{-1}\!\left(\tfrac{8}{6\sqrt{2}}\right) \approx 43.3^\circ$

Final answer:

The diagonal makes about 43.3° with the base.

Important: Two traps dominate this topic. Geometry: in $V = \tfrac{1}{3}\pi r² h$ only the radius is squared — never square h. And for an angle, find the right diagonal first (the adjacent side is often a base diagonal, not an edge). Calculator: on Paper 2 an inverse-trig answer is only right if the GDC is in DEGREE mode.

Tap each card to reveal the answer.

Exam Tips

Distance: subtract, square, add, root. Midpoint: add and halve. Missing end: B = 2M − A.
Volume/area formulas are in the booklet — pick the right one and read the correct r and h.
Composite solids: add the parts (subtract holes); count only exposed faces for surface area.
Angle problems are two-step — find the diagonal with Pythagoras first, then SOH-CAH-TOA.
On Paper 2, check the GDC is in DEGREE mode before any inverse-trig angle.

IB Math AA SL — 3D geometry

Key concepts in 3D geometry

📏 Distance & midpoint in 3D

🧊 Volume, surface area & angles

✏️ IB-style worked examples

IB-style question — distance & midpoint in 3D

IB-style question — volume, surface area and a composite solid

IB-style question — angle between a line and a plane (Paper 2)

Exam Tips

What you'll learn in Topic 3.1

Study resources — 3.1 3D geometry

Distance & midpoint (3D)

Volume & surface area

Angles in 3D

Ready to study 3D geometry?

Ready to practice?

IB Math AA SL — 3D geometry

Key concepts in 3D geometry

📏 Distance & midpoint in 3D

🧊 Volume, surface area & angles

✏️ IB-style worked examples

IB-style question — distance & midpoint in 3D

IB-style question — volume, surface area and a composite solid

IB-style question — angle between a line and a plane (Paper 2)

Exam Tips

What you'll learn in Topic 3.1

Study resources — 3.1 3D geometry

Distance & midpoint (3D)

Volume & surface area

Angles in 3D

Ready to study 3D geometry?

Ready to practice?