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IB Maths AI Topic 3 Notes

This page contains our IB Maths AI notes for topic 3. By reading each one of these notes, you will fully cover the content for IB Maths AI 'Geometry & Trigonometry'.

All Topic 3 Sub-topics3.1: Fundamentals Of Geometry3.2: Introduction To Trigonometry3.3: Applied Trigonometry3.4: Circles3.5: Perpendicular Bisectors3.6: Voronoi Diagrams (HL)3.7: Unit circle (HL)3.8: Geometric transformations (HL)3.9: Vectors (HL)3.10: Vector lines (HL)3.11: Graph theory (HL)3.12: Adjacency matrices (HL)3.13: Algorithms & problems

Chapters

3.1: Fundamentals of geometry

1. Distance in 3D

3.2: Introduction to trigonometry

1. Right-angled triangle trigonometry

3.3: Applied trigonometry

1. Bearings

3.4: Circles

1. Radians

3.5: Perpendicular bisectors

1. Perpendicular bisectors

3.6: Voronoi diagrams

1. Voronoi diagrams

3.7: Unit circle (HL)

1. The unit circle

3.8: Geometric transformations (HL)

1. Geometric transformations

3.9: Vectors (HL)

1. Scalars & Vectors

3.10: Vector lines (HL)

1. Equations of a line

3.11: Graph theory (HL)

1. Graph theory

3.12: Adjacency matrices (HL)

1. Adjacency matrices

3.13: Algorithms & problems (HL)

1. Eulerian trails & circuits
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Distance in 3D

If two points are A(x1,y1,z1)A(x_1,y_1,z_1) and B(x2,y2,z2)B(x_2,y_2,z_2), then the distance between them is:

d=(x2−x1)2+(y2−y1)2+(z2−z1)2d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Math Topic 3 subTopic 1 notes image 1

This is the three-dimensional version of Pythagoras' theorem.

Find the distance between A(0,6,0)A(0,6,0) and D(3,3,6)D(3,3,6).

Substitute the coordinates into the formula: d=(3−0)2+(3−6)2+(6−0)2d=\sqrt{(3-0)^2+(3-6)^2+(6-0)^2}.

d=32+(−3)2+62d=\sqrt{3^2+(-3)^2+6^2}.

d=9+9+36=54d=\sqrt{9+9+36}=\sqrt{54}.

d=36d=3\sqrt{6}.

So the distance between AA and DD is 363\sqrt{6}.

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