Mini-Lesson

Geometry & Trigonometry (AI HL)

8 missions. Guess, learn, then check. Your place is saved.

Mission 1What's the longest pencil that fits in a 20 × 12 × 9 cm box?
Mission 2How can you find a mountain's height without climbing it?
Mission 330 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?
Mission 4Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?
Mission 5Four phone masts: where in town is the signal weakest?
Mission 6One 2×2 matrix can rotate, stretch or flip. How can you tell which?
Mission 7How can one number tell you if two directions are at right angles?
Mission 8Two aircraft fly straight paths. Will they collide or just pass close?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Make a guess: What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Geometry · 3-D shapes

Distance and midpoint in 3-D

Points in space have three coordinates $(x, y, z)$ (x, y, z).
Distance between $(x_{1}, y_{1}, z_{1})$ (x₁, y₁, z₁) and $(x_{2}, y_{2}, z_{2})$ (x₂, y₂, z₂):

d = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}

Midpoint: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$ ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
In a cuboid $a \times b \times c$ a × b × c, Pythagoras on the base gives $a^{2} + b^{2}$ √(a²+b²); Pythagoras again with the height gives the space diagonal $a^{2} + b^{2} + c^{2}$ √(a²+b²+c²).

Worked example Geometry · 3-D shapes

Space diagonal and its angle with the base

Question

A cuboid measures 8 cm by 6 cm by 5 cm. Find the length of a space diagonal and the angle it makes with the base.

1
Base diagonal by Pythagoras: $8^{2} + 6^{2} = 10$ √(8²+6²)=10 cm
2
Space diagonal in the vertical right-angled triangle: $1 0^{2} + 5^{2} = 125 = 11.2$ √(10²+5²)=√125=11.2 cm (3 s.f.)
3
The angle with the base is between the diagonal and its projection, the base diagonal: $tan θ = \frac{5}{10}$ tan θ=5/10

Answer

11.2

cm and

θ = 26. 6^{\circ}

(3 s.f.)

⚠️ Watch out

the angle with the base uses the base diagonal (10 cm), not an edge;

tan^{- 1} \frac{5}{8}

gives the wrong

32. 0^{\circ}

.

Calculate

?

A

is

(2, - 1, 5)

and

B

is

(- 4, 1, 1)

.

M

is the midpoint of

A B

. Find the distance from

M

to

C (1, 6, 0)

. Give an exact answer.

Geometry · 3-D shapes

Angles in 3-D

The angle between a line and a plane is the angle between the line and its projection on the plane (the "shadow" straight below it).
Drop a perpendicular from a point on the line to the plane: this makes a right-angled triangle containing the angle.
The angle between two intersecting lines is found in the triangle that both lines lie in.
At SL, 3-D angle questions use right-angled trigonometry: sketch the triangle flat, label it, then use SOH CAH TOA.

Quick check

?

V A BC D

is a right pyramid on a square base

A BC D

.

M

is the centre of the base, directly below

V

, and

N

is the midpoint of

A B

. Which angle is the angle between the edge

V A

and the base?

Geometry · 3-D shapes

Volume and surface area

Solid	Volume	Surface area
Right pyramid	$\frac{1}{3} \times$ 1/3× base area $\times$ × height	base + triangular faces
Right cone	$\frac{1}{3} π r^{2} h$ 1/3π r²h	curved: $π r l$ π rl ( $l$ l = slant height)
Sphere	$\frac{4}{3} π r^{3}$ 4/3π r³	$4 π r^{2}$ 4π r²
Hemisphere	$\frac{2}{3} π r^{3}$ 2/3π r³	curved: $2 π r^{2}$ 2π r²

These formulas are in the IB formula booklet. For a combined solid, add the volumes, but count only the surfaces on the outside.
Slant height of a cone: $l = r^{2} + h^{2}$ l=√(r²+h²).

Worked example Geometry · 3-D shapes

A cone on a hemisphere

Question

A toy is a cone of radius 3 cm and height 4 cm fixed to a hemisphere of radius 3 cm. Find its volume and its total surface area.

1
Cone volume: $\frac{1}{3} π (3^{2}) (4) = 12 π$ 1/3π(3²)(4)=12π
2
Hemisphere volume: $\frac{2}{3} π (3^{3}) = 18 π$ 2/3π(3³)=18π, so total volume $= 30 π = 94.2$ =30π=94.2 cm³ (3 s.f.)
3
Slant height: $l = 3^{2} + 4^{2} = 5$ l=√(3²+4²)=5, so the curved cone surface is $π (3) (5) = 15 π$ π(3)(5)=15π
4
Curved hemisphere surface: $2 π (3^{2}) = 18 π$ 2π(3²)=18π; the flat circles are joined inside, so they don't count

Answer

volume

94.2

cm³, surface area

33 π = 104

cm² (3 s.f.)

⚠️ Watch out

using the whole sphere's

4 π r^{2}

gives

51 π

; only half the sphere is there.

Calculate

?A right pyramid has a square base of side 10 cm and vertical height 12 cm. Find the angle between a sloping edge and the base. Give your answer in degrees to 3 significant figures.

Calculate

?A grain silo is a cylinder of radius 4 m and height 10 m with a hemisphere of radius 4 m on top. Find the volume of the silo to 3 significant figures.

m³

Mission check

?What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Mission 2

How can you find a mountain's height without climbing it?

Make a guess: How can you find a mountain's height without climbing it?

Geometry · trigonometry

Right-angled trigonometry, elevation and bearings

In a right-angled triangle: $sin θ = \frac{opp}{hyp}$ sin θ=opp/hyp, $cos θ = \frac{adj}{hyp}$ cos θ=adj/hyp, $tan θ = \frac{opp}{adj}$ tan θ=opp/adj. Use $sin^{- 1}$ sin⁻¹, $cos^{- 1}$ cos⁻¹, $tan^{- 1}$ tan⁻¹ to find angles.
Angle of elevation: measured up from the horizontal. Angle of depression: measured down from the horizontal.
Bearings are measured clockwise from north, written with three figures (e.g. 040°).
Always sketch and label a diagram from the words first.

Quick check

?From the top of a vertical cliff 40 m high, the angle of depression of a boat is 25°. How far is the boat from the foot of the cliff, to 3 significant figures?

Geometry · trigonometry

Sine rule, cosine rule and area

Label the side opposite angle $A$ A as $a$ a, and so on.
Sine rule: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ a/(sin A)=b/(sin B)=c/(sin C). Use it with a side and its opposite angle known.
Cosine rule: $c^{2} = a^{2} + b^{2} - 2 ab cos C$ c²=a²+b²−2abcos C, or $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ cos C=(a²+b²−c²)/2ab. Use it with two sides and the angle between them, or with all three sides.
Area $= \frac{1}{2} ab sin C$ =1/2absin C ( $C$ C is the angle between sides $a$ a and $b$ b).
Finding an angle with the sine rule can give two possible triangles (the ambiguous case); you meet it in the unit circle mission.

Worked example Geometry · trigonometry

Cosine rule, then area

Question

In triangle

PQR

,

PQ = 8

cm,

PR = 11

cm and angle

QPR = 4 0^{\circ}

. Find

QR

and the area of the triangle.

1
Two sides and the included angle, so use the cosine rule: $Q R^{2} = 8^{2} + 1 1^{2} - 2 (8) (11) cos 4 0^{\circ}$ QR²=8²+11²−2(8)(11)cos 40°
2
Evaluate: $Q R^{2} = 50.18 \dots$ QR²=50.18…, so $QR = 7.08$ QR=7.08 cm (3 s.f.)
3
Area with the same two sides and included angle: $\frac{1}{2} (8) (11) sin 4 0^{\circ}$ 1/2(8)(11)sin 40°

Answer

QR = 7.08

cm, area

= 28.3

cm² (3 s.f.)

⚠️ Watch out

with your GDC in radian mode,

cos 40

gives

QR = 17.4

cm; set degree mode for degree questions.

Sort it

?Which rule do you use first?

Tap a statement, then tap the group it belongs in. Then press Check.

Worked example Geometry · trigonometry

Height from two angles of elevation

Question

From

A

, the angle of elevation of the top

T

of a tower is 20°. From

B

, 500 m closer on level ground, it is 32°. Find the height of the tower.

1
In triangle $A BT$ ABT: angle $A BT = 18 0^{\circ} - 3 2^{\circ} = 14 8^{\circ}$ ABT=180°−32°=148°, so angle $A TB = 18 0^{\circ} - 2 0^{\circ} - 14 8^{\circ} = 1 2^{\circ}$ ATB=180°−20°−148°=12°
2
Sine rule: $\frac{BT}{sin 2 0 ^{\circ}} = \frac{500}{sin 1 2 ^{\circ}}$ BT/(sin 20°)=500/(sin 12°), so $BT = 822.5$ BT=822.5 m
3
Right-angled triangle below $T$ T: height $= BT sin 3 2^{\circ}$ =BTsin 32°

Answer

height

= 436

m (3 s.f.)

⚠️ Watch out

the angle at

T

is the difference of the two elevations,

3 2^{\circ} - 2 0^{\circ}

, not their sum.

Calculate

?A ship sails 15 km from port

P

on a bearing of 040° to

Q

, then 9 km on a bearing of 110° to

R

. Find the distance

PR

. Give your answer to 3 significant figures.

km

Calculate

?A triangular field has sides 50 m, 70 m and 90 m. Find its area. Give your answer to the nearest square metre.

m²

Mission check

?How can you find a mountain's height without climbing it?

Mission 3

30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Make a guess: 30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Geometry · circles

Arc length and sector area

A sector is a "pizza slice" of a circle; its curved edge is an arc.
A sector with angle $θ^{\circ}$ θ° is the fraction $\frac{θ}{360}$ θ/360 of the whole circle, so

arc length = \frac{θ}{360} \times 2 π r sector area = \frac{θ}{360} \times π r^{2}

Perimeter of a sector $=$ = arc $+ 2 r$ + 2r (two radii).

Worked example Geometry · circles

A sector in degrees

Question

A sector has radius 9 cm and angle 140°. Find its arc length, area and perimeter.

1
Fraction of the circle: $\frac{140}{360} = \frac{7}{18}$ 140/360=7/18
2
Arc length: $\frac{7}{18} \times 2 π (9) = 7 π = 22.0$ 7/18× 2π(9)=7π=22.0 cm (3 s.f.)
3
Area: $\frac{7}{18} \times π (9^{2}) = 31.5 π = 99.0$ 7/18× π(9²)=31.5π=99.0 cm² (3 s.f.)
4
Perimeter: $7 π + 2 (9) = 40.0$ 7π+2(9)=40.0 cm (3 s.f.)

Answer

arc

22.0

cm, area

99.0

cm², perimeter

40.0

cm

⚠️ Watch out

the perimeter includes both straight radii, not just the arc.

Calculate

?A sector of a circle of radius 6 cm has area 30 cm². Find the angle of the sector in degrees, to 3 significant figures.

Geometry · circles

Radians

One radian is the angle at the centre when the arc length equals the radius.
A full turn is $2 π$ 2π radians, so $π rad = 18 0^{\circ}$ π rad=180°.
Degrees to radians: multiply by $\frac{π}{180}$ π/180. Radians to degrees: multiply by $\frac{180}{π}$ 180/π.
Learn: $3 0^{\circ} = \frac{π}{6}$ 30°=π/6, $4 5^{\circ} = \frac{π}{4}$ 45°=π/4, $6 0^{\circ} = \frac{π}{3}$ 60°=π/3, $9 0^{\circ} = \frac{π}{2}$ 90°=π/2.
With $θ$ θ in radians: arc $l = r θ$ l=rθ and sector area $A = \frac{1}{2} r^{2} θ$ A=1/2r²θ. Set your GDC to radian mode.

Worked example Geometry · circles

A sector in radians

Question

A sector has radius 8 cm and angle 1.2 radians. Find its arc length, area and perimeter.

1
Arc length: $l = r θ = 8 (1.2) = 9.6$ l=rθ=8(1.2)=9.6 cm
2
Area: $A = \frac{1}{2} r^{2} θ = \frac{1}{2} (8^{2}) (1.2) = 38.4$ A=1/2r²θ=1/2(8²)(1.2)=38.4 cm²
3
Perimeter: $9.6 + 2 (8) = 25.6$ 9.6+2(8)=25.6 cm

Answer

arc

9.6

cm, area

38.4

cm², perimeter

25.6

cm

⚠️ Watch out

l = r θ

only works with

θ

in radians; 1.2 rad is about

68. 8^{\circ}

, not

1. 2^{\circ}

.

Quick check

?Write

13 5^{\circ}

in radians.

Geometry · circles

Segments and compound shapes

A segment is the region between a chord and its arc.
Segment area $=$ = sector area $-$ − triangle area, with the triangle area $\frac{1}{2} r^{2} sin θ$ 1/2r²sin θ (two radii and the angle between them).
For a shape made of sectors, rectangles and triangles: find each piece, then add or subtract.
Work with unrounded values and round only the final answer to 3 significant figures.

Calculate

?A circle has centre

O

and radius 10 cm. Points

A

and

B

lie on the circle with angle

A OB = 1.4

radians. Find the area of the segment between chord

A B

and the minor arc

A B

. Give your answer to 3 significant figures.

cm²

Calculate

?A fan-shaped badge is a sector of radius 12 cm. Its perimeter is 40 cm. Find its area.

cm²

Mission check

?30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Mission 4

Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?

Make a guess: Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?

Trigonometry · unit circle

Sine and cosine on the unit circle

The unit circle has centre $O$ O and radius 1. Turn a radius $θ$ θ anticlockwise from the positive $x$ x-axis.
Its end point is $(cos θ, sin θ)$ (cos θ, sin θ): cosine is the $x$ x-coordinate, sine is the height.
$tan θ = \frac{sin θ}{cos θ}$ tan θ=(sin θ)/(cos θ), the gradient of the radius; the line is $y = x tan θ$ y=xtan θ.
This works for any angle, so signs follow the quadrant: in the 2nd quadrant $sin θ > 0$ sin θ>0 and $cos θ < 0$ cos θ<0.

Trigonometry · unit circle

Angles in different quadrants

Reflect in the $y$ y-axis: $sin (18 0^{\circ} - θ) = sin θ$ sin (180°−θ)=sin θ, $cos (18 0^{\circ} - θ) = - cos θ$ cos (180°−θ)=−cos θ.
Reflect in the $x$ x-axis: $sin (- θ) = - sin θ$ sin (−θ)=−sin θ, $cos (- θ) = cos θ$ cos (−θ)=cos θ.
Half a turn: $sin (18 0^{\circ} + θ) = - sin θ$ sin (180°+θ)=−sin θ, $cos (18 0^{\circ} + θ) = - cos θ$ cos (180°+θ)=−cos θ.
In radians, $18 0^{\circ}$ 180° is $π$ π: for example $sin (π - x) = sin x$ sin (π−x)=sin x.

Calculate

?The angle

θ

is obtuse and

sin θ = sin 5 0^{\circ}

. Find

θ

in degrees.

Quick check

?Which angle has the same cosine as

7 0^{\circ}

?

Trigonometry · unit circle

Identities

From the unit circle and Pythagoras: $cos^{2} θ + sin^{2} θ = 1$ cos²θ+sin²θ=1.
Use it to find one ratio from another; the quadrant decides the sign of the square root.

Worked example Trigonometry · unit circle

One ratio from another

Question

sin θ = \frac{3}{5}

and

θ

is obtuse. Find

cos θ

and

tan θ

.

1
Pythagorean identity: $cos^{2} θ = 1 - (\frac{3}{5})^{2} = \frac{16}{25}$ cos²θ=1−(3/5)²=16/25
2
$θ$ θ is in the 2nd quadrant, so cosine is negative: $cos θ = - \frac{4}{5}$ cos θ=−4/5
3
$tan θ = \frac{sin θ}{cos θ} = \frac{3/5}{- 4/5}$ tan θ=(sin θ)/(cos θ)=(3/5)/(−4/5)

Answer

cos θ = - \frac{4}{5}

,

tan θ = - \frac{3}{4}

⚠️ Watch out

cos^{2} θ = \frac{16}{25}

gives

\pm \frac{4}{5}

; taking

+ \frac{4}{5}

ignores that

θ

is obtuse.

Trigonometry · unit circle

Solving equations with graphs

Graph $y = sin x$ y=sin x (built from the heights on the unit circle) and the line $y = k$ y=k on your GDC over the given interval.
Each intersection is a solution: count them all, not just the first one the GDC finds.
The symmetry of the curve checks your answers: if $x_{1}$ x₁ solves $sin x = k$ sin x=k, so does $π - x_{1}$ π−x₁.
Radians are assumed on HL papers unless degrees are stated: set the mode first.

Calculate

?Use your GDC to solve

sin x = 0.4

for

0 \leq x \leq 2 π

. Give the larger solution to 3 significant figures.

Worked example Trigonometry · unit circle

The ambiguous case of the sine rule

Question

In triangle

A BC

,

A B = 12

cm,

BC = 8

cm and angle

A = 3 0^{\circ}

. Find the possible sizes of angle

C

.

1
Sine rule: $\frac{sin C}{12} = \frac{sin 3 0 ^{\circ}}{8}$ (sin C)/12=(sin 30°)/8, so $sin C = 0.75$ sin C=0.75
2
Inverse sine: $C = 48. 6^{\circ}$ C=48.6°; the unit circle gives a second angle, $18 0^{\circ} - 48. 6^{\circ} = 131. 4^{\circ}$ 180°−48.6°=131.4°
3
Test each: $3 0^{\circ} + 131. 4^{\circ} < 18 0^{\circ}$ 30°+131.4°<180°, so both triangles exist

Answer

C = 48. 6^{\circ}

or

C = 131. 4^{\circ}

(1 d.p.)

⚠️ Watch out

when the known angle is opposite the shorter given side, check

18 0^{\circ}

minus your answer; a GDC shows only the acute one.

Mission check

?Why is

sin 15 0^{\circ}

exactly equal to

sin 3 0^{\circ}

?

Mission 5

Four phone masts: where in town is the signal weakest?

Make a guess: Four phone masts: where in town is the signal weakest?

Geometry · Voronoi diagrams

Perpendicular bisectors

The perpendicular bisector of $A B$ AB is the set of points equally distant from $A$ A and $B$ B.
It passes through the midpoint $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ ((x₁ + x₂)/2, (y₁ + y₂)/2).
Its gradient is the negative reciprocal of the gradient of $A B$ AB: if $A B$ AB has gradient $m$ m, the bisector has gradient $- \frac{1}{m}$ −1/m.
Then use $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁) through the midpoint. If $A B$ AB is horizontal, the bisector is vertical ( $x =$ x = constant), and vice versa.

Worked example Geometry · Voronoi diagrams

The boundary between two masts

Question

Masts are at

A (1, 1)

and

D (8, 8)

, in km. Find the equation of the perpendicular bisector of

A D

.

1
Midpoint: $(\frac{1 + 8}{2}, \frac{1 + 8}{2}) = (4.5, 4.5)$ ((1 + 8)/2, (1 + 8)/2) = (4.5, 4.5)
2
Gradient of $A D$ AD: $\frac{8 - 1}{8 - 1} = 1$ (8 − 1)/(8 − 1) = 1
3
Perpendicular gradient: $- \frac{1}{1} = - 1$ −1/1 = −1
4
Line through the midpoint: $y - 4.5 = - 1 (x - 4.5)$ y − 4.5 = −1(x − 4.5)

Answer

y = - x + 9

⚠️ Watch out

the bisector goes through the midpoint, not through

A

or

D

. Using

(1, 1)

gives

y = - x + 2

, which is wrong.

Calculate

?A line segment lies on the line

y = 3 x - 4

and has midpoint

(2, 2)

. Its perpendicular bisector meets the

x

-axis at

(k, 0)

. Find

k

.

Geometry · Voronoi diagrams

Voronoi diagrams

A Voronoi diagram splits a region into cells, one around each site. Every point in a cell is nearer its own site than any other.
Edges are parts of perpendicular bisectors between pairs of sites. A vertex is where three edges meet: it is equidistant from three sites.
Nearest neighbour interpolation: every point in a cell is given the same value as its site (for example, the rainfall measured there).
Exams give the site coordinates; you will not construct bisectors.

Quick check

?Rain gauges at the four sites

A (1, 1)

,

B (11, 1)

,

C (1, 7)

and

D (8, 8)

record 12 mm, 20 mm, 16 mm and 9 mm. Use nearest neighbour interpolation to estimate the rainfall at

H (4, 6)

.

Geometry · Voronoi diagrams

Adding a site; the toxic waste dump

Adding a site: draw the bisectors between the new site and its neighbours; the new cell is carved from theirs.
Toxic waste dump problem: find the point in the region as far as possible from every site (a dump, or the weakest signal).
Within the region, the farthest point from every site is at a Voronoi vertex or on the region's boundary; exams put it at a vertex. Find each vertex's distance to one of its sites: the largest wins.

Worked example Geometry · Voronoi diagrams

Where is the weakest signal?

Question

In the town

0 \leq x \leq 12

,

0 \leq y \leq 10

(km), masts

A (1, 1)

,

B (11, 1)

,

C (1, 7)

and

D (8, 8)

give a Voronoi diagram with vertices

P (5, 4)

and

Q

. The edge between cells

A

and

B

is

x = 6

, and between

A

and

D

it is

y = - x + 9

. Find

Q

, and the point of the town farthest from every mast.

1
Solve $x = 6$ x = 6 with $y = - x + 9$ y = −x + 9: $Q = (6, 3)$ Q = (6, 3), equidistant from $A$ A, $B$ B and $D$ D
2
Distance from $Q$ Q to $A$ A: $5^{2} + 2^{2} = 29 = 5.39$ √(5² + 2²) = √29 = 5.39 km
3
Distance from $P$ P to $A$ A (one of $A$ A, $C$ C, $D$ D): $4^{2} + 3^{2} = 5$ √(4² + 3²) = 5 km
4
Compare: $5.39 > 5$ 5.39 > 5

Answer

Q (6, 3)

, 5.39 km from its three nearest masts.

⚠️ Watch out

measure from the vertex to one of its own three sites.

Q

to

C

is 6.40 km, but

A

,

B

and

D

are nearer.

Calculate

?For the town and masts above, the cell of mast

A

has vertices

(0, 0)

,

(6, 0)

,

(6, 3)

,

(5, 4)

and

(0, 4)

. Find the area of the cell of

A

.

km²

Calculate

?A fifth mast

E

is built at

(11, 6)

. The boundary between the new cells of

D (8, 8)

and

E

has equation

y = m x + c

. Find

c

as a decimal.

Mission check

?Four phone masts: where in town is the signal weakest?

Mission 6

One 2×2 matrix can rotate, stretch or flip. How can you tell which?

Make a guess: One 2×2 matrix can rotate, stretch or flip. How can you tell which?

Geometry · matrix transformations

Columns are images

A point $(x, y)$ (x, y) is written as a column vector. Its image is $(a c b d) (x y) + (e f)$ ( a , b ; c , d )( x ; y ) + ( e ; f ); the added vector is a translation.
$(a c b d) (10) = (a c)$ ( a , b ; c , d )( 1 ; 0 ) = ( a ; c ) and $(a c b d) (01) = (b d)$ ( a , b ; c , d )( 0 ; 1 ) = ( b ; d ).
So the first column is the image of $(1, 0)$ (1, 0) and the second column is the image of $(0, 1)$ (0, 1). To find a matrix, ask where those two points go.

Worked example Geometry · matrix transformations

Building a rotation matrix

Question

Find the matrix for a rotation of

9 0^{\circ}

anticlockwise about the origin, and the image of

(3, 1)

.

1
$(1, 0)$ (1, 0) rotates to $(0, 1)$ (0, 1): first column $(01)$ ( 0 ; 1 )
2
$(0, 1)$ (0, 1) rotates to $(- 1, 0)$ (−1, 0): second column $(- 1 0)$ ( −1 ; 0 )
3
Matrix: $(01 - 1 0)$ ( 0 , −1 ; 1 , 0 )
4
Multiply: $(01 - 1 0) (31) = (- 1 3)$ ( 0 , −1 ; 1 , 0 )( 3 ; 1 ) = ( −1 ; 3 )

Answer

(01 - 1 0)

; the image is

(- 1, 3)

.

⚠️ Watch out

writing the images as rows instead of columns gives

(0 - 1 10)

, a clockwise rotation, which sends

(3, 1)

to

(1, - 3)

.

Quick check

?Which matrix represents a reflection in the

y

-axis?

Geometry · matrix transformations

The standard matrices

Transformation	Matrix
rotation $θ$ θ anticlockwise about $O$ O	$(cos θ sin θ - sin θ cos θ)$ ( cos θ , −sin θ ; sin θ , cos θ )
reflection in $y = (tan θ) x$ y = (tan θ)x	$(cos 2 θ sin 2 θ sin 2 θ - cos 2 θ)$ ( cos 2θ , sin 2θ ; sin 2θ , −cos 2θ )
horizontal stretch, scale factor $k$ k	$(k 0 01)$ ( k , 0 ; 0 , 1 )
vertical stretch, scale factor $k$ k	$(10 0 k)$ ( 1 , 0 ; 0 , k )
enlargement, scale factor $k$ k, centre $O$ O	$(k 0 0 k)$ ( k , 0 ; 0 , k )
translation by $(e f)$ ( e ; f )	add $(e f)$ ( e ; f )

A clockwise rotation by $θ$ θ uses $- θ$ −θ. The reflection matrix uses $2 θ$ 2θ, where $θ$ θ is the line's angle with the $x$ x-axis.

Calculate

?The point

P (4, 0)

is reflected in the line

y = (tan 3 0^{\circ}) x

and the image is then translated by

(1 - 2)

. Use your GDC to find the

y

-coordinate of the final image, to 3 significant figures.

Geometry · matrix transformations

Composites and the determinant

"Transformation $A$ A, then transformation $B$ B" has matrix $B A$ BA: the first one sits next to the vector, $B (A x) = (B A) x$ B(Ax) = (BA)x.
Order matters: in general $B A \neq = A B$ BA ≠ AB.
With translations, apply step by step: $x \mapsto A x + b$ x mapsto Ax + b, then $\mapsto B (A x + b) + c$ mapsto B(Ax + b) + c.
Area of image $= ∣ det A ∣ \times$ = |det A| × area of object, where $det (a c b d) = a d - b c$ det ( a , b ; c , d ) = ad − bc.

Worked example Geometry · matrix transformations

A reflection then a rotation

Question

Triangle

T

is reflected in the

x

-axis, then rotated

9 0^{\circ}

anticlockwise about the origin. Find the single matrix for the combined transformation and describe it.

1
Reflection in the $x$ x-axis: $X = (10 0 - 1)$ X = ( 1 , 0 ; 0 , −1 ); rotation: $R = (01 - 1 0)$ R = ( 0 , −1 ; 1 , 0 )
2
$X$ X is applied first, so the combined matrix is $RX$ RX
3
$RX = (01 - 1 0) (10 0 - 1) = (0110)$ RX = ( 0 , −1 ; 1 , 0 )( 1 , 0 ; 0 , −1 ) = ( 0 , 1 ; 1 , 0 )
4
Read the columns: $(1, 0) \mapsto (0, 1)$ (1, 0) mapsto (0, 1) and $(0, 1) \mapsto (1, 0)$ (0, 1) mapsto (1, 0), so $x$ x and $y$ y swap

Answer

(0110)

, a reflection in the line

y = x

.

⚠️ Watch out

XR

is a different transformation. Write the matrices right to left in the order they happen.

Quick check

?A shape is rotated

9 0^{\circ}

anticlockwise about the origin, then reflected in the

x

-axis. Which single matrix does this?

Calculate

?Triangle

T

has vertices

(0, 0)

,

(4, 0)

and

(1, 3)

. It is transformed by

M = (32 1 - 1)

. Find the area of the image.

square units

Mission check

?One 2×2 matrix can rotate, stretch or flip. How can you tell which?

Mission 7

How can one number tell you if two directions are at right angles?

Make a guess: How can one number tell you if two directions are at right angles?

Vectors · scalar product

Components, magnitude and unit vectors

A vector has size and direction: $v = v_{1} v_{2} v_{3} = v_{1} i + v_{2} j + v_{3} k$ v = ( v₁ ; v₂ ; v₃ ) = v₁i + v₂j + v₃k.
Position vector $O A = a$ OA→ = a. Displacement $A B = b - a$ AB→ = b − a (end minus start).
Magnitude: $∣ v ∣ = v_{12} + v_{22} + v_{32}$ |v| = √(v₁² + v₂² + v₃²), so the distance $A B$ AB is $∣ A B ∣$ |AB→|.
A unit vector has magnitude 1: $\frac{v}{∣ v ∣}$ v/(|v|). Vectors are parallel if one is a scalar multiple of the other, $w = k v$ w = kv.

Calculate

?Points

A

and

B

have coordinates

(1, - 2, 3)

and

(3, 4, 0)

. Find the distance

A B

.

Worked example Vectors · scalar product

A velocity with a given speed

Question

A drone flies at a constant speed of 15 m s⁻¹ in the direction of

2 i - 2 j + k

. Find its velocity vector.

1
Find the magnitude of the direction vector: $2^{2} + (- 2)^{2} + 1^{2} = 9 = 3$ √(2² + (−2)² + 1²) = √9 = 3
2
Normalise to get a unit vector: $\frac{1}{3} 2 - 2 1$ 1/3( 2 ; −2 ; 1 )
3
Rescale to the speed: $15 \times \frac{1}{3} 2 - 2 1 = 10 - 10 5$ 15 × 1/3( 2 ; −2 ; 1 ) = ( 10 ; −10 ; 5 )

Answer

velocity

= 10 i - 10 j + 5 k

m s⁻¹. Check:

100 + 100 + 25 = 15

.

⚠️ Watch out

15 (2 i - 2 j + k)

has magnitude 45, not 15. Normalise first.

Vectors · scalar product

The scalar product

The scalar product (dot product) of $v$ v and $w$ w is a number, not a vector:

v \cdot w = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} = ∣ v ∣∣ w ∣ cos θ

$θ$ θ is the angle between the vectors when they point away from the same point. So $cos θ = \frac{v \cdot w}{∣ v ∣∣ w ∣}$ cos θ = (v·w)/(|v||w|).
For non-zero vectors: $v \cdot w > 0$ v·w > 0 means $θ$ θ is acute, $v \cdot w = 0$ v·w = 0 means perpendicular, and $v \cdot w < 0$ v·w < 0 means obtuse.

Worked example Vectors · scalar product

The angle between two vectors

Question

Find the angle between

p = 21 - 2

and

q = 304

, in degrees.

1
Scalar product: $p \cdot q = 6 + 0 - 8 = - 2$ p·q = 6 + 0 − 8 = −2
2
Magnitudes: $∣ p ∣ = 4 + 1 + 4 = 3$ |p| = √(4 + 1 + 4) = 3 and $∣ q ∣ = 9 + 0 + 16 = 5$ |q| = √(9 + 0 + 16) = 5
3
Cosine: $cos θ = \frac{- 2}{3 \times 5} = - \frac{2}{15}$ cos θ = (−2)/(3 × 5) = −2/15
4
Inverse cosine: $θ = arccos (- \frac{2}{15})$ θ = arccos (−2/15)

Answer

θ \approx 97. 7^{\circ}

(3 s.f.), an obtuse angle.

⚠️ Watch out

keep the minus sign. Dropping it gives

82. 3^{\circ}

, the supplement of the true angle.

Calculate

?Points

A (1, 0, 2)

,

B (3, 1, 0)

and

C (4, k, 5)

are such that angle

A B^C = 9 0^{\circ}

. Find the value of

k

.

Sort it

?Use the scalar product (and scalar multiples) to classify each pair of vectors.

Tap a statement, then tap the group it belongs in. Then press Check.

Vectors · scalar product

The vector product and components

The vector product $v \times w$ v×w is a vector perpendicular to both $v$ v and $w$ w:

v \times w = v_{2} w_{3} - v_{3} w_{2} v_{3} w_{1} - v_{1} w_{3} v_{1} w_{2} - v_{2} w_{1}

Its direction follows the right-hand screw rule, and $∣ v \times w ∣ = ∣ v ∣∣ w ∣ sin θ$ |v×w| = |v||w|sin θ.
$∣ v \times w ∣$ |v×w| is the area of the parallelogram with sides $v$ v and $w$ w. A triangle is half of that.
The component of $a$ a in the direction of $b$ b is $\frac{a \cdot b}{∣ b ∣} = ∣ a ∣ cos θ$ (a·b)/(|b|) = |a|cos θ. Perpendicular to $b$ b it is $\frac{∣ a \times b ∣}{∣ b ∣} = ∣ a ∣ sin θ$ (|a×b|)/(|b|) = |a|sin θ.

Calculate

?Find the area of the triangle with vertices

P (1, 0, 0)

,

Q (0, 2, 0)

and

R (0, 0, 3)

.

Mission check

?How can one number tell you if two directions are at right angles? Take

u = 2 - 3 1

and

v = 42 - 2

.

Mission 8

Two aircraft fly straight paths. Will they collide or just pass close?

Make a guess: Two aircraft fly straight paths. Will they collide or just pass close?

Vectors · lines

The vector equation of a line

A line through the point with position vector $a$ a, in the direction of $b$ b, is

r = a + λ b, λ \in R

$a$ a fixes where the line is; the direction vector $b$ b fixes which way it points. Each value of $λ$ λ gives one point on the line.
Parametric form: write each component separately, $x = x_{0} + λ l$ x = x₀ + λ l, $y = y_{0} + λm$ y = y₀ + λ m, $z = z_{0} + λn$ z = z₀ + λ n.

Quick check

?Which point lies on the line

r = 12 - 1 + λ 2 - 1 3

?

Vectors · lines

Parallel, intersecting or skew

Compare the direction vectors first. If one is a multiple of the other, the lines are parallel (or coincident, if they share a point).
If not, set the two equations equal: $a_{1} + λ b_{1} = a_{2} + μ b_{2}$ a₁ + λb₁ = a₂ + μb₂. Use a different letter for each parameter.
Solve two of the component equations for $λ$ λ and $μ$ μ, then test the third.
Third equation true: the lines intersect. False: in 3-D the lines are skew; they are not parallel and never meet.

Worked example Vectors · lines

Finding a point of intersection

Question

Show that the lines

r = 1 - 1 2 + λ 21 - 1

and

r = 3 - 3 2 + μ 12 - 1

intersect, and find the point of intersection.

1
Direction vectors are not multiples of each other, so the lines are not parallel
2
Equate $x$ x and $y$ y: $1 + 2 λ = 3 + μ$ 1 + 2λ = 3 + μ and $- 1 + λ = - 3 + 2 μ$ −1 + λ = −3 + 2μ
3
Substitute $μ = 2 λ - 2$ μ = 2λ − 2 into the second: $- 1 + λ = - 7 + 4 λ$ −1 + λ = −7 + 4λ, so $λ = 2$ λ = 2 and $μ = 2$ μ = 2
4
Test $z$ z: $2 - λ = 0$ 2 − λ = 0 and $2 - μ = 0$ 2 − μ = 0. Both are 0, so the lines intersect
5
Substitute $λ = 2$ λ = 2 into the first line: $(1 + 4, - 1 + 2, 2 - 2)$ (1 + 4, −1 + 2, 2 − 2)

Answer

the lines intersect at

(5, 1, 0)

.

⚠️ Watch out

you must test the third equation. If the second line started at

(3, - 3, 5)

, the same

λ

and

μ

would fail the

z

test and the lines would be skew.

Calculate

?Find the acute angle between the lines

r = 014 + λ 122

and

r = 30 - 1 + μ 2 - 1 - 2

. Give your answer in degrees to 3 significant figures.

Vectors · lines

Lines as motion

An object moving with constant velocity $v$ v from position $r_{0}$ r₀ has position $r = r_{0} + t v$ r = r₀ + tv at time $t$ t. The parameter is now time.
The speed is $∣ v ∣$ |v|. The relative position of $B$ B from $A$ A is $A B = r_{B} - r_{A}$ AB→ = r_B − r_A.
Two paths can cross with no collision. A collision needs the same point at the same time: solve $r_{A} (t) = r_{B} (t)$ r_A(t) = r_B(t) with one $t$ t.
For the closest approach, find when $∣ A B ∣$ |AB→| is smallest: minimise $∣ A B ∣^{2}$ |AB→|² by calculus, or use your GDC.

Worked example Vectors · lines

Crossing paths

Question

Positions in km relative to a control tower,

t

minutes after noon, are

r_{A} = 123 + t 210

and

r_{B} = 8 - 2 0 + t - 1 21

. Do the paths cross, and do the aircraft collide?

1
Paths: use different parameters, $1 + 2 s = 8 - u$ 1 + 2s = 8 − u and $2 + s = - 2 + 2 u$ 2 + s = −2 + 2u, giving $s = 2$ s = 2, $u = 3$ u = 3
2
Test $z$ z: $3 = 0 + 3$ 3 = 0 + 3, true. The paths cross at $(5, 4, 3)$ (5, 4, 3)
3
Times: $A$ A is there when $t = 2$ t = 2, but $B$ B only when $t = 3$ t = 3
4
Collision test: $x$ x needs $1 + 2 t = 8 - t$ 1 + 2t = 8 − t, so $t = \frac{7}{3}$ t = 7/3, but $z$ z needs $3 = t$ 3 = t. No single $t$ t works

Answer

the paths cross at

(5, 4, 3)

, but the aircraft do not collide:

B

arrives 1 minute after

A

has passed.

⚠️ Watch out

using the same letter for both parameters in step 1 would wrongly suggest the paths never meet.

Calculate

?Two drones move with positions

r_{P} = 120 + t 2 - 1 1

and

r_{Q} = 7 - 4 9 + t 01 - 2

, in metres,

t

seconds after launch. They collide. Find the value of

t

when they collide.

Calculate

?For the aircraft

A

and

B

in the crossing-paths example,

A B = 7 - 3 t - 4 + t - 3 + t

. Find the minimum distance between the aircraft, in km to 3 significant figures. Use your GDC or calculus.

Mission check

?Two aircraft fly straight paths. Will they collide or just pass close? Take

r_{A} = 012 + t 110

and

r_{B} = 100 + t 011

.

Recap

Big ideas

3-D distance: $(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}$ √((x₁−x₂)²+(y₁−y₂)²+(z₁−z₂)²); a cuboid's space diagonal is $a^{2} + b^{2} + c^{2}$ √(a²+b²+c²).
The angle between a line and a plane is the angle with its projection: find the right-angled triangle and use trigonometry.
Combined solids: add volumes; count only the outside surfaces.
Right-angled triangles: SOH CAH TOA. Elevation is up from the horizontal; depression is down; bearings are clockwise from north.
Sine rule needs a side with its opposite angle; cosine rule needs two sides and the included angle, or three sides. Area $= \frac{1}{2} ab sin C$ =1/2absin C.
Break a real problem into triangles and sketch a labelled diagram first.
Arc $= \frac{θ}{360} \times 2 π r$ =θ/360× 2π r and sector $= \frac{θ}{360} \times π r^{2}$ =θ/360×π r²; perimeter adds two radii.
In radians: $π = 18 0^{\circ}$ π=180°, arc $l = r θ$ l=rθ, sector $A = \frac{1}{2} r^{2} θ$ A=1/2r²θ.
Segment $=$ = sector $-$ − triangle, with the triangle $\frac{1}{2} r^{2} sin θ$ 1/2r²sin θ.
On the unit circle, the point at angle $θ$ θ is $(cos θ, sin θ)$ (cos θ, sin θ) and $tan θ = \frac{s i n θ}{c o s θ}$ tan θ=(sin θ)/(cos θ); the quadrant fixes the signs.
$sin (18 0^{\circ} - θ) = sin θ$ sin (180°−θ)=sin θ, so the sine rule can give two angles: test both.
$cos^{2} θ + sin^{2} θ = 1$ cos²θ+sin²θ=1; solve trig equations on a GDC by counting every intersection in the interval.
The perpendicular bisector of $A B$ AB passes through the midpoint with gradient $- \frac{1}{m}$ −1/m; every point on it is equidistant from $A$ A and $B$ B.
Voronoi cells hold the points nearest each site; edges lie on bisectors; vertices are equidistant from three sites. Nearest neighbour interpolation gives a cell its site's value.
Toxic waste dump: test the vertices and choose the one farthest from its three sites.
The columns of a transformation matrix are the images of $(1, 0)$ (1, 0) and $(0, 1)$ (0, 1).
" $A$ A then $B$ B" is $B A$ BA; order matters. Translations are added as a vector.
Area of image $= ∣ det A ∣ \times$ = |det A| × area of object.
$A B = b - a$ AB→ = b − a, $∣ v ∣ = v_{12} + v_{22} + v_{32}$ |v| = √(v₁² + v₂² + v₃²), and a unit vector is $\frac{v}{∣ v ∣}$ v/(|v|): normalise, then rescale.
$v \cdot w = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} = ∣ v ∣∣ w ∣ cos θ$ v·w = v₁w₁ + v₂w₂ + v₃w₃ = |v||w|cos θ gives the angle between vectors; zero means perpendicular.
$v \times w$ v×w is perpendicular to both vectors, and $∣ v \times w ∣$ |v×w| is the area of the parallelogram they form.
A line is $r = a + λ b$ r = a + λb: a point plus multiples of a direction; the angle between lines comes from the direction vectors.
To compare two lines: parallel if the directions are multiples; otherwise solve with two parameters and test the third equation (intersect or skew).
With $r = r_{0} + t v$ r = r₀ + tv, a collision needs the same point at the same time; crossing paths alone don't collide.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Geometry & Trigonometry (AI HL) for IB Diploma Mathematics: Applications & Interpretation HL. 🎉

Mission 1What's the longest pencil that fits in a 20 × 12 × 9 cm box?
Mission 2How can you find a mountain's height without climbing it?
Mission 330 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?
Mission 4Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?
Mission 5Four phone masts: where in town is the signal weakest?
Mission 6One 2×2 matrix can rotate, stretch or flip. How can you tell which?
Mission 7How can one number tell you if two directions are at right angles?
Mission 8Two aircraft fly straight paths. Will they collide or just pass close?

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Geometry & Trigonometry (AI HL)

What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Distance and midpoint in 3-D

Space diagonal and its angle with the base

Angles in 3-D

Volume and surface area

A cone on a hemisphere

How can you find a mountain's height without climbing it?

Right-angled trigonometry, elevation and bearings

Sine rule, cosine rule and area

Cosine rule, then area

Height from two angles of elevation

30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Arc length and sector area

A sector in degrees

Radians

A sector in radians

Segments and compound shapes

Why is sin150∘sin 150° exactly equal to sin30∘sin 30°?

Sine and cosine on the unit circle

Angles in different quadrants

Identities

One ratio from another

Solving equations with graphs

The ambiguous case of the sine rule

Four phone masts: where in town is the signal weakest?

Perpendicular bisectors

The boundary between two masts

Voronoi diagrams

Adding a site; the toxic waste dump

Where is the weakest signal?

One 2×2 matrix can rotate, stretch or flip. How can you tell which?

Columns are images

Building a rotation matrix

The standard matrices

Composites and the determinant

A reflection then a rotation

How can one number tell you if two directions are at right angles?

Components, magnitude and unit vectors

A velocity with a given speed

The scalar product

The angle between two vectors

The vector product and components

Two aircraft fly straight paths. Will they collide or just pass close?

The vector equation of a line

Parallel, intersecting or skew

Finding a point of intersection

Lines as motion

Crossing paths

Big ideas

Mini-lesson complete!

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Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?