Vectors in Geometric Problems — Edexcel A-level Mathematics (Pure)

SPEC 10.6 • VECTORS

Modelling with Vectors

Use vectors to solve problems involving velocity, displacement, force, and bearings

Vector vs Scalar Quantities

In mechanics, we distinguish between vector quantities (magnitude AND direction) and scalar quantities (magnitude only).

Key Relationships

Vector Quantities:

Velocity

Displacement

Force

Acceleration

Scalar Quantities:

Speed = |velocity|

Distance = |displacement|

Mass

Time

Critical Distinction

Speed is the magnitude of the velocity vector.

Distance in a straight line between A and B is the magnitude of displacement vector

A B

.

Unit Vector Convention

When modelling with vectors in mechanics, we use standard unit vectors:

Standard Convention

i

= unit vector due East (positive x-direction)

j

= unit vector due North (positive y-direction)

DirectionVector FormExample (5 units)

East

i

5 i

North

j

5 j

West

- i

- 5 i

South

- j

- 5 j

Memory Aid

i = East (horizontal), j = North (vertical up)

01Worked Example

A girl walks 2 km due east from a fixed point O to A, and then 3 km due south from A to B. Find:

(a) the total distance travelled

(b) the position vector of B relative to O

(c) $∣ OB ∣$

a)

Total distance = 2 km + 3 km = 5 km

b)

East =

i

, so 2 km east =

2 i

•

South =

- j

, so 3 km south =

- 3 j

•

OB = (2 i - 3 j)

km

c)

∣ OB ∣ = 2^{2} + 3^{2} = 13 = 3.61

km (3 s.f.)

Note: Distance travelled (5 km) ≠ Distance from start (√13 km)

Speed and Velocity

A particle's velocity tells us how fast it's moving and in which direction. Speed is the magnitude of velocity.

Key Formulas

Speed =

∣ v ∣ = v_{x}^{2} + v_{y}^{2}

Distance = Speed × Time

Example

If velocity

v = (3 i + 4 j)

m s⁻¹

Speed =

3^{2} + 4^{2} = 25 = 5

m s⁻¹

Distance in 10 seconds = 5 × 10 = 50 m

Common Error

Don't just add components! Distance is NOT |3| + |4| = 7.

Use Pythagoras:

3^{2} + 4^{2} = 5

02Worked Example

A particle moves with velocity $v = (5 i - 12 j)$ m s⁻¹.

(a) Find the speed of the particle.

(b) Find the distance travelled in 8 seconds.

a)

Speed =

∣ v ∣ = 5^{2} + (- 12)^{2}

•

=

25 + 144 = 169 = 13

m s⁻¹

b)

Distance = Speed × Time

•

= 13 × 8 = 104 m

Bearings to Vectors

Bearings measure angles clockwise from North. To convert to i-j form:

Conversion Formula

For bearing θ° at speed/distance r:

v = r sin θ ° i + r cos θ ° j

East component = r sin θ, North component = r cos θ

Bearingsin θcos θVector (speed v)

030°0.50.866

0.5 v i + 0.866 v j

120°0.866−0.5

0.866 v i - 0.5 v j

240°−0.866−0.5

- 0.866 v i - 0.5 v j

Important

For bearings: i-component uses SIN, j-component uses COS (opposite to standard angles!)

Worked Example03

A cadet walks 15 km on a bearing of 120° from O to A. Find the position vector of A relative to O.

1

Bearing 120° means 120° clockwise from North

2

The angle from East (for the triangle) = 120° − 90° = 30°

3

East component:

15 sin 120° = 15 \times \frac{3}{2} = 13.0

km

4

North component:

15 cos 120° = 15 \times (- 0.5) = - 7.5

km

5

O A = (15 cos 30° i - 15 sin 30° j)

km

$O A = (13.0 i - 7.5 j)$ km

Finding Bearings from Vectors

To find the bearing of B from A, given position vector

A B

:

Method

Draw a diagram showing the vector from A to B

Find the angle θ using

tan θ = \frac{East component}{North component}

Convert to a three-figure bearing (measured clockwise from North)

Quadrant Check

i-componentj-componentQuadrantBearing++NEθ+−SE180° − θ−−SW180° + θ−+NW360° − θ

Worked Example04

Given $OB = (2 i - 3 j)$ km, find the bearing of B from O.

1

i-component = 2 (East), j-component = −3 (South)

2

This is in the SE quadrant (+ East, − North)

3

tan θ = \frac{3}{2} = 1.5

4

θ = tan^{- 1} (1.5) = 56.3°

5

Bearing = 90° + 56.3° = 146° (3 s.f.)

Remember: Bearings are always 3 figures, so write 146°

Force, Mass and Acceleration

Newton's Second Law applies to vectors: F = ma

Vector Form of F = ma

F = m a

Where F is resultant force (N), m is mass (kg), a is acceleration (m s⁻²)

Applications

Given F and m, find a:

a = \frac{F}{m}

Given a and m, find F:

F = m a

Magnitude of force:

∣ F ∣ = m ∣ a ∣

Resultant Force

If multiple forces act:

R = F_{1} + F_{2} + F_{3} + ...

Then

R = m a

05Worked Example

A particle P of mass 0.3 kg has acceleration $a = (5 i + 7 j)$ m s⁻².

(a) Find the angle between the acceleration and the vector i.

(b) Find the magnitude of the force F acting on P.

a)

Angle with i:

tan θ = \frac{7}{5}

•

θ = tan^{- 1} (1.4) = 54.5°

b)

Using

F = m a

:

•

∣ F ∣ = m ∣ a ∣ = 0.3 \times 5^{2} + 7^{2}

•

= 0.3 \times 74 = 0.3 \times 8.60...

|F| = 2.58 N (3 s.f.)

Acceleration from Velocities

For constant acceleration, we can find acceleration from initial and final velocities:

Acceleration Formula

a = \frac{v - u}{t}

Where u = initial velocity, v = final velocity, t = time taken

Vector SUVAT

The SUVAT equations work with vectors:

v = u + a t

s = u t + \frac{1}{2} a t^{2}

s = \frac{1}{2} (u + v) t

Finding Distance

For constant acceleration from rest:

s = \frac{1}{2} ∣ a ∣ t^{2}

06Worked Example

A particle P accelerates at a constant rate. At t = 0, P has velocity $u = (2 i + 3 j)$ m s⁻¹. At t = 5 s, P has velocity $v = (16 i - 5 j)$ m s⁻¹.

Find the acceleration of P.

1

Use

a = \frac{v - u}{t}

2

v - u = (16 i - 5 j) - (2 i + 3 j)

•

= (14 i - 8 j)

m s⁻¹

3

a = \frac{14 i - 8 j}{5}

$a = (2.8 i - 1.6 j)$ m s⁻²

Resultant Forces

When multiple forces act on a particle, we find the resultant by vector addition.

Resultant Force

R = F_{1} + F_{2} + F_{3} + ...

Example

Three forces act on a particle:

F_{1} = (2 i - j + 2 k)

N

F_{2} = (- i + 3 j - 3 k)

N

F_{3} = (4 i - 3 j - 2 k)

N

Resultant:

R = (5 i - j - 3 k)

N

Column Vectors

It's often easier to add vectors in column form:

2 - 1 2 + - 1 3 - 3 + 4 - 3 - 2 = 5 - 1 - 3

07Worked Example

A particle of mass 0.5 kg is acted on by three forces:

$F_{1} = (2 i - j + 2 k)$ N, $F_{2} = (- i + 3 j - 3 k)$ N, $F_{3} = (4 i - 3 j - 2 k)$ N

(a) Find the resultant force. (b) Find the acceleration. (c) Find |a|.

a)

R = (2 - 1 + 4) i + (- 1 + 3 - 3) j + (2 - 3 - 2) k

•

R = (5 i - j - 3 k)

N

b)

a = \frac{R}{m} = \frac{5 i - j - 3 k}{0.5}

•

a = (10 i - 2 j - 6 k)

m s⁻²

c)

∣ a ∣ = 100 + 4 + 36 = 140

m s⁻²

Practice

Question 1: Find the speed of a particle moving with these velocities:

(a)

v = (3 i + 4 j)

m s⁻¹

(b)

v = (24 i - 7 j)

km h⁻¹

(a) Speed =

3^{2} + 4^{2} = 9 + 16 = 25 =

5 m s⁻¹

(b) Speed =

2 4^{2} + (- 7)^{2} = 576 + 49 = 625 =

25 km h⁻¹

Practice

Question 2: A boat has a position vector of

(2 i + j)

km and a buoy has position vector

(6 i - 4 j)

km, relative to a fixed origin O.

(a) Find the distance of the boat from the buoy.

(b) Find the bearing of the boat from the buoy.

(a) Displacement from buoy to boat:

(2 - 6) i + (1 - (- 4)) j = (- 4 i + 5 j)

km

Distance =

16 + 25 = 41 =

6.40 km (3 s.f.)

(b)

tan θ = \frac{4}{5}

,

θ = 38.7°

Boat is NW of buoy, so bearing = 360° − 38.7° = 321° (3 s.f.)

Practice

Question 3: A particle, initially at rest, is acted upon by a force that causes it to accelerate at

(4 i - 2 j + 3 k)

m s⁻² for 2 seconds.

Work out the distance travelled by the particle.

Since particle starts at rest and moves in straight line with constant acceleration:

s = \frac{1}{2} ∣ a ∣ t^{2}

∣ a ∣ = 16 + 4 + 9 = 29

m s⁻²

s = \frac{1}{2} \times 29 \times 4 = 229

Distance = 10.8 m (3 s.f.)

Summary

Key Formulas

ConceptFormula

Speed from velocity

Speed = ∣ v ∣ = v_{x}^{2} + v_{y}^{2}

Bearing to vector

r sin θ ° i + r cos θ ° j

Newton's Second Law

F = m a

Acceleration

a = \frac{v - u}{t}

Remember

i = East, j = North (standard convention)

Speed is magnitude of velocity (scalar)

Distance is magnitude of displacement (scalar)

Bearings use sin for i-component, cos for j-component

Always give bearings as three figures