Position and Displacement Vectors

Vectors provide a useful way to describe where an object is and how its position changes.

Two important ideas are:

• position vector — describes the location of a point relative to an origin
• displacement vector — describes the change in position from one point to another

These ideas are used throughout mathematics, physics, navigation, engineering, robotics, and computer graphics.

Position Vectors

A position vector describes the location of a point relative to an origin.

In a two-dimensional coordinate system, the origin is:

O = (0, 0)

Suppose point A is located at:

A = (4, 3)

The position vector of A describes the movement from the origin to A:

4 units right and 3 units up

We can represent the position vector simply as:

A = (4, 3)

For these notes, we will continue using ordinary capital letters for vectors rather than vector-arrow symbols so that the notation copies cleanly into Moodle.

Position Vectors and Coordinates

A point's coordinates give us its position vector directly.

For example:

The signs of the components indicate direction.

For:

B = (−2, 4)

the point is:

• 2 units left of the origin
• 4 units above the origin

For:

C = (5, −3)

the point is:

• 5 units right of the origin
• 3 units below the origin

Magnitude of a Position Vector

The magnitude of a position vector gives the straight-line distance between the point and the origin.

Suppose:

A = (3, 4)

The magnitude is:

|A| = √(3² + 4²)

|A| = √(9 + 16)

|A| = √25

|A| = 5

Therefore, point A is 5 units from the origin.

The vertical bars mean:

|A| = magnitude of A

Position Is Not the Same as Displacement

A position vector tells us where something is relative to the origin.

A displacement vector tells us how position changes.

Suppose an object starts at:

A = (2, 3)

and finishes at:

B = (7, 5)

The position vectors tell us where A and B are.

The displacement tells us how to get from A to B.

Calculating a Displacement Vector

To calculate displacement:

displacement = final position − initial position

Therefore:

D = B − A

For:

A = (2, 3)

B = (7, 5)

we calculate:

D = (7, 5) − (2, 3)

D = (7 − 2, 5 − 3)

D = (5, 2)

The displacement is:

• 5 units right
• 2 units up

The Order Matters

When calculating displacement, always use:

final − initial

Suppose:

A = (2, 3)

B = (7, 5)

From A to B:

B − A = (5, 2)

But from B to A:

A − B = (−5, −2)

These vectors have the same magnitude but opposite directions.

Therefore:

displacement from A to B ≠ displacement from B to A

Direction matters.

Worked Example: Displacement

An object moves from:

P = (−3, 2)

to:

Q = (4, 6)

Find the displacement.

Use:

D = Q − P

D = (4, 6) − (−3, 2)

D = (4 − (−3), 6 − 2)

D = (7, 4)

Therefore:

D = (7, 4)

The object moves:

• 7 units right
• 4 units up

Finding the Magnitude of Displacement

The magnitude of a displacement vector gives the straight-line distance between the initial and final positions.

For:

D = (7, 4)

the magnitude is:

|D| = √(7² + 4²)

|D| = √(49 + 16)

|D| = √65

|D| ≈ 8.06

Therefore, the two points are approximately 8.06 units apart.

Distance Between Two Points

The vector method leads directly to the familiar distance formula.

Suppose:

A = (x₁, y₁)

and:

B = (x₂, y₂)

The displacement from A to B is:

D = (x₂ − x₁, y₂ − y₁)

The magnitude is:

|D| = √[(x₂ − x₁)² + (y₂ − y₁)²]

Therefore, the distance between the two points is:

distance = √[(x₂ − x₁)² + (y₂ − y₁)²]

The distance formula is really an application of vector magnitude.

Worked Example: Distance Between Points

Find the distance between:

A = (2, −1)

and:

B = (8, 7)

First find the displacement:

D = B − A

D = (8 − 2, 7 − (−1))

D = (6, 8)

Now find its magnitude:

|D| = √(6² + 8²)

|D| = √(36 + 64)

|D| = √100

|D| = 10

Therefore, the points are 10 units apart.

Distance and Displacement Are Different

Suppose a person walks:

• 4 m east
• 3 m north
• 4 m west

The total distance travelled is:

4 + 3 + 4 = 11 m

But the person finishes directly 3 m north of the starting position.

Therefore, the displacement is:

D = (0, 3) m

and:

|D| = 3 m

So:

distance = 11 m

displacement magnitude = 3 m

Distance depends on the path travelled.

Displacement depends only on the starting and finishing positions.

Finding a Final Position

The displacement relationship can be rearranged.

Since:

D = final position − initial position

we can write:

final position = initial position + displacement

or:

B = A + D

Example

An object begins at:

A = (3, 2)

and undergoes displacement:

D = (5, −4)

Find its final position.

B = A + D

B = (3, 2) + (5, −4)

B = (8, −2)

Therefore:

B = (8, −2)

Finding an Initial Position

We can also rearrange:

D = B − A

to find the initial position:

A = B − D

Example

An object finishes at:

B = (10, 6)

after displacement:

D = (4, 9)

Find its starting position.

A = B − D

A = (10, 6) − (4, 9)

A = (6, −3)

Therefore:

A = (6, −3)

Direction of a Displacement

A displacement vector contains information about both magnitude and direction.

Suppose:

D = (6, 8)

Magnitude:

|D| = √(6² + 8²)

|D| = 10

To find the angle:

tan θ = 8/6

θ = tan⁻¹(8/6)

θ ≈ 53.1°

Since both components are positive, the displacement points up and to the right.

Therefore:

D = 10 units at 53.1° above the positive x-axis

Negative Components

Suppose:

D = (−5, 3)

This means:

• 5 units left
• 3 units up

The magnitude is:

|D| = √[(−5)² + 3²]

|D| = √34

|D| ≈ 5.83

The negative x-component affects the direction, but its square is positive when calculating magnitude.

Position and Displacement in Navigation

Vectors are particularly useful in navigation.

Suppose a boat begins at:

A = (2, 1) km

and travels with displacement:

D₁ = (5, 3) km

Its new position is:

A + D₁

= (2, 1) + (5, 3)

= (7, 4)

It then undergoes another displacement:

D₂ = (−2, 6) km

Final position:

(7, 4) + (−2, 6)

= (5, 10)

The boat's final position relative to the origin is:

(5, 10) km

Multiple Displacements

When an object undergoes several displacements, the overall displacement is found by adding them.

Suppose:

D₁ = (4, 3)

D₂ = (−2, 5)

D₃ = (6, −4)

The resultant displacement is:

R = D₁ + D₂ + D₃

R = (4 − 2 + 6, 3 + 5 − 4)

R = (8, 4)

Magnitude:

|R| = √(8² + 4²)

|R| = √80

|R| ≈ 8.94

The object finishes approximately 8.94 units from its starting point.

Geometric Problems with Vectors

Vectors are useful for solving geometric problems because points can be represented using position vectors.

Suppose:

A = (2, 1)

B = (8, 5)

The vector from A to B is:

AB = B − A

AB = (8 − 2, 5 − 1)

AB = (6, 4)

Here, AB means the displacement from point A to point B.

Similarly:

BA = A − B

BA = (2 − 8, 1 − 5)

BA = (−6, −4)

Therefore:

BA = −AB

The vectors have equal magnitudes but opposite directions.

Midpoints Using Vectors

Vectors can also help determine the midpoint between two points.

Suppose:

A = (2, 4)

B = (8, 10)

The midpoint M is halfway between A and B.

We can calculate:

M = (A + B)/2

Therefore:

M = ((2 + 8)/2, (4 + 10)/2)

M = (10/2, 14/2)

M = (5, 7)

So the midpoint is:

M = (5, 7)

Why the Midpoint Formula Works

The displacement from A to B is:

AB = B − A

To move halfway from A toward B:

M = A + ½(B − A)

For:

A = (2, 4)

B = (8, 10)

B − A = (6, 6)

Half of this displacement is:

½(6, 6) = (3, 3)

Therefore:

M = (2, 4) + (3, 3)

M = (5, 7)

This shows how the midpoint formula follows naturally from vectors.

Dividing a Line Segment

Vectors can locate points that divide a line segment in other ratios.

Suppose:

A = (2, 3)

B = (10, 7)

Find the point P that is one-quarter of the way from A to B.

First:

AB = B − A

AB = (10 − 2, 7 − 3)

AB = (8, 4)

One-quarter of this vector is:

¼AB = (2, 1)

Add this to A:

P = A + ¼AB

P = (2, 3) + (2, 1)

P = (4, 4)

Showing Lines Are Parallel

Vectors can be used to determine whether line segments are parallel.

Suppose:

A = (1, 2)

B = (4, 6)

C = (3, 1)

D = (9, 9)

Find:

AB = B − A

AB = (3, 4)

Now:

CD = D − C

CD = (6, 8)

Notice:

CD = 2AB

Since one vector is a scalar multiple of the other, the vectors point in the same direction.

Therefore:

AB is parallel to CD.

Showing Points Are Collinear

Points are collinear if they lie on the same straight line.

Suppose:

A = (1, 2)

B = (3, 5)

C = (7, 11)

Calculate:

AB = B − A

AB = (2, 3)

Now calculate:

AC = C − A

AC = (6, 9)

Notice:

AC = 3AB

Because AC is a scalar multiple of AB, the two vectors have the same direction.

Therefore:

A, B, and C are collinear.

Geometric Problem: Is It a Parallelogram?

Suppose a quadrilateral has vertices:

A = (1, 1)

B = (5, 2)

C = (7, 6)

D = (3, 5)

Find opposite side vectors.

AB = B − A

AB = (4, 1)

DC = C − D

DC = (4, 1)

Now:

BC = C − B

BC = (2, 4)

AD = D − A

AD = (2, 4)

Therefore:

AB = DC

and:

BC = AD

Both pairs of opposite sides have equal vectors.

Therefore, ABCD is a parallelogram.

Geometric Problem: Is It a Rectangle?

Vectors can also help test whether sides are perpendicular.

Suppose:

A = (1, 1)

B = (5, 1)

C = (5, 4)

D = (1, 4)

AB = (4, 0)

BC = (0, 3)

AB is horizontal.

BC is vertical.

Therefore, AB and BC are perpendicular.

Opposite sides also have equal lengths and directions.

Therefore, the quadrilateral is a rectangle.

More advanced vector methods can use the dot product to test perpendicularity.

Three-Dimensional Position Vectors

The same ideas work in three dimensions.

Suppose:

A = (2, 3, 4)

B = (7, 5, 10)

The displacement from A to B is:

AB = B − A

AB = (7 − 2, 5 − 3, 10 − 4)

AB = (5, 2, 6)

The distance between the points is:

|AB| = √(5² + 2² + 6²)

|AB| = √(25 + 4 + 36)

|AB| = √65

|AB| ≈ 8.06

Position Vectors in Physics

Suppose an object's position at one moment is:

A = (2, 5) m

and later its position is:

B = (9, 1) m

Its displacement is:

D = B − A

D = (9 − 2, 1 − 5)

D = (7, −4) m

Magnitude:

|D| = √(7² + (−4)²)

|D| = √65

|D| ≈ 8.06 m

The object has moved 7 m right and 4 m down relative to its starting position.

Position Vectors in Computer Graphics

Computer graphics uses position vectors to specify the locations of objects and points on a screen or in three-dimensional space.

For example:

P = (100, 200)

could represent the position of an object on a two-dimensional screen.

If the object moves by:

D = (30, −20)

its new position is:

P_new = P + D

P_new = (100, 200) + (30, −20)

P_new = (130, 180)

Similar calculations are used in:

• animation
• video games
• simulations
• robotics
• 3D modelling

Position Vectors in Navigation

Navigation systems also use vector ideas.

A location can be represented relative to a chosen reference point.

Movement is represented by displacement vectors.

The new position is then calculated using:

new position = old position + displacement

This basic relationship appears in GPS calculations, aircraft navigation, ship navigation, robotics, and autonomous vehicles.

A General Problem-Solving Strategy

When solving position and displacement problems:

Identify the initial position.

Write its coordinates clearly.

Identify the final position.

Make sure the order is correct.

Calculate displacement.

Use:

D = final − initial

Calculate distance if required.

Use:

|D| = √(Dₓ² + Dᵧ²)

Find direction if required.

Use trigonometry and check the signs of the components.

For several movements, add the displacement vectors.

For geometry problems, compare vectors.

Equal or scalar-multiple vectors can reveal important geometric relationships.

Common Mistakes

Subtracting in the wrong order

For displacement from A to B:

AB = B − A

not A − B.

Confusing position with displacement

Position is measured relative to the origin.

Displacement describes a change from one position to another.

Confusing distance and displacement

Distance is a scalar.

Displacement is a vector.

Adding coordinates when finding displacement

To move from A to B, subtract:

B − A

Ignoring negative coordinates

For:

7 − (−3)

the result is:

10

not 4.

Adding displacement magnitudes instead of vectors

When several displacements occur in different directions, add their components first.

Assuming equal magnitudes mean equal vectors

Two vectors can have the same magnitude but different directions.

Key Terms

Position — the location of a point relative to a chosen reference point.

Position vector — a vector from the origin to a point that describes its position.

Origin — the reference point (0, 0) in a two-dimensional coordinate system.

Displacement vector — a vector describing the change from an initial position to a final position.

Magnitude — the size or length of a vector.

Distance — a scalar measurement of length.

Resultant displacement — the single displacement representing several combined displacements.

Component — the part of a vector acting along a coordinate axis.

Midpoint — the point halfway between two endpoints.

Collinear — describing points that lie on the same straight line.

Scalar multiple — a vector produced by multiplying another vector by a number.

Parallel vectors — vectors whose components are proportional.

Key Takeaways

• A position vector describes where a point is relative to the origin.
• A displacement vector describes the change between two positions.
• Displacement is calculated using final position − initial position.
• From A to B, AB = B − A.
• Reversing the order reverses the direction: BA = −AB.
• The magnitude of displacement gives the straight-line distance between the initial and final positions.
• For D = (Dₓ, Dᵧ), |D| = √(Dₓ² + Dᵧ²).
• Several displacement vectors can be added to determine the overall displacement.
• The final position can be found using final = initial + displacement.
• Position vectors can be used to find midpoints and divide line segments.
• Scalar multiples of vectors can be used to identify parallel lines and collinear points.
• Vector methods provide a powerful connection between algebra, geometry, and real-world motion.

Check Your Understanding

1. Write the position vector for the point A(4, 7).

2. Find the magnitude of the position vector:

   A = (6, 8)

3. An object moves from:

   A = (2, 3)

   to:

   B = (9, 7)

   Find its displacement vector.

4. Find the distance between the points in Question 3.

5. An object moves from:

   P = (−4, 2)

   to:

   Q = (3, −3)

   Find the displacement vector and its magnitude.

6. An object begins at:

   A = (5, −2)

   and undergoes displacement:

   D = (−3, 7)

   Find its final position.

7. An object finishes at:

   B = (10, 8)

   after displacement:

   D = (6, −2)

   Find its initial position.

8. Find the midpoint between:

   A = (−2, 4)

   and:

   B = (8, 10)

9. Given:

   A = (1, 3)

   B = (4, 7)

   C = (7, 11)

   Use vectors to determine whether A, B, and C are collinear.

10. An object undergoes three displacements:

D₁ = (5, 2)

D₂ = (−3, 6)

D₃ = (4, −5)

Find the resultant displacement and its magnitude.

11. A point P is one-quarter of the way from A = (2, 4) to B = (10, 12). Find the coordinates of P.
12. Explain, using an example, why distance travelled and the magnitude of displacement are not necessarily equal.

Applications of Vectors

Vectors are especially useful when a real-world quantity has both a magnitude and a direction.

They are used to describe:

• displacement
• velocity
• acceleration
• force
• momentum
• wind and water currents
• aircraft and ship motion

Vector mathematics allows us to combine these quantities and determine their overall effect.

Vectors in Navigation

Navigation involves both distance and direction, so vectors provide a natural way to represent movement.

Suppose a boat travels:

8 km east

and then:

6 km north

Using components:

D₁ = (8, 0) km

D₂ = (0, 6) km

The resultant displacement is:

R = D₁ + D₂

R = (8, 0) + (0, 6)

R = (8, 6) km

Its magnitude is:

|R| = √(8² + 6²)

|R| = √100

|R| = 10 km

The boat finishes 10 km from its starting position.

Finding the Direction

For:

R = (8, 6)

use:

tan θ = 6/8

θ = tan⁻¹(6/8)

θ ≈ 36.9°

Therefore, the boat's displacement is:

10 km at 36.9° north of east

Notice that the total distance travelled was:

8 + 6 = 14 km

but the displacement magnitude was only:

10 km

Multiple Navigation Vectors

Suppose a hiker travels:

• 5 km east
• 4 km north
• 2 km west
• 1 km south

Choose:

east = +x

north = +y

Horizontal displacement:

Rₓ = 5 − 2

Rₓ = 3 km

Vertical displacement:

Rᵧ = 4 − 1

Rᵧ = 3 km

Therefore:

R = (3, 3) km

Magnitude:

|R| = √(3² + 3²)

|R| = √18

|R| ≈ 4.24 km

Direction:

tan θ = 3/3

θ = 45°

Therefore, the hiker finishes approximately:

4.24 km at 45° north of east

from the starting position.

Bearings

Navigation often describes direction using bearings.

A bearing is normally measured:

• clockwise
• from north
• using three digits

Examples:

000° = north

090° = east

180° = south

270° = west

Therefore, a direction such as:

045°

means northeast, halfway between north and east.

Vectors can be resolved into components when a bearing is known.

Resolving a Navigation Vector

Suppose a boat travels 20 km on a bearing of 060°.

Because bearings are measured from north:

east component:

Dₓ = 20 sin 60°

Dₓ ≈ 17.3 km

north component:

Dᵧ = 20 cos 60°

Dᵧ = 10.0 km

Therefore:

D ≈ (17.3, 10.0) km

This is an important difference from angles measured from the horizontal axis.

For an angle measured from the positive x-axis, we normally use:

x = R cos θ

y = R sin θ

For a bearing measured from north, the sine and cosine positions are reversed.

Aircraft and Wind

An aircraft's motion relative to the ground depends on both:

• its velocity through the air
• the velocity of the wind

These velocities must be added as vectors.

Suppose an aircraft flies east at:

200 km/h

while a wind blows north at:

50 km/h

Aircraft velocity:

V₁ = (200, 0) km/h

Wind velocity:

V₂ = (0, 50) km/h

Ground velocity:

V = V₁ + V₂

V = (200, 50) km/h

Finding the Aircraft's Ground Speed

The magnitude of the ground velocity is:

|V| = √(200² + 50²)

|V| = √42500

|V| ≈ 206 km/h

The direction is:

tan θ = 50/200

θ = tan⁻¹(0.25)

θ ≈ 14.0°

Therefore, the aircraft's ground velocity is approximately:

206 km/h at 14.0° north of east

Even though the pilot points the aircraft directly east, the wind causes the aircraft to drift north.

Correcting for Wind

Pilots may need to point the aircraft slightly into the wind so that its actual path over the ground follows the desired direction.

For example, if the wind pushes an aircraft north but the desired ground path is east, the aircraft may need to point slightly south of east.

This is called a wind correction.

The same vector principle applies to:

• ships affected by ocean currents
• swimmers crossing rivers
• drones flying in wind
• moving objects on conveyor belts

Boats and Water Currents

Suppose a boat travels north through the water at:

6 m/s

while a river flows east at:

8 m/s

Boat velocity:

V₁ = (0, 6) m/s

River velocity:

V₂ = (8, 0) m/s

Resultant velocity:

V = (8, 6) m/s

Magnitude:

|V| = √(8² + 6²)

|V| = 10 m/s

The boat's actual motion relative to the shore is therefore 10 m/s diagonally downstream.

Crossing a River

Suppose a river is 120 m wide.

A swimmer can swim directly across the river at:

3 m/s

while the current moves downstream at:

2 m/s

The time required to cross depends on the across-river component, not the total resultant speed.

Time:

t = distance / speed

t = 120 / 3

t = 40 s

During those 40 seconds, the river moves the swimmer downstream:

d = vt

d = 2 × 40

d = 80 m

Therefore, the swimmer reaches the opposite bank 80 m downstream from the point directly opposite the starting position.

Vectors and Forces

Force is a vector because it has both:

• magnitude
• direction

If several forces act on an object, their combined effect is called the resultant force or net force.

For example, suppose:

• 15 N acts east
• 9 N acts west

Choose east as positive.

Net horizontal force:

Fₓ = 15 − 9

Fₓ = 6 N east

Because the forces act along the same line, their effects can be combined directly.

Forces in Two Dimensions

Suppose an object experiences:

• 12 N east
• 5 N west
• 10 N north
• 4 N south

Horizontal component:

Fₓ = 12 − 5

Fₓ = 7 N

Vertical component:

Fᵧ = 10 − 4

Fᵧ = 6 N

Therefore:

F = (7, 6) N

Magnitude:

|F| = √(7² + 6²)

|F| = √85

|F| ≈ 9.22 N

Direction:

tan θ = 6/7

θ ≈ 40.6°

Therefore, the resultant force is:

9.22 N at 40.6° north of east

Resolving Forces into Components

Forces do not always act horizontally or vertically.

Suppose a force of 50 N acts at 30° above the horizontal.

Horizontal component:

Fₓ = 50 cos 30°

Fₓ ≈ 43.3 N

Vertical component:

Fᵧ = 50 sin 30°

Fᵧ = 25.0 N

Therefore:

F ≈ (43.3, 25.0) N

Resolving forces into components allows angled forces to be combined with other forces.

Worked Force Problem

Three forces act on an object:

• 20 N east
• 12 N north
• 10 N at 30° south of west

For the third force:

horizontal component:

F₃ₓ = −10 cos 30°

F₃ₓ ≈ −8.66 N

vertical component:

F₃ᵧ = −10 sin 30°

F₃ᵧ = −5.00 N

Now add the components.

Horizontal:

Rₓ = 20 − 8.66

Rₓ = 11.34 N

Vertical:

Rᵧ = 12 − 5

Rᵧ = 7.00 N

Resultant:

R = (11.34, 7.00) N

Magnitude:

|R| = √(11.34² + 7²)

|R| ≈ 13.3 N

Direction:

tan θ = 7/11.34

θ ≈ 31.7°

Therefore, the resultant is approximately:

13.3 N at 31.7° north of east

Balanced Forces

If the vector sum of all forces is zero:

F_net = (0, 0)

then the forces are balanced.

For example:

• 10 N east
• 10 N west
• 6 N north
• 6 N south

The components are:

Fₓ = 10 − 10 = 0

Fᵧ = 6 − 6 = 0

Therefore:

F_net = (0, 0)

The object has zero acceleration.

This does not necessarily mean the object is stationary. It could be moving at constant velocity.

Vectors and Newton's Second Law

Newton's Second Law is fundamentally a vector equation:

F_net = ma

Acceleration occurs in the same direction as the net force.

Suppose:

F_net = (12, 16) N

and:

m = 4 kg

Then:

a = F_net / m

a = (12/4, 16/4)

a = (3, 4) m/s²

Magnitude:

|a| = √(3² + 4²)

|a| = 5 m/s²

The acceleration has the same direction as the resultant force.

Velocity as a Vector

Velocity is different from speed.

Speed tells us how fast something moves.

Velocity tells us how fast and in what direction.

For example:

20 m/s

is a speed.

20 m/s east

is a velocity.

Velocity can therefore be represented using vector components.

Combining Velocities

Suppose:

V₁ = (12, 5) m/s

and:

V₂ = (−3, 4) m/s

Then:

V = V₁ + V₂

V = (12 − 3, 5 + 4)

V = (9, 9) m/s

Magnitude:

|V| = √(9² + 9²)

|V| = √162

|V| ≈ 12.7 m/s

Direction:

45° above the positive x-axis

Relative Velocity

Vectors are also used to describe motion relative to another moving object.

Suppose:

Car A travels east at:

25 m/s

Car B travels east at:

18 m/s

From Car A's perspective, Car B appears to move backward at:

18 − 25 = −7 m/s

The relative velocity of B compared with A is:

V_BA = V_B − V_A

This concept is important in:

• road traffic
• aircraft navigation
• spacecraft
• ships
• sports

Vectors and Acceleration

Acceleration is also a vector.

An object accelerates whenever its velocity changes.

This can happen because:

• its speed increases
• its speed decreases
• its direction changes
• both speed and direction change

For example, an object moving around a circular path may maintain constant speed while continually changing velocity because its direction changes.

Therefore, it is accelerating.

Vectors and Momentum

Momentum is defined as:

p = mv

Because velocity is a vector, momentum is also a vector.

Suppose a 5 kg object moves east at 8 m/s.

Its momentum is:

p = 5 × 8

p = 40 kg·m/s east

If it moved west instead, its momentum would point west.

Momentum conservation therefore requires us to account for direction.

Collision Example

Suppose:

• a 2 kg cart moves east at 5 m/s
• a 3 kg cart moves west at 2 m/s

Choose east as positive.

Cart 1:

p₁ = 2(5)

p₁ = +10 kg·m/s

Cart 2:

p₂ = 3(−2)

p₂ = −6 kg·m/s

Total momentum:

p_total = 10 − 6

p_total = +4 kg·m/s

The total momentum is:

4 kg·m/s east

The signs allow opposite directions to be included correctly.

Vector Applications in Engineering

Engineers regularly use vectors to analyse:

• forces on bridges
• forces in cables
• loads on buildings
• aircraft motion
• fluid flow
• vehicle dynamics
• robotic motion
• mechanical systems

For example, several cables may support a load from different directions.

Each cable produces a tension force vector.

The vector sum determines whether the load is:

• in equilibrium
• accelerating
• moving in a particular direction

Vector Applications in Sports

Vectors can describe the motion of:

• footballs
• basketballs
• golf balls
• runners
• swimmers
• cyclists

Suppose a football is kicked at 20 m/s at 35° above the horizontal.

Horizontal velocity:

Vₓ = 20 cos 35°

Vₓ ≈ 16.4 m/s

Vertical velocity:

Vᵧ = 20 sin 35°

Vᵧ ≈ 11.5 m/s

These components can then be analysed separately when studying projectile motion.

Vector Applications in Robotics

A robot may need to move from:

A = (2, 3)

to:

B = (10, 8)

Required displacement:

D = B − A

D = (10 − 2, 8 − 3)

D = (8, 5)

Magnitude:

|D| = √(8² + 5²)

|D| = √89

|D| ≈ 9.43 units

Vector calculations can help the robot determine both the required distance and direction.

Vector Applications in Space Travel

Spacecraft motion depends heavily on vector calculations.

Engineers must consider:

• spacecraft velocity
• gravitational forces
• thrust
• orbital velocity
• changes in direction
• relative motion between planets and spacecraft

A rocket's engines produce thrust vectors.

Changing the direction or magnitude of the thrust changes the spacecraft's velocity.

Because objects in space are constantly moving relative to one another, vector analysis is essential for planning trajectories.

Choosing a Coordinate System

Before solving an applied vector problem, choose positive directions.

A common choice is:

east/right = +x

west/left = −x

north/up = +y

south/down = −y

This makes the signs of the components meaningful.

However, another coordinate system can be used if it makes the problem easier.

The important thing is to remain consistent.

General Strategy for Vector Applications

When solving an authentic vector problem:

Identify the vector quantities.

Determine which quantities have both magnitude and direction.

Choose coordinate directions.

Clearly define positive x and positive y.

Convert vectors into components.

For an angle θ measured from the positive x-axis:

Vₓ = V cos θ

Vᵧ = V sin θ

Add or subtract the components.

Rₓ = sum of x-components

Rᵧ = sum of y-components

Find the resultant magnitude.

|R| = √(Rₓ² + Rᵧ²)

Find the direction.

Use:

tan θ = Rᵧ/Rₓ

Then check the signs of the components to identify the correct direction.

Finally, interpret the result in context.

Include:

• units
• direction
• physical meaning

Worked Authentic Problem: Rescue Boat

A rescue boat can travel at 12 m/s through still water. It heads directly north across a channel while the water current flows east at 5 m/s.

Boat velocity:

V_b = (0, 12) m/s

Current velocity:

V_c = (5, 0) m/s

Ground velocity:

V = V_b + V_c

V = (5, 12) m/s

Magnitude:

|V| = √(5² + 12²)

|V| = √169

|V| = 13 m/s

Direction relative to north:

tan θ = 5/12

θ ≈ 22.6°

Therefore, the boat actually moves:

13 m/s at 22.6° east of north

This is a good example of why direction cannot be ignored.

Worked Authentic Problem: Drone

A drone flies at 15 m/s east relative to the air.

Wind blows at 8 m/s north.

Resultant velocity:

V = (15, 8) m/s

Magnitude:

|V| = √(15² + 8²)

|V| = √289

|V| = 17 m/s

Direction:

tan θ = 8/15

θ ≈ 28.1°

Therefore, the drone moves relative to the ground at:

17 m/s at 28.1° north of east

If the drone needs to travel directly east, its flight controller must compensate for the wind.

Worked Authentic Problem: Towing a Vehicle

Two cables pull a vehicle.

Cable A pulls with 600 N east.

Cable B pulls with 400 N at 60° north of east.

Resolve Cable B:

Bₓ = 400 cos 60°

Bₓ = 200 N

Bᵧ = 400 sin 60°

Bᵧ ≈ 346.4 N

Add components:

Rₓ = 600 + 200

Rₓ = 800 N

Rᵧ = 346.4 N

Magnitude:

|R| = √(800² + 346.4²)

|R| ≈ 872 N

Direction:

tan θ = 346.4/800

θ ≈ 23.4°

Therefore, the combined pulling force is approximately:

872 N at 23.4° north of east

Checking Whether an Answer Makes Sense

Vector calculations should always be checked physically.

For example, if two perpendicular velocities are:

6 m/s

and:

8 m/s

the resultant should be:

• greater than 8 m/s
• less than 14 m/s

The calculated result:

10 m/s

is reasonable.

If your calculation produced 25 m/s, you should immediately suspect an error.

Estimating before calculating is a useful habit.

Common Mistakes

Treating vectors as ordinary numbers

Directions must be considered.

Adding magnitudes instead of components

For vectors acting in different directions, resolve them into components before adding.

Confusing speed and velocity

Speed is scalar. Velocity is vector.

Confusing distance and displacement

Distance describes the total path travelled. Displacement describes the change in position.

Ignoring wind or current

The motion relative to the ground may differ from motion relative to the air or water.

Using sine and cosine incorrectly

Check where the angle is measured from.

Forgetting that bearings are measured from north

Standard mathematical angles are usually measured from the positive x-axis, while bearings are measured clockwise from north.

Ignoring negative components

Negative values usually indicate movement or force in the opposite coordinate direction.

Giving only a magnitude

A complete vector answer often requires both magnitude and direction.

Key Terms

Vector — a quantity with both magnitude and direction.

Scalar — a quantity with magnitude but no direction.

Resultant vector — the single vector representing the combined effect of several vectors.

Component — the part of a vector acting along a particular coordinate direction.

Displacement — the vector change in position.

Velocity — the rate of change of displacement, including direction.

Speed — the scalar rate of distance travelled.

Force — a vector interaction capable of changing an object's motion.

Net force — the vector sum of all forces acting on an object.

Acceleration — the vector rate of change of velocity.

Momentum — the vector quantity p = mv.

Relative velocity — the velocity of one object measured relative to another.

Bearing — a navigation direction measured clockwise from north.

Ground velocity — the actual velocity of an aircraft or other moving object relative to the ground.

Wind correction — adjusting the direction of travel to compensate for wind.

Key Takeaways

• Vectors are essential for modelling quantities that have both magnitude and direction.
• Navigation problems often involve adding displacement or velocity vectors.
• Wind and water currents can change the actual direction and speed of moving objects.
• Forces combine according to vector addition.
• The vector sum of all forces gives the net force.
• If the net force is zero, acceleration is zero.
• Velocity and acceleration are vector quantities.
• Momentum is also a vector because it depends on velocity.
• Angled vectors can be resolved using x = R cos θ and y = R sin θ when θ is measured from the x-axis.
• Resultant magnitude is calculated using |R| = √(Rₓ² + Rᵧ²).
• Navigation bearings require careful attention because they are measured clockwise from north.
• Vector methods are widely used in engineering, aviation, shipping, robotics, sports, and space travel.
• Authentic vector problems should end with an interpretation that includes both magnitude and direction.

Check Your Understanding

1. A person walks 6 km east and then 8 km north. Determine the magnitude and direction of the resultant displacement.
2. A boat moves north at 12 m/s while a current flows east at 5 m/s. Determine the boat's velocity relative to the shore.
3. An aircraft flies east at 300 km/h while wind blows north at 80 km/h. Calculate its ground speed and direction.
4. A swimmer crosses a 90 m wide river at 3 m/s directly across the water. The current flows downstream at 1.5 m/s. Determine:
   • the crossing time
   • the downstream displacement
5. An object experiences forces of 20 N east, 8 N west, 15 N north, and 6 N south. Determine the resultant force.
6. A force of 60 N acts at 40° above the horizontal. Calculate its horizontal and vertical components.

7. A 5 kg object experiences a net force:

   F = (15, 20) N

   Determine its acceleration vector and the magnitude of its acceleration.

8. A 4 kg object moves east at 6 m/s while a 2 kg object moves west at 5 m/s. Determine the total momentum of the two-object system.
9. A drone flies east at 18 m/s while wind blows south at 7 m/s. Determine the drone's ground speed and direction.
10. Two cables pull an object. One produces a force of 500 N east. The other produces a force of 300 N at 45° north of east. Determine the magnitude and direction of the resultant force.
11. A rescue boat can travel at 10 m/s relative to the water. A current of 4 m/s flows east. Explain why the boat cannot simply point directly at a destination directly north if it needs to travel along a straight northward path.
12. Choose a real situation involving navigation, forces, velocity, sports, engineering, or transportation. Identify the vectors involved and explain how vector operations could be used to solve the problem.