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.