Position and displacement vectors | Year 11

01 / LEARN

The idea, step by step

Position uses a chosen origin

A position vector gives a point's directed location from a fixed origin. If the position vectors of A and B are $a$ and $b$ , their displacement from A to B is $b - a$ .

Changing the origin changes both position vectors, but it does not change their difference. This is why physical displacements and geometric relationships can be found without a special origin.

\vec{A B} = b - a

Points along a segment

Starting at A and travelling a fraction t of the displacement toward B gives position $a + t (b - a)$ . For $0 \leq t \leq 1$ the point is on the closed segment; t outside this interval gives a point on its extension.

The midpoint uses $t = 1 / 2$ . For internal division AP:PB=m:n, use $t = m / (m + n)$ . The fraction describes the distance travelled from A, so reversing the endpoints changes how the ratio is used.

r = a + t (b - a)

A common trap

Using $a + b$ as the displacement from A to B.

Instead: A displacement is final position minus initial position: $b - a$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Moving between locations

A has position $(2, - 3)$ m and B has position $(- 4, 5)$ m. Find the displacement A to B and its length.

Subtract start from end: $(- 4, 5) - (2, - 3) = (- 6, 8)$ .
The length is $\sqrt{36 + 64} = 10$ m.

(- 6, 8)

m; length

10

Example 2

A ratio on a segment

A=(1,2), B=(9,6). Find P if AP:PB=3:1.

P is $3 / 4$ of the way from A to B.
$\vec{A B} = (8, 4)$ , so $\vec{A P} = (6, 3)$ .
Add the displacement to A.

P = (7, 5)

Check your understanding

Find the midpoint of $(- 2, 5)$ and $(6, - 1)$ .

Show answer

$(2, 2)$ .

Average the two x-coordinates and the two y-coordinates.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A=(3,1), B=(-2,7). Find $\vec{A B}$ .
2 marks
A particle starts at $(4, - 3)$ and is displaced by $(- 6, 5)$ . Find its final position.
2 marks
Find the midpoint of $(7, - 4)$ and $(- 1, 6)$ .
2 marks
A=(0,2), B=(12,8). Find P for AP:PB=1:2.
3 marks
A=(2,1), B=(5,3). Find the point with $t = 2$ in $a + t (b - a)$ and explain its position.
3 marks
An origin is shifted by vector $c$ . Prove that the displacement between A and B stays unchanged.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A=(3,1), B=(-2,7). Find $\vec{A B}$ .
2 marks
Subtract A's coordinates from B's coordinates.
$(- 5, 6)$ .
A particle starts at $(4, - 3)$ and is displaced by $(- 6, 5)$ . Find its final position.
2 marks
Add position and displacement.
$(- 2, 2)$ .
Find the midpoint of $(7, - 4)$ and $(- 1, 6)$ .
2 marks
Average corresponding coordinates.
$(3, 1)$ .
A=(0,2), B=(12,8). Find P for AP:PB=1:2.
3 marks
Use one third of $(12, 6)$ from A.
$(0, 2) + (4, 2) = (4, 4)$ .
$(4, 4)$ .
A=(2,1), B=(5,3). Find the point with $t = 2$ in $a + t (b - a)$ and explain its position.
3 marks
The displacement is $(3, 2)$ .
$A + 2 (3, 2) = (8, 5)$ .
$(8, 5)$ , beyond B on the line from A through B.
An origin is shifted by vector $c$ . Prove that the displacement between A and B stays unchanged.
3 marks
Subtract the shifted position vectors.
The two copies of the origin shift cancel.
$(b - c) - (a - c) = b - a$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: vectors in the plane. Covers magnitude, components, geometry, scalar products, projections and proofs; follows the approved Specialist preparation structure.

Official curriculum source ↗

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