Position and displacement vectors | Year 11

01 / LEARN

Understand the idea.

Position uses a chosen origin

A position vector starts at a chosen origin and ends at a point. A displacement starts at the first location and ends at the second. Following $O$ to $A$ and then $A$ to $B$ gives $a + \vec{A B} = b$ .

Rearranging gives end position minus start position. If the origin changes, the same vector is subtracted from both positions, so it cancels in their difference. A displacement therefore describes the movement independently of the chosen origin.

\vec{A B} = b - a

Points along a segment

Write a point on the line through distinct points $A$ and $B$ as $a + t (b - a)$ . The parameter $t$ specifies how much of the full displacement has been travelled: $t = 0$ gives $A$ and $t = 1$ gives $B$ .

For an internal ratio $A P : P B = m : n$ with $m, n > 0$ , the whole segment contains $m + n$ parts, so use $t = \frac{m}{m + n}$ . Values between zero and one lie on the segment; other values lie on its extensions.

r = a + t (b - a)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

1
Subtract the initial position from the final position.
$\vec{A B} = (- 4, 5) - (2, - 3) = (- 6, 8) m$
The displacement describes the change required to move from A to B.
2
Calculate its length.
$∣ \vec{A B} ∣ = \sqrt{{(- 6)}^{2} + 8^{2}} = \sqrt{100} = 10 m$
Magnitude uses the squared component changes.
3
Check that the displacement reaches B.
$(2, - 3) + (- 6, 8) = (- 4, 5)$
Adding the displacement to the starting position must recover the destination.

Answer

$(- 6, 8)$ m; length $10$ m.

EXAMPLE 02

A ratio on a segment

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

Worked solution

1
Convert the ratio to a fraction of the whole segment.
$t = \frac{3}{3 + 1} = \frac{3}{4}$
The distance from A to P occupies three of the four equal parts of AB.
2
Find the displacement from A to P.
$\begin{aligned} \vec{A B} & = (9, 6) - (1, 2) = (8, 4), \\ \vec{A P} & = \frac{3}{4} (8, 4) = (6, 3) . \end{aligned}$
Scaling the full displacement by three quarters reaches the required division point.
3
Add this displacement to A.
$P = (1, 2) + (6, 3) = (7, 5)$
The displacement is measured from A, rather than from the origin.
4
Check the remaining part.
$\vec{P B} = (9, 6) - (7, 5) = (2, 1), \vec{A P} = 3 \vec{P B}$
The two displacements have the same direction and the required length ratio.

Answer

$P = (7, 5)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$(2, 2)$ .

At the midpoint, $t = \frac{1}{2}$ . Average corresponding coordinates: $(\frac{- 2 + 6}{2}, \frac{5 + (- 1)}{2}) = (2, 2)$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 $A P : P B = 1 : 2$ .
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 $A P : P B = 1 : 2$ .
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 ↗

In this lesson

Understand the idea.

Position uses a chosen origin

Points along a segment

See the method in action.

Moving between locations

Subtract the initial position from the final position.

Calculate its length.

Check that the displacement reaches B.

A ratio on a segment

Convert the ratio to a fraction of the whole segment.

Find the displacement from A to P.

Add this displacement to A.

Check the remaining part.

Try it yourself.

Check your understanding

Keep practising