Parallelism and collinearity | Year 11

01 / LEARN

The idea, step by step

Parallel directions are scalar multiples

Two non-zero vectors are parallel when one is a real scalar multiple of the other. A positive multiple has the same direction; a negative multiple points in the opposite direction. Neither possibility changes the fact that the directions are parallel.

Solve for a single multiplier that works for every component. Comparing only one component is insufficient. The zero vector has no direction, so avoid using it as the sole evidence for a geometric parallelism claim.

v = k u for one real k

Collinearity needs a common reference point

Three distinct points A, B and C are collinear exactly when the displacements AB and AC are parallel. These displacements share A as their initial point, so a scalar-multiple relation places both other points on the same line.

A multiplier between zero and one puts C between A and B; a multiplier greater than one puts it beyond B. A negative multiplier places it on the extension through A in the opposite direction.

A common trap

Comparing position vectors OA and OB to decide whether A, B and C are collinear.

Instead: Use two displacements from the same one of the three points, such as AB and AC.

02 / WORKED EXAMPLES

See the thinking

Example 1

An unknown component

Find k if $(2, 3)$ is parallel to $(8, k)$ .

The first component is multiplied by $4$ .
The second component must use the same multiplier: $k = 4 \times 3$ .

k = 12

Example 2

Proving collinearity

Show that A=(1,2), B=(3,5), C=(7,11) are collinear.

$\vec{A B} = (2, 3)$ and $\vec{A C} = (6, 9)$ .
$\vec{A C} = 3 \vec{A B}$ .
The two displacements from A are parallel, so A, B and C lie on one line.

The points are collinear, with C beyond B.

Check your understanding

Are $(3, - 6)$ and $(- 2, 4)$ parallel?

Show answer

Yes; the second is $- 2 / 3$ times the first.

The same multiplier changes both components and reverses direction.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Are $(2, 5)$ and $(6, 15)$ parallel? State a multiplier.
2 marks
Are $(4, - 2)$ and $(8, - 5)$ parallel? Explain.
2 marks
Find $a$ if $(a, 6)$ is parallel to $(2, 3)$ .
2 marks
Show that A=(-1,2), B=(2,8), C=(4,12) are collinear.
3 marks
A=(1,1), B=(5,3), C=(-1,0). Express AC as a multiple of AB and locate C.
3 marks
Find t for which $(2 + t, 1 - 2 t)$ is parallel to $(3, - 1)$ .
3 marks

Read the worked solutions

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

Are $(2, 5)$ and $(6, 15)$ parallel? State a multiplier.
2 marks
Both components are multiplied by three.
Yes; multiplier $3$ .
Are $(4, - 2)$ and $(8, - 5)$ parallel? Explain.
2 marks
The first components suggest multiplier $2$ , but $2 (- 2) = - 4$ , not $- 5$ .
No.
Find $a$ if $(a, 6)$ is parallel to $(2, 3)$ .
2 marks
The y-component doubles, so the x-component must also double.
$a = 4$ .
Show that A=(-1,2), B=(2,8), C=(4,12) are collinear.
3 marks
AB= $(3, 6)$ and AC= $(5, 10)$ .
These are scalar multiples from a common starting point.
$\vec{A C} = \frac{5}{3} \vec{A B}$ .
A=(1,1), B=(5,3), C=(-1,0). Express AC as a multiple of AB and locate C.
3 marks
AB= $(4, 2)$ and AC= $(- 2, - 1)$ .
The negative multiplier reverses direction.
$\vec{A C} = - \frac{1}{2} \vec{A B}$ ; C is beyond A away from B.
Find t for which $(2 + t, 1 - 2 t)$ is parallel to $(3, - 1)$ .
3 marks
Require $1 - 2 t = - (2 + t) / 3$ .
$3 - 6 t = - 2 - t$ , hence $t = 1$ .
$t = 1$ .

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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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