Parallelism and collinearity | Year 11

01 / LEARN

Understand the idea.

Parallel directions are scalar multiples

Non-zero vectors are parallel when one is a single real scalar multiple of the other. The multiplier must work for every component. A positive multiplier gives the same direction; a negative multiplier gives opposite directions along parallel lines.

Find the multiplier from a non-zero component, then test the remaining component. Avoid dividing by zero when a component vanishes. The zero vector has no direction and cannot establish a geometric claim that two directions are parallel.

v = k u for one real k

Collinearity needs a common reference point

To test whether three distinct points are collinear, form two displacements from one of them, such as $\vec{A B}$ and $\vec{A C}$ . A scalar-multiple relation places B and C on the same line through A.

If $\vec{A C} = t \vec{A B}$ , the multiplier also locates C: $0 < t < 1$ puts it between A and B, $t > 1$ beyond B, and $t < 0$ beyond A in the reverse direction.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An unknown component

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

Worked solution

1
Name one common multiplier.
$(8, k) = t (2, 3)$
Parallel non-zero vectors must be related by one scalar for both components.
2
Use the horizontal component.
$8 = 2 t \Rightarrow t = 4$
The non-zero first component determines the multiplier.
3
Apply it to the vertical component.
$k = 3 t = 3 (4) = 12$
The same factor must scale the second component; $(8, 12) = 4 (2, 3)$ .

Answer

$k = 12$ .

EXAMPLE 02

Proving collinearity

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

Worked solution

1
Form two displacements from A.
$\begin{aligned} \vec{A B} & = (3 - 1, 5 - 2) = (2, 3), \\ \vec{A C} & = (7 - 1, 11 - 2) = (6, 9) . \end{aligned}$
Sharing the initial point allows a parallelism statement to establish collinearity.
2
Show one is a scalar multiple of the other.
$\vec{A C} = 3 (2, 3) = 3 \vec{A B}$
Both components have the same multiplier.
3
Interpret the vector equation.
The non-zero displacements share A and are parallel, so A, B and C lie on one line. The multiplier $3 > 1$ places C beyond B.

Answer

The points are collinear, with C beyond B.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

$- \frac{2}{3} (3, - 6) = (- 2, 4)$ . Both vectors are non-zero, so this single negative multiplier proves parallelism and shows that their directions are opposite.

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 $A C$ as a multiple of $A B$ 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
$A B = (3, 6)$ and $A C = (5, 10)$ .
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 $A C$ as a multiple of $A B$ and locate $C$ .
3 marks
$A B = (4, 2)$ and $A C = (- 2, - 1)$ .
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 $1 - 2 t = - \frac{2 + t}{3}$ .
$3 - 6 t = - 2 - t$ , 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 ↗

In this lesson

Understand the idea.

Parallel directions are scalar multiples

Collinearity needs a common reference point

See the method in action.

An unknown component

Name one common multiplier.

Use the horizontal component.

Apply it to the vertical component.

Proving collinearity

Form two displacements from A.

Show one is a scalar multiple of the other.

Interpret the vector equation.

Try it yourself.

Check your understanding

Keep practising