Vector operations | Year 12

01 / LEARN

Understand the idea.

Operate component by component

Vector addition combines independent coordinate displacements. Add matching components, subtract matching components, and distribute any scalar to every component. A free vector can be moved parallel to itself without changing its magnitude or direction.

The point $p = a + t (b - a)$ starts at $A$ when $t = 0$ and reaches $B$ when $t = 1$ . For $0 < t < 1$ , it lies between them; $t = 1 / 2$ gives the midpoint.

Connect algebra and geometry

Two nonzero vectors are parallel when one is a scalar multiple of the other. A positive multiplier preserves direction; a negative multiplier reverses it. To test whether points are collinear, compare displacements from a common starting point.

For internal division $A P : P B = m : n$ , the fraction of the whole journey is $m / (m + n)$ . Thus $p = a + \frac{m}{m + n} (b - a)$ . This also explains the opposite endpoint weights in the usual section formula.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

For $a = (1, 2, - 1)$ and $b = (3, - 1, 4)$ , find $2 a - b$ .

Worked solution

1
Multiply every component of a by two.
$2 a = (2, 4, - 2)$
Scalar multiplication applies to all coordinates.
2
Subtract the corresponding components of b.
$2 a - b = (2 - 3, 4 - (- 1), - 2 - 4)$
Subtraction acts independently along each axis.
3
Simplify the vector.
$2 a - b = (- 1, 5, - 6)$
Subtracting the negative second component increases that coordinate.

Answer

$(- 1, 5, - 6)$

EXAMPLE 02

Worked example 2

A=(1,0,2), B=(7,3,-1). Find P with AP:PB=1:2 internally.

Worked solution

1
Find the full displacement.
$\vec{A B} = (7 - 1, 3 - 0, - 1 - 2) = (6, 3, - 3)$
The endpoints determine the direction and length of the segment.
2
Use the fraction of the full segment.
$\vec{A P} = \frac{1}{1 + 2} \vec{A B} = (2, 1, - 1)$
The ratio $1 : 2$ divides the whole journey into three equal parts, with one part before $P$ .
3
Add the displacement to the starting point.
$p = (1, 0, 2) + (2, 1, - 1) = (3, 1, 1)$
A position vector equals the starting position plus the displacement.
4
Check the remaining section.
$\vec{P B} = (7, 3, - 1) - (3, 1, 1) = (4, 2, - 2) = 2 \vec{A P}$
The remaining displacement is twice as long in the same direction, confirming internal ratio $1 : 2$ .

Answer

$(3, 1, 1)$

03 / YOUR TURN

Try it yourself.

Check your understanding

Are $(2, - 4, 6)$ and $(- 1, 2, - 3)$ parallel?

Show answer & reasoning

Yes.

The equality $(2, - 4, 6) = - 2 (- 1, 2, - 3)$ holds in every component. The nonzero scalar multiple proves parallelism; its negative sign means opposite directions.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

For $a = (2, - 1, 3)$ , $b = (- 1, 4, 2)$ , find $a + b$ and $3 a - 2 b$ .
2 marks
Find the midpoint of $(2, 4, - 2)$ and $(6, - 2, 8)$ .
2 marks
Find $P$ one quarter of the way from $(0, 1, 2)$ to $(8, 5, - 2)$ .
2 marks
Are $A (1, 1, 0)$ , $B (3, 5, 2)$ , $C (4, 7, 3)$ collinear?
2 marks
Solve $2 p + (1, - 1, 3) = (7, 5, - 1)$ .
2 marks
For parallelogram $A B C D$ in order, $A (1, 0, 1)$ , $B (3, 2, 0)$ , $C (6, 1, 4)$ , find $D$ .
2 marks

Read the worked solutions

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

For $a = (2, - 1, 3)$ , $b = (- 1, 4, 2)$ , find $a + b$ and $3 a - 2 b$ .
2 marks
Add components.
For the second result subtract $(- 2, 8, 4)$ from $(6, - 3, 9)$ .
$(1, 3, 5)$ and $(8, - 11, 5)$
Find the midpoint of $(2, 4, - 2)$ and $(6, - 2, 8)$ .
2 marks
Add coordinates to get $(8, 2, 6)$ .
Divide by 2.
$(4, 1, 3)$
Find $P$ one quarter of the way from $(0, 1, 2)$ to $(8, 5, - 2)$ .
2 marks
One quarter of displacement $(8, 4, - 4)$ is $(2, 1, - 1)$ .
Add this to the first point.
$(2, 2, 1)$
Are $A (1, 1, 0)$ , $B (3, 5, 2)$ , $C (4, 7, 3)$ collinear?
2 marks
$\vec{A B} = (2, 4, 2) = 2 (1, 2, 1)$ .
$\vec{A C} = 3 (1, 2, 1)$ , so the displacements are parallel.
Yes.
Solve $2 p + (1, - 1, 3) = (7, 5, - 1)$ .
2 marks
Subtract the fixed vector to obtain $2 p = (6, 6, - 4)$ .
Divide by 2.
$p = (3, 3, - 2)$
For parallelogram $A B C D$ in order, $A (1, 0, 1)$ , $B (3, 2, 0)$ , $C (6, 1, 4)$ , find $D$ .
2 marks
$\vec{A D} = \vec{B C}$ , so $d = a + c - b$ .
Substitute and combine components.
$(4, - 1, 5)$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics vector algebra, vector/Cartesian equations and linear systems. Original explanations and tasks.

Official curriculum source ↗

In this lesson

Understand the idea.

Operate component by component

Connect algebra and geometry

See the method in action.

Worked example 1

Multiply every component of a by two.

Subtract the corresponding components of b.

Simplify the vector.

Worked example 2

Find the full displacement.

Use the fraction of the full segment.

Add the displacement to the starting point.

Check the remaining section.

Try it yourself.

Check your understanding

Keep practising