Component and unit-vector notation | Year 11

01 / LEARN

The idea, step by step

A fixed pair of directions

The standard basis vectors are $i = (1, 0)$ and $j = (0, 1)$ . They are perpendicular unit vectors. The statement $u = 3 i - 2 j$ means a horizontal change of three and a vertical change of negative two.

An ordered pair describes the same vector as basis notation: $(a, b) = a i + b j$ . Do not confuse the vector $i$ with the imaginary unit used in complex-number work; context determines the meaning.

u = (a, b) = a i + b j

Unit vectors point rather than measure distance

A unit vector has magnitude one. To normalise a non-zero vector, divide every component by the vector's magnitude. The denominator must be the length, not the sum of the components.

Exactly two unit vectors are parallel to a given non-zero vector: one in the same direction and one in the opposite direction. A problem asking for a unit vector 'in the direction of' the given vector means the same direction.

\hat{u} = \frac{u}{∣ u ∣} (u \neq 0)

A common trap

Calling $u / 2$ a unit vector simply because $u$ has two components.

Instead: Divide by the magnitude; the resulting squared components must add to one.

02 / WORKED EXAMPLES

See the thinking

Example 1

Changing notation

Write $a = 4 j - 3 i$ as an ordered pair, then find its magnitude.

Collect the i- and j-components in that order: $a = - 3 i + 4 j$ .
Thus $a = (- 3, 4)$ and its magnitude is $\sqrt{9 + 16}$ .

(- 3, 4)

; magnitude

5

Example 2

Normalising a vector

Find a unit vector in the direction of $v = 2 i - 2 j$ .

$∣ v ∣ = \sqrt{4 + 4} = 2 \sqrt{2}$ .
Divide each component by $2 \sqrt{2}$ .
Check: ${(1 / \sqrt{2})}^{2} + {(- 1 / \sqrt{2})}^{2} = 1$ .

\frac{1}{\sqrt{2}} i - \frac{1}{\sqrt{2}} j

Check your understanding

Find the unit vector in the direction of $(0, - 9)$ .

Show answer

$(0, - 1)$ .

Its magnitude is $9$ , so divide the two components by $9$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Write $(5, - 7)$ in basis-vector notation.
2 marks
Write $6 j + 2 i$ as an ordered pair.
2 marks
Find a unit vector in the direction of $(8, 6)$ .
2 marks
Find both unit vectors parallel to $(1, - 1)$ .
3 marks
Find $k$ if $(3 / 5, k)$ is a unit vector.
3 marks
Explain why normalising the zero vector is impossible.
2 marks

Read the worked solutions

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

Write $(5, - 7)$ in basis-vector notation.
2 marks
The first component multiplies i and the second multiplies j.
$5 i - 7 j$ .
Write $6 j + 2 i$ as an ordered pair.
2 marks
Read coefficients in x-then-y order.
$(2, 6)$ .
Find a unit vector in the direction of $(8, 6)$ .
2 marks
The magnitude is $10$ .
Divide both components by $10$ .
$(4 / 5, 3 / 5)$ .
Find both unit vectors parallel to $(1, - 1)$ .
3 marks
The vector has magnitude $\sqrt{2}$ .
Normalise and then take its opposite.
$\pm (1 / \sqrt{2}, - 1 / \sqrt{2})$ .
Find $k$ if $(3 / 5, k)$ is a unit vector.
3 marks
$9 / 25 + k^{2} = 1$ gives $k^{2} = 16 / 25$ .
Both vertical signs give length one.
$k = 4 / 5$ or $k = - 4 / 5$ .
Explain why normalising the zero vector is impossible.
2 marks
The magnitude of the zero vector is zero, so the normalisation formula is undefined.
It would require division by zero and there is no unique direction.

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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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