Magnitude and direction | Year 11

01 / LEARN

Understand the idea.

A directed change

A vector describes a directed change. Moving three metres east and four metres north gives components $(3, 4)$ : the signs record the horizontal and vertical directions. Its magnitude measures the straight-line displacement, rather than the distance travelled along a route.

The component triangle is right-angled, so its hypotenuse has length $\sqrt{3^{2} + 4^{2}} = 5$ . In general, square both components, add, then take the non-negative square root. Opposite vectors have the same magnitude.

∣ (a, b) ∣ = \sqrt{a^{2} + b^{2}}

Choose the quadrant before using a calculator

First use the component signs to locate the quadrant. Then use trigonometry to find a reference angle and place it in that quadrant. Unless stated otherwise, measure direction anticlockwise from the positive $x$ -axis, between $0^{\circ}$ and $360^{\circ}$ .

On an axis, read the direction directly instead of dividing by a zero component. The zero vector has no unique direction. Keep magnitudes exact where possible; round an angle only at the final step.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Magnitude and a first-quadrant direction

Find the magnitude and direction of $(3, 4)$ ; give the direction to ${0.1}^{\circ}$ .

Worked solution

1
Find the length.
$∣ (3, 4) ∣ = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$
The components are perpendicular sides of a right triangle.
2
Locate the direction.
$3 > 0, 4 > 0$
Positive horizontal and vertical components place the vector in quadrant I.
3
Calculate and round the angle.
$θ = {tan}^{- 1} (\frac{4}{3}) = 53.130^{\dots \circ} \approx {53.1}^{\circ}$
In quadrant I, the reference angle is the direction from the positive horizontal axis.

Answer

Magnitude $5$ ; direction ${53.1}^{\circ}$ .

EXAMPLE 02

A north-west displacement

A walker has displacement $(- 6, 6)$ km. Give its exact magnitude and direction from the positive x-axis.

Worked solution

1
Calculate the exact magnitude.
$∣ (- 6, 6) ∣ = \sqrt{{(- 6)}^{2} + 6^{2}} = \sqrt{72} = 6 \sqrt{2} km$
Squaring removes the component signs when calculating length.
2
Find the reference angle.
$α = {tan}^{- 1} (\frac{∣ 6 ∣}{∣ - 6 ∣}) = 45^{\circ}$
The absolute horizontal and vertical changes are equal.
3
Place the angle in quadrant II.
$θ = 180^{\circ} - 45^{\circ} = 135^{\circ}$
The negative horizontal component points west and the positive vertical component points north.

Answer

$6 \sqrt{2}$ km at $135^{\circ}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

What are the magnitude and direction of $(0, - 7)$ ?

Show answer & reasoning

Magnitude $7$ ; direction $270^{\circ}$ .

$\sqrt{0^{2} + {(- 7)}^{2}} = 7$ . A zero horizontal component and negative vertical component point straight down the negative $y$ -axis, whose anticlockwise direction is $270^{\circ}$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Find the magnitude of $(5, 12)$ .
2 marks
Find the exact magnitude and direction of $(- 4, 4)$ .
2 marks
A displacement is $(8, - 6)$ m. Find its magnitude and its direction south of east to ${0.1}^{\circ}$ .
3 marks
A non-zero vector has magnitude $10$ and direction $60^{\circ}$ . Find its exact components.
3 marks
Find $k$ if $(k, 4)$ has magnitude $5$ .
3 marks
Explain why $(0, 0)$ cannot be assigned a unique direction angle.
2 marks

Read the worked solutions

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

Find the magnitude of $(5, 12)$ .
2 marks
$\sqrt{25 + 144} = 13$ .
$13$
Find the exact magnitude and direction of $(- 4, 4)$ .
2 marks
The magnitude is $\sqrt{32}$ .
A quadrant-II arrow with equal component sizes makes angle $135^{\circ}$ .
$4 \sqrt{2}$ and $135^{\circ}$ .
A displacement is $(8, - 6)$ m. Find its magnitude and its direction south of east to ${0.1}^{\circ}$ .
3 marks
$\sqrt{64 + 36} = 10$ .
${tan}^{- 1} (\frac{6}{8}) = 36.869^{\dots \circ}$ .
$10$ m; ${36.9}^{\circ}$ south of east.
A non-zero vector has magnitude $10$ and direction $60^{\circ}$ . Find its exact components.
3 marks
Horizontal component: $10 cos 60^{\circ} = 5$ .
Vertical component: $10 sin 60^{\circ} = 5 \sqrt{3}$ .
$(5, 5 \sqrt{3})$ .
Find $k$ if $(k, 4)$ has magnitude $5$ .
3 marks
$k^{2} + 16 = 25$ , so $k^{2} = 9$ .
Both horizontal directions have the required magnitude.
$k = 3$ or $k = - 3$ .
Explain why $(0, 0)$ cannot be assigned a unique direction angle.
2 marks
Every angle would produce the same zero displacement when the magnitude is zero.
It represents no directed movement, so no direction is distinguished.

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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.

A directed change

Choose the quadrant before using a calculator

See the method in action.

Magnitude and a first-quadrant direction

Find the length.

Locate the direction.

Calculate and round the angle.

A north-west displacement

Calculate the exact magnitude.

Find the reference angle.

Place the angle in quadrant II.

Try it yourself.

Check your understanding

Keep practising