Three-dimensional coordinates, magnitude and unit vectors | Year 12

01 / LEARN

Understand the idea.

From points to displacements

A point gives a location; a displacement vector gives a change of location. To find $\vec{A B}$ , subtract the starting coordinates from the finishing coordinates: $b - a$ . Reversing the journey reverses every component.

The three axes are mutually perpendicular, so Pythagoras extends to $∣ v ∣ = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}$ . Components may be negative because they record direction. Magnitude is a length and is always non-negative.

Schematic three-dimensional coordinate axes x,y,z with origin O, point P, position vector OP, and dashed vertical and floor projections. A consistent orthographic projection of perpendicular spatial axes is used. No numerical point coordinates are asserted; this is a sketch, not to scale.

Separate length from direction

A unit vector has magnitude one. For nonzero $v$ , divide every component by $∣ v ∣$ to keep its direction while removing its length. The zero vector has no unit direction because this division would be by zero.

To construct a vector of length $k > 0$ in the same direction, use $k v / ∣ v ∣$ . Keep exact fractions and radicals, then check the result's magnitude and its positive scalar relationship to the original vector.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find the magnitude and a unit direction for $(2, - 3, 6)$ .

Worked solution

1
Calculate the squared magnitude.
$∣ v ∣^{2} = 2^{2} + {(- 3)}^{2} + 6^{2} = 49$
Perpendicular coordinate components combine by Pythagoras.
2
Take the non-negative square root.
$∣ v ∣ = 7$
Magnitude is a non-negative length.
3
Divide each component by the magnitude.
$\hat{v} = \frac{1}{7} (2, - 3, 6) = (\frac{2}{7}, - \frac{3}{7}, \frac{6}{7})$
A positive division changes length to one while preserving direction.

Answer

Magnitude $7$ ; unit vector $(2 / 7, - 3 / 7, 6 / 7)$ .

EXAMPLE 02

Worked example 2

A=(1,2,-1), B=(4,6,11). Find AB and a length-10 vector towards B.

Worked solution

1
Find the displacement from A to B.
$\vec{A B} = (4 - 1, 6 - 2, 11 - (- 1)) = (3, 4, 12)$
The displacement is finishing position minus starting position.
2
Find the distance between the points.
$A B = \sqrt{3^{2} + 4^{2} + 12^{2}} = \sqrt{169} = 13$
The length of the displacement is the straight-line distance.
3
Scale its unit direction to length ten.
$u = 10 \frac{(3, 4, 12)}{13} = (\frac{30}{13}, \frac{40}{13}, \frac{120}{13})$
Normalising gives length one, then multiplying by $10$ gives the requested length toward $B$ .
4
Check the constructed length.
$∣ u ∣ = \frac{10}{13} \sqrt{169} = 10$
The scale factor is positive, so the vector also retains the direction from $A$ toward $B$ .

Answer

$A B = 13$ ; vector $(30 / 13, 40 / 13, 120 / 13)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Can the zero vector be given a unit direction?

Show answer & reasoning

No.

The zero vector has magnitude zero and no preferred direction. The expression $0 / ∣ 0 ∣$ would divide by zero, so it cannot define a unit vector.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the magnitude of $(1, 2, 2)$ .
2 marks
Find a unit vector along $(0, - 3, 4)$ .
2 marks
Find the distance from $(2, - 1, 3)$ to $(5, 3, 15)$ .
2 marks
Find a length-6 vector opposite to $(1, 2, 2)$ .
2 marks
Find positive $k$ when $∣ (k, 2, 2) ∣ = 5$ .
2 marks
A drone moves 6 m east, 3 m north and 2 m up. Find its displacement length.
2 marks

Read the worked solutions

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

Find the magnitude of $(1, 2, 2)$ .
2 marks
Add squares: $1 + 4 + 4 = 9$ .
Take the positive square root.
$3$
Find a unit vector along $(0, - 3, 4)$ .
2 marks
The length is 5.
Divide every component by 5.
$(0, - \frac{3}{5}, \frac{4}{5})$
Find the distance from $(2, - 1, 3)$ to $(5, 3, 15)$ .
2 marks
The displacement is $(3, 4, 12)$ .
Its squared length is $9 + 16 + 144 = 169$ .
$13$
Find a length-6 vector opposite to $(1, 2, 2)$ .
2 marks
The original length is 3.
Multiply by $- 2$ to reverse direction and double length.
$(- 2, - 4, - 4)$
Find positive $k$ when $∣ (k, 2, 2) ∣ = 5$ .
2 marks
$k^{2} + 8 = 25$ .
Hence $k^{2} = 17$ ; use the positive condition.
$k = \sqrt{17}$
A drone moves 6 m east, 3 m north and 2 m up. Find its displacement length.
2 marks
The directions are perpendicular.
$\sqrt{36 + 9 + 4} = 7$ .
$7 m$

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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.

From points to displacements

Separate length from direction

See the method in action.

Worked example 1

Calculate the squared magnitude.

Take the non-negative square root.

Divide each component by the magnitude.

Worked example 2

Find the displacement from A to B.

Find the distance between the points.

Scale its unit direction to length ten.

Check the constructed length.

Try it yourself.

Check your understanding

Keep practising