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

01 / LEARN

The idea, step by step

From points to displacements

A point $P (x, y, z)$ names a location along three perpendicular axes. Its position vector is $\vec{O P} = (x, y, z)$ . To travel from A to B, subtract the coordinates of A from those of B. Reversing the endpoints reverses the displacement.

Pythagoras gives $∣ v ∣ = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}$ . Magnitude is non-negative. Negative components record direction along an axis, not a negative length.

Separate length from direction

For nonzero $v$ , the unit vector $\hat{v} = v / ∣ v ∣$ has the same direction and length one. To construct length $k > 0$ , multiply this unit direction by k. The zero vector cannot be normalised.

Check a unit answer by adding its squared components: the result must be one. Preserve exact radicals until the final rounding step, especially when the unit vector is used in a later calculation.

A common trap

Divide by the sum of components to find a unit vector.

Instead: Divide by the magnitude, not the component sum.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

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

The squared magnitude is $4 + 9 + 36 = 49$ .
Take its positive square root and divide all components by 7.

Magnitude

7

; unit vector

(2 / 7, - 3 / 7, 6 / 7)

Example 2

Worked example 2

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

$\vec{A B} = (3, 4, 12)$ has squared magnitude 169.
Normalise by 13 and multiply by 10.

A B = 13

; vector

(30 / 13, 40 / 13, 120 / 13)

Check your understanding

Can the zero vector be given a unit direction?

Show answer

No.

Its magnitude is zero, so normalising would divide by zero.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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.
$(0, - 3 / 5, 4 / 5)$
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 ↗

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