Year 12 Maths Extension 1 (2027) Vectors

3D vectors (component form, magnitude, operations)

Q: How do you write a 3D vector?

As x i plus y j plus z k, the triple x comma y comma z, or a column vector.

Q: What is the magnitude of a 3D vector?

The square root of x squared plus y squared plus z squared.

Q: How do you show three points are collinear?

Show AB equals k times BC, parallel and sharing a point.

Q: Are 3D operations different from 2D?

No, adding, scaling and testing parallel all work componentwise, just with an extra z.

20 practice questions 2 video lessons Theory + worked examples

Extend your skills to 3D vectors in NSW Year 12 Mathematics Extension 1. Adding the k unit vector lets you describe points and directions in three-dimensional space using component form, an ordered triple or a column vector.

You will learn to form the vector between two points in space, calculate its magnitude and midpoint, and operate with three-dimensional vectors — extending the component form methods of the plane into 3D problems in the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

Vectors : Magnitude of a vector 3D : ExamSolutions Watch
Collinear Points in 3D (Ch1 Pr18) Watch

Theory

3D vectors add a third component $z$ along $k$ : written $x i + y j + z k$ with magnitude $\sqrt{x^{2} + y^{2} + z^{2}}$ . Every 2D technique carries over componentwise. This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-02.

Three-dimensional vectors work exactly like 2D vectors, with a third component $z$ along the unit vector $k$ . A 3D vector is written $x i + y j + z k$ , as $(x, y, z)$ , or as a column vector.

From $A (x, y, z)$ to $B (u, v, w)$ : $\vec{A B} = (u - x) i + (v - y) j + (w - z) k$ . The magnitude is $\sqrt{x^{2} + y^{2} + z^{2}}$ , and $\hat{a} = \frac{1}{| a |} a$ .

Adding, subtracting, scaling and testing parallel all work componentwise, exactly as in 2D. On the $x y$ -plane $z = 0$ ; on the $y$ -axis $x = z = 0$ .

NESA link. Part of the Year 12 Introduction to vectors focus area, outcome ME1-12-02 ("operates with 2D and 3D vectors and uses 2D vectors to solve problems involving motion in two dimensions") with MAO-WM-01.

The unit vectors

i

j

k

along the axes.

a = x i + y j + z k

, the diagonal of its component box.

| x i + y j + z k | = \sqrt{x^{2} + y^{2} + z^{2}}, \hat{a} = \frac{1}{| a |} a .

magnitude = sqrt(x^2 + y^2 + z^2); unit vector = a / |a|

Points $A, B, C$ are collinear if $\vec{A B} = k \vec{B C}$ (parallel and sharing a point).

collinear if AB = k BC

Three squared terms. The magnitude adds $z^{2}$ ; otherwise every 2D technique carries straight over.

How to work in 3D

Magnitude: $\sqrt{x^{2} + y^{2} + z^{2}}$ .
Vector between points: subtract the tail's coordinates from the head's.
Unit vector: divide each component by the magnitude.
Collinearity: check $\vec{A B} = k \vec{B C}$ .

Example 1 — Magnitude

Find

| a |

for

a = i + 4 j + 8 k

Solution

| a |

=

\sqrt{1^{2} + 4^{2} + 8^{2}} = \sqrt{81} = 9

|a| = 9

$| a | = 9$ .

Example 2 — Between points

Given

A (0, 1, 2)

and

B (4, 5, 4)

, find

\vec{A B}

and

| \vec{A B} |

Solution

$\vec{A B}$	$=$	$4 i + 4 j + 2 k$
$\| \vec{A B} \|$	$=$	$\sqrt{16 + 16 + 4} = 6$

AB = 4i + 4j + 2k, |AB| = 6

$\vec{A B} = 4 i + 4 j + 2 k$ , $| \vec{A B} | = 6$ .

Example 3 — Unit vector

Find the unit vector for

a = 2 i - 6 j + 3 k

Solution

$| a | = \sqrt{4 + 36 + 9} = 7$ .

\hat{a}

=

\frac{1}{7} (2 i - 6 j + 3 k)

a-hat = (1/7)(2i - 6j + 3k)

$\hat{a} = \frac{1}{7} (2 i - 6 j + 3 k)$ .

Example 4 — Collinearity

Show

A (1, 0, 2)

B (3, 3, 5)

C (7, 9, 11)

are collinear.

Solution

$\vec{A B}$	$=$	$2 i + 3 j + 3 k$
$\vec{B C}$	$=$	$4 i + 6 j + 6 k = 2 \vec{A B}$

BC = 2 AB and they share B, so A, B, C collinear

Since $\vec{B C} = 2 \vec{A B}$ and they share $B$ : $A$ , $B$ , $C$ are collinear.

Common pitfalls

Missing the $z$ term. The magnitude has three squares:

\sqrt{x^{2} + y^{2} + z^{2}}

Coordinate planes. On the

x y

-plane

z = 0

; on the

y

-axis

x = z = 0

Collinearity needs a shared point. Parallel vectors alone are not enough — they must share a point.

Same techniques. Everything else is exactly as in 2D — don't overcomplicate.

Frequently asked questions

How do you write a 3D vector?

As $x i + y j + z k$ , the triple $(x, y, z)$ , or a column vector.

What is the magnitude of a 3D vector?

$\sqrt{x^{2} + y^{2} + z^{2}}$ .

How do you find a 3D unit vector?

Divide the vector by its magnitude.

How do you show three points are collinear?

Show $\vec{A B} = k \vec{B C}$ — parallel and sharing a point.

Are 3D operations different from 2D?

No — adding, scaling and testing parallel all work componentwise, just with an extra $z$ .

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Operating with vectors in two dimensions

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Scalar (dot) product – 2D & 3D

3D vectors (component form, magnitude, operations)

📖 Theory

How to work in 3D

Common pitfalls

Frequently asked questions

Theory