Year 12 Maths Extension 1 (2027) Vectors

Scalar (dot) product – 2D & 3D

Q: What is the scalar (dot) product?

A number equal to x1 x2 plus y1 y2 (plus z1 z2), which also equals the product of the magnitudes and cosine of the angle.

Q: How do you find the angle between two vectors?

Use cos theta equals the dot product divided by the product of the magnitudes.

Q: What does a zero dot product mean?

The vectors are perpendicular, for non-zero vectors.

Q: How is the dot product linked to magnitude?

The dot product of a vector with itself equals the square of its magnitude.

20 practice questions 2 video lessons Theory + worked examples

Master the scalar (dot) product for NSW Year 12 Mathematics Extension 1, in both two and three dimensions. The dot product combines two vectors into a single number and connects directly to the angle between them.

You will learn to compute the dot product from components, find the angle between two vectors, and test when vectors are perpendicular — a versatile tool for angle and geometry problems throughout the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

Vectors - Scalar Products (by Class Mathematics) Watch
Dot Product and Angle Between 3D Vectors Watch

Theory

The scalar (dot) product is $a \cdot b = x_{1} x_{2} + y_{1} y_{2} (+ z_{1} z_{2}) = | a | | b | \cos θ$ , giving the angle between vectors and testing perpendicularity via $a \cdot b = 0$ . This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-02.

The scalar (dot) product multiplies two vectors to give a number. It measures how much the vectors point the same way, and unlocks the angle between them.

Algebraic form: $a \cdot b = x_{1} x_{2} + y_{1} y_{2}$ in 2D (add $z_{1} z_{2}$ in 3D). Geometric form: $a \cdot b = | a | | b | \cos θ$ , where $0^{\circ} \leq θ \leq 180^{\circ}$ .

Hence $\cos θ = \frac{a \cdot b}{| a | | b |}$ , $a \cdot a = | a |^{2}$ , and $a ⊥ b ⟺ a \cdot b = 0$ . The sign reveals the angle: positive is acute, zero is perpendicular, negative is obtuse.

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.

a \cdot b = | a | | b | \cos θ

uses the angle

θ

between them.

The sign of

a \cdot b

tells you acute, right, or obtuse.

a \cdot b = x_{1} x_{2} + y_{1} y_{2} (+ z_{1} z_{2}), a \cdot b = | a | | b | \cos θ .

dot product = x1 x2 + y1 y2 (+ z1 z2) = |a||b| cos theta

\cos θ = \frac{a \cdot b}{| a | | b |}, a \cdot a = | a |^{2}, a ⊥ b ⟺ a \cdot b = 0.

cos theta = (a.b)/(|a||b|); a.a = |a|^2; perpendicular iff a.b = 0

Scalar, not vector. The dot product is a number. A negative value means the angle is obtuse; zero means perpendicular.

How to use the dot product

Compute $a \cdot b$ by multiplying matching components and adding.
For an angle, divide by $| a | | b |$ to get $\cos θ$ .
For perpendicularity, set $a \cdot b = 0$ and solve.
For a magnitude, use $| a | = \sqrt{a \cdot a}$ .

Example 1 — Compute it

Find

a \cdot b

for

a = 3 i + 5 j

and

b = 4 i + 2 j

Solution

a \cdot b

=

(3) (4) + (5) (2) = 12 + 10 = 22

a.b = 22

$a \cdot b = 22$ .

Example 2 — Angle (2D)

Find the angle between

a = i + 2 j

and

b = 3 i + j

Solution

$a \cdot b = 5$ , $| a | = \sqrt{5}$ , $| b | = \sqrt{10}$ .

\cos θ

=

\frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}} \Rightarrow θ = 45^{\circ}

theta = 45 degrees

$θ = 45^{\circ}$ .

Example 3 — Perpendicular (3D)

Find

t

so that

a = 3 i + t j - 2 k

and

b = 2 i + j + 4 k

are perpendicular.

Solution

Set $a \cdot b = 0$ .

6 + t - 8

=

0 \Rightarrow t = 2

t = 2

$t = 2$ .

Example 4 — Obtuse angle (3D)

Find the angle between

a = i + 2 j - 2 k

and

b = 2 i - 2 j + k

Solution

$a \cdot b = 2 - 4 - 2 = - 4$ , $| a | = | b | = 3$ .

\cos θ

=

- \frac{4}{9} \Rightarrow θ \approx 116^{\circ}

theta approx 116 degrees (obtuse)

$θ \approx 116^{\circ}$ (obtuse).

Common pitfalls

Scalar result. The dot product is a number, not a vector.

Perpendicular test.

a \cdot b = 0

means perpendicular for non-zero vectors.

Sign of the angle. A negative dot product means the angle is obtuse; positive means acute.

Magnitude link.

a \cdot a = | a |^{2}

, so

| a | = \sqrt{a \cdot a}

Frequently asked questions

What is the scalar (dot) product?

A number $a \cdot b = x_{1} x_{2} + y_{1} y_{2} (+ z_{1} z_{2}) = | a | | b | \cos θ$ .

How do you find the angle between two vectors?

Use $\cos θ = \frac{a \cdot b}{| a | | b |}$ .

What does a zero dot product mean?

The vectors are perpendicular (for non-zero vectors).

What does the sign tell you?

Positive means an acute angle, zero perpendicular, negative obtuse.

How is the dot product linked to magnitude?

$a \cdot a = | a |^{2}$ .

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3D vectors (component form, magnitude, operations)

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Projection of a vector & perpendicular component

Scalar (dot) product – 2D & 3D

📖 Theory

How to use the dot product

Common pitfalls

Frequently asked questions

Theory