Dot product | Year 11 | Ascent Library

01 / LEARN

The idea, step by step

A product with a scalar answer

The dot product multiplies corresponding components and adds the products. It produces a scalar, not another vector. Geometrically it measures how strongly one direction aligns with another.

For non-zero vectors, the dot product is positive for an acute angle, zero for a right angle, and negative for an obtuse angle. This sign check is often useful before calculating the actual angle.

u \cdot v = u_{x} v_{x} + u_{y} v_{y} = ∣ u ∣ ∣ v ∣ cos θ

Useful algebraic rules

The dot product is commutative and distributes over addition. Scalars can be moved outside it. In particular, dotting a vector with itself gives its squared magnitude.

Use brackets to show exactly what is being dotted. For example $(u + v) \cdot w$ has a scalar result. An expression such as $(u \cdot v) w$ is a scalar multiple of a vector and is a different operation.

u \cdot u = ∣ u ∣^{2}

A common trap

Multiplying the x-component of one vector by the y-component of the other.

Instead: The dot product pairs matching components and adds: $u_{x} v_{x} + u_{y} v_{y}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Components and angle type

For $u = (3, - 2)$ and $v = (4, 5)$ , find the dot product and decide whether their angle is acute, right or obtuse.

Multiply corresponding components: $3 (4) + (- 2) (5)$ .
The result is $12 - 10 = 2 > 0$ .
For non-zero vectors, a positive dot product means a positive cosine.

u \cdot v = 2

; the angle is acute.

Example 2

Working without components

Suppose $∣ a ∣ = 3$ , $∣ b ∣ = 4$ and $a \cdot b = 5$ . Find $∣ a - b ∣^{2}$ .

Expand $(a - b) \cdot (a - b)$ .
This is $∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2}$ .
Substitute $9 - 10 + 16$ .

∣ a - b ∣^{2} = 15

Check your understanding

Find $(2, - 3) \cdot (4, 1)$ .

Show answer

$5$ .

$2 (4) + (- 3) (1) = 8 - 3 = 5$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Calculate $(5, 2) \cdot (- 1, 4)$ .
2 marks
Calculate $(3, - 4) \cdot (3, - 4)$ and interpret it.
2 marks
Classify the angle between $(1, 2)$ and $(- 3, 1)$ .
2 marks
Find k if $(k, 2) \cdot (3, - 1) = 10$ .
2 marks
If $a \cdot b = 7$ and $a \cdot c = - 2$ , find $a \cdot (2 b - 3 c)$ .
2 marks
Given $∣ a ∣ = 5$ , $∣ b ∣ = 2$ and $a \cdot b = - 3$ , find $∣ a + b ∣$ .
3 marks

Read the worked solutions

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

Calculate $(5, 2) \cdot (- 1, 4)$ .
2 marks
$5 (- 1) + 2 (4) = 3$ .
$3$ .
Calculate $(3, - 4) \cdot (3, - 4)$ and interpret it.
2 marks
$9 + 16 = 25$ ; the magnitude is $5$ .
$25$ , the squared magnitude.
Classify the angle between $(1, 2)$ and $(- 3, 1)$ .
2 marks
The dot product is $- 3 + 2 = - 1 < 0$ .
Obtuse.
Find k if $(k, 2) \cdot (3, - 1) = 10$ .
2 marks
$3 k - 2 = 10$ , so $3 k = 12$ .
$k = 4$ .
If $a \cdot b = 7$ and $a \cdot c = - 2$ , find $a \cdot (2 b - 3 c)$ .
2 marks
Distribute: $2 (7) - 3 (- 2) = 20$ .
$20$ .
Given $∣ a ∣ = 5$ , $∣ b ∣ = 2$ and $a \cdot b = - 3$ , find $∣ a + b ∣$ .
3 marks
Square the length using the dot product: $25 + 4 + 2 (- 3) = 23$ .
Take the non-negative square root.
$\sqrt{23}$ .

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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 ↗

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