Dot product | Year 11 | Ascent Library

01 / LEARN

Understand the idea.

A product with a scalar answer

The dot product combines two vectors into a scalar: multiply horizontal components together, multiply vertical components together, then add. It measures directional agreement as well as size, unlike adding vectors, which produces another vector.

For non-zero vectors, the product of magnitudes is positive, so the dot product has the sign of $cos θ$ . A positive result places the angle below $90^{\circ}$ ; zero gives a right angle; a negative result places it above $90^{\circ}$ , up to $180^{\circ}$ .

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

Useful algebraic rules

Dot products distribute over addition and subtraction, and their order can be reversed. In particular, $u \cdot u = u_{x}^{2} + u_{y}^{2} = ∣ u ∣^{2}$ . This turns an unknown length squared into an expression involving dot products you may already know.

Expand brackets term by term. Keep the dot symbol: $u \cdot v$ is a scalar, whereas $(u \cdot v) w$ is that scalar multiplying a vector. These expressions have different types of result.

u \cdot u = ∣ u ∣^{2}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

1
Multiply matching components and add.
$u \cdot v = 3 (4) + (- 2) (5) = 12 - 10 = 2$
The dot product pairs horizontal with horizontal and vertical with vertical.
2
Relate the sign to the cosine.
$cos θ = \frac{2}{\sqrt{13} \sqrt{41}} > 0$
Both magnitudes are positive because neither vector is zero.
3
Classify the angle.
$0^{\circ} < θ < 90^{\circ}$
A positive cosine gives an acute included angle here; the vectors are not parallel because their components are not scalar multiples.

Answer

$u \cdot v = 2$ ; the angle is acute.

EXAMPLE 02

Working without components

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

Worked solution

1
Express squared magnitude as a dot product.
$∣ a - b ∣^{2} = (a - b) \cdot (a - b)$
A vector dotted with itself equals its squared length.
2
Expand and combine the two mixed products.
$\begin{aligned} ∣ a - b ∣^{2} & = a \cdot a - a \cdot b - b \cdot a + b \cdot b \\ = ∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2} . \end{aligned}$
Distribution gives four terms, and the dot product is commutative.
3
Substitute the given values.
$∣ a - b ∣^{2} = 3^{2} - 2 (5) + 4^{2} = 9 - 10 + 16 = 15$
The question asks for the squared magnitude, so no square root is required.

Answer

$∣ a - b ∣^{2} = 15$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$5$ .

Pair the matching components: $(2, - 3) \cdot (4, 1) = 2 (4) + (- 3) (1) = 8 - 3 = 5$ . The answer is a scalar, so it is not written as an ordered pair.

Keep practising

6 worksheet questions · 13 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

A product with a scalar answer

Useful algebraic rules

See the method in action.

Components and angle type

Multiply matching components and add.

Relate the sign to the cosine.

Classify the angle.

Working without components

Express squared magnitude as a dot product.

Expand and combine the two mixed products.

Substitute the given values.

Try it yourself.

Check your understanding

Keep practising