Angle between vectors | Year 11

01 / LEARN

Understand the idea.

Rearranging the geometric dot product

To find the angle between two non-zero vectors, calculate their dot product and both magnitudes. Dividing the dot product by the product of those magnitudes removes their lengths, leaving only $cos θ$ . Apply inverse cosine to find the included angle.

Use the angle in $0^{\circ} \leq θ \leq 180^{\circ}$ , with the calculator in degree mode when degrees are required. Retain exact roots until the final calculation. The cosine ratio must lie between $- 1$ and $1$ .

cos θ = \frac{u \cdot v}{∣ u ∣ ∣ v ∣}

Equal-scale coordinate axes with u from (0,0) to (4,1) and v from (0,0) to (1,3). Tick intervals are one unit.

The arrow direction matters

Place the two arrows tail to tail when interpreting their included angle. Reversing one vector negates the dot product, so its angle becomes $180^{\circ} - θ$ . Reversing both preserves the original angle.

An unoriented pair of intersecting lines has supplementary angles, so its smaller angle may differ from the angle between chosen arrows. If either vector is zero, no angle is defined: its direction is absent and the angle formula would divide by zero.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An exact angle

Find the angle between $(1, 0)$ and $(1, \sqrt{3})$ .

Worked solution

1
Calculate the dot product.
$(1, 0) \cdot (1, \sqrt{3}) = 1 (1) + 0 (\sqrt{3}) = 1$
The dot product measures the component alignment.
2
Calculate both magnitudes.
$\begin{aligned} ∣ (1, 0) ∣ & = 1, \\ ∣ (1, \sqrt{3}) ∣ & = \sqrt{1 + 3} = 2. \end{aligned}$
The geometric dot-product formula uses the product of the lengths.
3
Find the cosine and select the included angle.
$cos θ = \frac{1}{1 (2)} = \frac{1}{2}, θ = 60^{\circ}$
The unique angle in the included-angle range with cosine one half is sixty degrees.

Answer

$θ = 60^{\circ}$ .

EXAMPLE 02

An obtuse angle

Find the angle between $(1, 1)$ and $(- 2, 0)$ .

Worked solution

1
Calculate the signed dot product.
$(1, 1) \cdot (- 2, 0) = 1 (- 2) + 1 (0) = - 2$
The negative result indicates an obtuse included angle.
2
Calculate the product of magnitudes.
$∣ (1, 1) ∣ ∣ (- 2, 0) ∣ = \sqrt{2} \times 2 = 2 \sqrt{2}$
Magnitudes are positive even when a vector has a negative component.
3
Find the included angle.
$cos θ = \frac{- 2}{2 \sqrt{2}} = - \frac{1}{\sqrt{2}}, θ = 135^{\circ}$
The negative cosine places the angle between ninety and one hundred eighty degrees.

Answer

$θ = 135^{\circ}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the angle between $(2, 0)$ and $(0, - 5)$ .

Show answer & reasoning

$90^{\circ}$ .

$(2, 0) \cdot (0, - 5) = 0$ and the magnitude product is $2 (5) = 10$ . Thus $cos θ = 0 / 10 = 0$ , giving $θ = 90^{\circ}$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Find the angle between $(1, 0)$ and $(1, 1)$ .
2 marks
Find the angle between $(1, 2)$ and $(2, - 1)$ .
2 marks
Find the angle between $(3, 4)$ and $(1, 0)$ to ${0.1}^{\circ}$ .
3 marks
If the angle between non-zero vectors $u$ and $v$ is $35^{\circ}$ , find the angle between $u$ and $- v$ .
2 marks
Can $a \cdot b = 20$ hold if $∣ a ∣ = 3$ and $∣ b ∣ = 5$ ? Explain.
3 marks
$A = (0, 0)$ , $B = (4, 0)$ , $C = (1, 3)$ . Find angle $B A C$ to ${0.1}^{\circ}$ .
3 marks

Read the worked solutions

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

Find the angle between $(1, 0)$ and $(1, 1)$ .
2 marks
$cos θ = \frac{1}{\sqrt{2}}$ .
$45^{\circ}$ .
Find the angle between $(1, 2)$ and $(2, - 1)$ .
2 marks
The dot product is $2 - 2 = 0$ .
$90^{\circ}$ .
Find the angle between $(3, 4)$ and $(1, 0)$ to ${0.1}^{\circ}$ .
3 marks
$cos θ = \frac{3}{5}$ .
Take inverse cosine.
${53.1}^{\circ}$ .
If the angle between non-zero vectors $u$ and $v$ is $35^{\circ}$ , find the angle between $u$ and $- v$ .
2 marks
Reversing $v$ changes cosine to its negative, giving $180^{\circ} - 35^{\circ}$ .
$145^{\circ}$ .
Can $a \cdot b = 20$ hold if $∣ a ∣ = 3$ and $∣ b ∣ = 5$ ? Explain.
3 marks
It would require $cos θ = \frac{20}{15} > 1$ .
The dot-product magnitude cannot exceed the product of the lengths.
No.
$A = (0, 0)$ , $B = (4, 0)$ , $C = (1, 3)$ . Find angle $B A C$ to ${0.1}^{\circ}$ .
3 marks
Use $A B = (4, 0)$ and $A C = (1, 3)$ .
$cos θ = \frac{4}{4 \sqrt{10}} = \frac{1}{\sqrt{10}}$ .
${71.6}^{\circ}$ .

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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.

Rearranging the geometric dot product

The arrow direction matters

See the method in action.

An exact angle

Calculate the dot product.

Calculate both magnitudes.

Find the cosine and select the included angle.

An obtuse angle

Calculate the signed dot product.

Calculate the product of magnitudes.

Find the included angle.

Try it yourself.

Check your understanding

Keep practising