Angle between vectors | Year 11

01 / LEARN

The idea, step by step

Rearranging the geometric dot product

For two non-zero vectors, divide the dot product by the product of their magnitudes to find the cosine of their smaller included angle. That angle is between zero and 180 degrees.

A negative cosine gives an obtuse angle. Keep exact values until the final inverse-cosine calculation. Check that the ratio is between negative one and one; a value outside this interval usually signals an arithmetic error.

cos θ = \frac{u \cdot v}{∣ u ∣ ∣ v ∣}

The arrow direction matters

Reversing one vector changes the sign of its dot product and replaces the angle by its supplement. The smaller angle between two unoriented lines may therefore differ from the angle between the specific directed vectors chosen along those lines.

If either vector is zero, the formula has zero in its denominator. A zero displacement has no defined direction, so its angle with another vector is undefined.

A common trap

Using the dot product alone as the cosine.

Instead: Divide by both magnitudes. The dot product changes if either vector is stretched, but the angle does not.

02 / WORKED EXAMPLES

See the thinking

Example 1

An exact angle

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

Their dot product is $1$ .
Their lengths are $1$ and $2$ .
Thus $cos θ = 1 / 2$ .

θ = 60^{\circ}

Example 2

An obtuse angle

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

The dot product is $- 2$ .
The magnitude product is $\sqrt{2} \times 2$ .
$cos θ = - 1 / \sqrt{2}$ .

θ = 135^{\circ}

Check your understanding

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

Show answer

$90^{\circ}$ .

The non-zero vectors have dot product zero, so their cosine is zero.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF 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 BAC 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
$cos θ = 1 / \sqrt{2}$ .
$45^{\circ}$ .
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
$cos θ = 3 / 5$ .
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 $cos θ = 20 / 15 > 1$ .
The dot-product magnitude cannot exceed the product of the lengths.
No.
A=(0,0), B=(4,0), C=(1,3). Find angle BAC to ${0.1}^{\circ}$ .
3 marks
Use AB= $(4, 0)$ and AC= $(1, 3)$ .
$cos θ = 4 / (4 \sqrt{10}) = 1 / \sqrt{10}$ .
${71.6}^{\circ}$ .

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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 ↗

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