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Grade 12 Math · Problem type

The dot product and the angle between vectors

Use this for “Find the angle between ⟨2, 1⟩ and ⟨1, 3⟩” or “Are these vectors perpendicular?”

Diagram

−1123−2−11234xyθ = 45°uv
u = ⟨2, 1⟩
|u|
=
√
5
≈
2.24.
v = ⟨1, 3⟩
|v|
=
√
10
≈
3.16.
u·v = 2 × 1 + 1 × 3 = 5.
u·v > 0, so the angle between them, 45°, is acute.
This is an example. Type your own values to start a new problem.
⟨
a
,
b
⟩
·
⟨
c
,
d
⟩
=
u · v
|u|
Magnitude of u
worked out (not typed)
|v|
Magnitude of v
worked out (not typed)
θ
Angle between u and v
worked out (not typed) · °

Formulas

u
·
v
=
a
×
c
+
b
×
d
5 = 2 × 1 + 1 × 3
|u|
=
√
a
2
+
b
2
2.2361
=
√
2
2
+
1
2
|v|
=
√
c
2
+
d
2
3.1623
=
√
1
2
+
3
2
θ
=
cos⁻¹(
u · v
|u| × |v|
)
45
=
cos⁻¹(
5
2.2361 × 3.1623
)

Assumptions

•
u · v = ac + bd: multiply matching components and add.
•
cos θ = u · v ÷ (|u||v|), with θ from 0° to 180°.
•
A dot product of 0 means perpendicular: ⟨3, 4⟩ · ⟨4, −3⟩ = 0.

Step-by-step

WE KNOW
a = 2, b = 1, c = 1, d = 3
FIND
dot product (u · v), magnitude of u (|u|), magnitude of v (|v|), angle between u and v (θ)
Step 1 · Find dot product (u · v)
Use u · v = a × c + b × d
Multiply matching components and add.
u · v = 2 × 1 + 1 × 3
u · v = 5
Step 2 · Find magnitude of u (|u|)
Use |u| = √(a² + b²)
The Pythagorean theorem on u’s components.
|u|
=
√
2
2
+
1
2
|u|
=
√
4 + 1
|u|
=
√
5
|u| = 2.2361
Step 3 · Find magnitude of v (|v|)
Use |v| = √(c² + d²)
The Pythagorean theorem on v’s components.
|v|
=
√
1
2
+
3
2
|v|
=
√
1 + 9
|v|
=
√
10
|v| = 3.1623
Step 4 · Find angle between u and v (θ)
Use θ = cos⁻¹((u · v) ÷ (|u| × |v|))
The cosine of the angle is the dot product over the product of the magnitudes.
θ
=
cos⁻¹(
5
2.2361 × 3.1623
)
θ
=
cos⁻¹(
5
7.0711
)
θ
=
cos⁻¹(0.7071)
θ
=
45°
Check
Put the numbers back in. Both sides should match.
5 = 2 × 1 + 1 × 3 ✓
2.2361
≈
√
2
2
+
1
2
✓
3.1623
≈
√
1
2
+
3
2
✓
45
=
cos⁻¹(
5
2.2361 × 3.1623
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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