SME
Grade 12 Math · Precalculus
Vectors in three dimensions: dot and cross products
Use this for “Find u · v and the angle between u = ⟨1, 2, 2⟩ and v = ⟨4, 0, 3⟩.”
Diagram
u = ⟨1, 2, 2⟩, with length |u| = 3.
v = ⟨4, 0, 3⟩, with length |v| = 5.
u·v = 1 × 4 + 2 × 0 + 2 × 3 = 10.
cos
θ
=
10
3 × 5
,
so
θ
=
48.19°:
acute.
This is an example. Type your own values to start a new problem.
⟨
u₁
,
u₂
,
u₃
⟩
·
⟨
v₁
,
v₂
,
v₃
⟩
=
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
=
u₁
×
v₁
+
u₂
×
v₂
+
u₃
×
v₃
10 = 1 × 4 + 2 × 0 + 2 × 3
|u|
=
√
u₁
2
+
u₂
2
+
u₃
2
3
=
√
1
2
+
2
2
+
2
2
|v|
=
√
v₁
2
+
v₂
2
+
v₃
2
5
=
√
4
2
+
0
2
+
3
2
θ
=
cos⁻¹(
u · v
|u| × |v|
)
48.1897
=
cos⁻¹(
10
3 × 5
)
Assumptions
•
A vector in space has three components, ⟨x, y, z⟩; u · v multiplies matching components and adds.
•
cos θ = u · v ÷ (|u||v|), with θ from 0° to 180°.
•
A dot product of 0 means the vectors are perpendicular.
Step-by-step
WE KNOW
u₁ = 1, u₂ = 2, u₃ = 2, v₁ = 4, v₂ = 0, v₃ = 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 = u₁ × v₁ + u₂ × v₂ + u₃ × v₃
Multiply matching components and add.
u · v = 1 × 4 + 2 × 0 + 2 × 3
u · v = 4 + 6
u · v = 10
Step 2 · Find magnitude of u (|u|)
Use |u| = √(u₁² + u₂² + u₃²)
The Pythagorean theorem twice: across the floor, then up. That is u’s length.
|u|
=
√
1
2
+
2
2
+
2
2
|u|
=
√
1 + 4 + 4
|u|
=
√
9
|u| = 3
Step 3 · Find magnitude of v (|v|)
Use |v| = √(v₁² + v₂² + v₃²)
The Pythagorean theorem twice: across the floor, then up. That is v’s length.
|v|
=
√
4
2
+
0
2
+
3
2
|v|
=
√
16 + 9
|v|
=
√
25
|v| = 5
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(
10
3 × 5
)
θ
=
cos⁻¹(
10
15
)
θ
=
cos⁻¹(0.6667)
θ
=
48.1897°
Check
Put the numbers back in. Both sides should match.
10 = 1 × 4 + 2 × 0 + 2 × 3 ✓
3
=
√
1
2
+
2
2
+
2
2
✓
5
=
√
4
2
+
0
2
+
3
2
✓
48.1897
=
cos⁻¹(
10
3 × 5
)
✓
Topic check
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