Consider the vectors U and V (with respective magnitudes |U | and |V |). If those vectors enclose an angle θ then the dot product of those vectors can be written as:
Calculating bond angles of a symmetrical tetrahedral molecule such as methane using a dot product
U
⋅
V
=
|
U
|
|
V
|
cos
(
θ
)
{\displaystyle \mathbf {U} \cdot \mathbf {V} =|\mathbf {U} ||\mathbf {V} |\cos(\theta )}
If the vectors can be written as:
U
=
(
U
x
,
U
y
,
U
z
)
{\displaystyle \mathbf {U} =(U_{x},U_{y},U_{z})}
V
=
(
V
x
,
V
y
,
V
z
)
{\displaystyle \mathbf {V} =(V_{x},V_{y},V_{z})}
then the dot product is given by:
U
⋅
V
=
U
x
V
x
+
U
y
V
y
+
U
z
V
z
{\displaystyle \mathbf {U} \cdot \mathbf {V} =U_{x}V_{x}+U_{y}V_{y}+U_{z}V_{z}}
For example,
(
1
,
2
,
3
)
⋅
(
2
,
2
,
2
)
=
1
(
2
)
+
2
(
2
)
+
3
(
2
)
=
12.
{\displaystyle (1,2,3)\cdot (2,2,2)=1(2)+2(2)+3(2)=12.}
and
(
0
,
5
,
0
)
⋅
(
4
,
0
,
0
)
=
0.
{\displaystyle (0,5,0)\cdot (4,0,0)=0.}
We can interpret the last case by noting that the product is zero because the angle between the two vectors is 90 degrees.
Since
|
U
|
=
U
x
2
+
U
y
2
+
U
z
2
{\displaystyle |\mathbf {U} |={\sqrt {U_{x}^{2}+U_{y}^{2}+U_{z}^{2}}}}
and
|
V
|
=
V
x
2
+
V
y
2
+
V
z
2
{\displaystyle |\mathbf {V} |={\sqrt {V_{x}^{2}+V_{y}^{2}+V_{z}^{2}}}}
this means that
cos
(
θ
)
=
U
x
V
x
+
U
y
V
y
+
U
z
V
z
U
x
2
+
U
y
2
+
U
z
2
V
x
2
+
V
y
2
+
V
z
2
{\displaystyle \cos(\theta )={\frac {U_{x}V_{x}+U_{y}V_{y}+U_{z}V_{z}}{{\sqrt {U_{x}^{2}+U_{y}^{2}+U_{z}^{2}}}{\sqrt {V_{x}^{2}+V_{y}^{2}+V_{z}^{2}}}}}}