β
γ
ω
2
=
m
2
m
3
+
n
2
n
3
+
p
2
p
3
{\displaystyle {\frac {\beta \gamma }{\omega ^{2}}}=m_{2}m_{3}+n_{2}n_{3}+p_{2}p_{3}}
γ
α
ω
2
=
m
3
m
1
+
n
3
n
1
+
p
3
p
1
{\displaystyle {\frac {\gamma \alpha }{\omega ^{2}}}=m_{3}m_{1}+n_{3}n_{1}+p_{3}p_{1}}
α
β
ω
2
=
m
1
m
2
+
n
1
n
2
+
p
1
p
2
{\displaystyle {\frac {\alpha \beta }{\omega ^{2}}}=m_{1}m_{2}+n_{1}n_{2}+p_{1}p_{2}}
4)
α
ω
2
=
a
m
1
+
b
n
1
+
c
p
1
{\displaystyle {\frac {\alpha }{\omega ^{2}}}=am_{1}+bn_{1}+cp_{1}}
β
ω
2
=
a
m
2
+
b
n
2
+
c
p
2
{\displaystyle {\frac {\beta }{\omega ^{2}}}=am_{2}+bn_{2}+cp_{2}}
γ
ω
2
=
a
m
3
+
b
n
3
+
c
p
3
{\displaystyle {\frac {\gamma }{\omega ^{2}}}=am_{3}+bn_{3}+cp_{3}}
5)
If we take
α
β
γ
{\displaystyle \alpha \beta \gamma }
as given, then we have 12 constants available, so we can arbitrarily use three of them.
The solution is most comfortable when we use a temporary co-ordinate system X 1 , Y 1 , Z 1 , for which β and γ disappear in equations (2), α is equal to ϰ, that is, a co-ordinate system whose X1 -axis falls in the direction, of which the direction cosine is proportional to X, Y, Z with α, β, γ.
Furthermore, it should be set
m
h
2
+
n
h
2
+
p
h
2
=
q
h
2
,
m
h
/
q
h
=
μ
h
,
n
h
/
q
h
=
ν
h
,
p
h
/
q
h
=
π
h
a
2
+
b
2
+
c
2
=
d
2
,
a
/
d
=
μ
,
b
/
d
=
ν
,
c
/
d
=
π
,
{\displaystyle {\begin{array}{clcclcclcclc}m_{h}^{2}+n_{h}^{2}+p_{h}^{2}&=&q_{h}^{2},&m_{h}/q_{h}&=&\mu _{h},&n_{h}/q_{h}&=&\nu _{h},&p_{h}/q_{h}&=&\pi _{h}\\\\a^{2}+b^{2}+c^{2}&=&d^{2},&a/d&=&\mu ,&b/d&=&\nu ,&c/d&=&\pi ,\end{array}}}
then μ, ν, π are the direction cosines of 4 directions, which we will denote by δ1 , δ2 , δ3 and δ, against the system X1 , Y1 , Z1 .
By these introductions our equations (3), (4) and (5) will be:
1
−
ω
2
d
2
=
q
1
2
−
ϰ
2
ω
2
=
q
2
2
=
q
3
2
{\displaystyle 1-\omega ^{2}d^{2}=q_{1}^{2}-{\frac {\varkappa ^{2}}{\omega ^{2}}}=q_{2}^{2}=q_{3}^{2}}
3')
μ
2
μ
3
+
ν
2
ν
3
+
π
2
π
3
=
μ
3
μ
1
+
ν
3
ν
1
+
π
3
π
1
=
μ
1
μ
2
+
ν
1
ν
2
+
π
1
π
2
=
0
{\displaystyle \mu _{2}\mu _{3}+\nu _{2}\nu _{3}+\pi _{2}\pi _{3}=\mu _{3}\mu _{1}+\nu _{3}\nu _{1}+\pi _{3}\pi _{1}=\mu _{1}\mu _{2}+\nu _{1}\nu _{2}+\pi _{1}\pi _{2}=0}
that is
cos
(
δ
2
,
δ
3
)
=
cos
(
δ
3
,
δ
1
)
=
cos
(
δ
1
,
δ
2
)
=
0
{\displaystyle {\text{that is }}\cos(\delta _{2},\delta _{3})=\cos(\delta _{3},\delta _{1})=\cos(\delta _{1},\delta _{2})=0}
4')
μ
μ
1
+
ν
ν
1
+
π
π
1
=
ϰ
ω
2
q
1
d
,
μ
μ
2
+
ν
ν
2
+
π
π
2
+
μ
μ
3
+
ν
ν
3
+
π
π
3
=
0
{\displaystyle \mu \mu _{1}+\nu \nu _{1}+\pi \pi _{1}={\frac {\varkappa }{\omega ^{2}q_{1}d}},\ \mu \mu _{2}+\nu \nu _{2}+\pi \pi _{2}+\mu \mu _{3}+\nu \nu _{3}+\pi \pi _{3}=0}
that is
cos
(
δ
,
δ
1
)
=
ϰ
ω
2
q
1
d
,
cos
(
δ
,
δ
2
)
=
cos
(
δ
,
δ
3
)
=
0.
{\displaystyle {\text{that is }}\cos(\delta ,\delta _{1})={\frac {\varkappa }{\omega ^{2}q_{1}d}},\ \cos(\delta ,\delta _{2})=\cos(\delta ,\delta _{3})=0.}
5')
According to (4'), the three directions δ1 , δ2 , δ3 are perpendicular to each other, according to (5')
δ
1
{\displaystyle \delta _{1}}
falls into δ, then it must be:
μ
=
μ
1
,
ν
=
ν
1
,
π
=
π
1
and
ϰ
ω
2
q
1
d
=
1.
{\displaystyle \mu =\mu _{1},\ \nu =\nu _{1},\ \pi =\pi _{1}\ {\text{ and }}\ {\frac {\varkappa }{\omega ^{2}q_{1}d}}=1.}
6)