Development
Three spheres of known centers A, B, C and known radii AD, BD, CD intersect at two points D and D' . Similarly, three sticks of known length are planted in the ground at known coordinates . The other ends meet at a calculated apex . Calculating D and D' , the projection[ 1] AD onto AB and AC , and the projection of BD onto BC results in,
M
A
B
=
A
+
A
D
cos
(
∠
B
A
D
)
A
B
‖
A
B
‖
=
A
+
[
(
A
D
)
2
+
(
A
B
)
2
−
(
B
D
)
2
2
(
A
B
)
2
]
A
B
{\displaystyle \mathbf {M_{AB}} =\mathbf {A} +AD\cos(\angle {BAD}){\dfrac {\mathbf {AB} }{\left\Vert \mathbf {AB} \right\|}}=\mathbf {A} +\left[{\dfrac {(AD)^{2}+(AB)^{2}-(BD)^{2}}{2(AB)^{2}}}\right]\mathbf {AB} }
M
A
C
=
A
+
A
D
cos
(
∠
C
A
D
)
A
C
‖
A
C
‖
=
A
+
[
(
A
D
)
2
+
(
A
C
)
2
−
(
C
D
)
2
2
(
A
C
)
2
]
A
C
{\displaystyle \mathbf {M_{AC}} =\mathbf {A} +AD\cos(\angle {CAD}){\dfrac {\mathbf {AC} }{\left\Vert \mathbf {AC} \right\|}}=\mathbf {A} +\left[{\dfrac {(AD)^{2}+(AC)^{2}-(CD)^{2}}{2(AC)^{2}}}\right]\mathbf {AC} }
M
B
C
=
B
+
B
D
cos
(
∠
C
B
D
)
B
C
‖
B
C
‖
=
B
+
[
(
B
D
)
2
+
(
B
C
)
2
−
(
C
D
)
2
2
(
B
C
)
2
]
B
C
{\displaystyle \mathbf {M_{BC}} =\mathbf {B} +BD\cos(\angle {CBD}){\dfrac {\mathbf {BC} }{\left\Vert \mathbf {BC} \right\|}}=\mathbf {B} +\left[{\dfrac {(BD)^{2}+(BC)^{2}-(CD)^{2}}{2(BC)^{2}}}\right]\mathbf {BC} }
(
B
D
)
2
=
(
A
B
)
2
+
(
A
D
)
2
−
2
(
A
B
)
(
A
D
)
cos
(
∠
B
A
D
)
{\displaystyle (BD)^{2}=(AB)^{2}+(AD)^{2}-2(AB)(AD)\cos(\angle {BAD})}
(
C
D
)
2
=
(
A
C
)
2
+
(
A
D
)
2
−
2
(
A
C
)
(
A
D
)
cos
(
∠
C
A
D
)
{\displaystyle (CD)^{2}=(AC)^{2}+(AD)^{2}-2(AC)(AD)\cos(\angle {CAD})}
(
C
D
)
2
=
(
B
C
)
2
+
(
B
D
)
2
−
2
(
B
C
)
(
B
D
)
cos
(
∠
C
B
D
)
{\displaystyle (CD)^{2}=(BC)^{2}+(BD)^{2}-2(BC)(BD)\cos(\angle {CBD})}
By the law of cosines .
The three unit normals to AB , AC and BC in the plane of ABC are:
N
A
B
=
A
C
−
A
C
∙
A
B
(
A
B
)
2
A
B
‖
A
C
−
A
C
∙
A
B
(
A
B
)
2
A
B
‖
{\displaystyle \mathbf {N_{AB}} ={\cfrac {\mathbf {AC} -{\cfrac {\mathbf {AC} \bullet \mathbf {AB} }{(AB)^{2}}}\mathbf {AB} }{\left\Vert {\mathbf {AC} -{\cfrac {\mathbf {AC} \bullet \mathbf {AB} }{(AB)^{2}}}\mathbf {AB} }\right\|}}}
N
A
C
=
A
B
−
A
B
∙
A
C
(
A
C
)
2
A
C
‖
A
B
−
A
B
∙
A
C
(
A
C
)
2
A
C
‖
{\displaystyle \mathbf {N_{AC}} ={\cfrac {\mathbf {AB} -{\cfrac {\mathbf {AB} \bullet \mathbf {AC} }{(AC)^{2}}}\mathbf {AC} }{\left\Vert {\mathbf {AB} -{\cfrac {\mathbf {AB} \bullet \mathbf {AC} }{(AC)^{2}}}\mathbf {AC} }\right\|}}}
N
B
C
=
A
B
−
A
B
∙
B
C
(
B
C
)
2
B
C
‖
A
B
−
A
B
∙
B
C
(
B
C
)
2
B
C
‖
{\displaystyle \mathbf {N_{BC}} ={\cfrac {\mathbf {AB} -{\cfrac {\mathbf {AB} \bullet \mathbf {BC} }{(BC)^{2}}}\mathbf {BC} }{\left\Vert {\mathbf {AB} -{\cfrac {\mathbf {AB} \bullet \mathbf {BC} }{(BC)^{2}}}\mathbf {BC} }\right\|}}}
Then the three vectors intersect at a common point:
M
A
B
+
m
A
B
N
A
B
=
M
A
C
+
m
A
C
N
A
C
=
M
B
C
+
m
B
C
N
B
C
{\displaystyle \mathbf {M_{AB}} +m_{AB}\mathbf {N_{AB}} =\mathbf {M_{AC}} +m_{AC}\mathbf {N_{AC}} =\mathbf {M_{BC}} +m_{BC}\mathbf {N_{BC}} }
mAB , mAC and mBC
|
m
A
B
m
A
C
m
B
C
|
=
(
H
T
H
)
−
1
H
T
g
{\displaystyle {\begin{vmatrix}m_{AB}\\m_{AC}\\m_{BC}\\\end{vmatrix}}=(H^{T}H)^{-1}H^{T}\mathbf {g} }
A spreadsheet command for calculating this is,
PRODUCT(PRODUCT(MINVERSE(PRODUCT(TRANSPOSE H, H)), TRANSPOSE H), g ) An example of a spreadsheet that does complete calculations of this entire problem is given at the External links section at the end of this article.
g in this least squares solution[ 2]
H
=
|
N
A
B
x
−
N
A
C
x
0
N
A
B
y
−
N
A
C
y
0
N
A
B
z
−
N
A
C
z
0
0
N
A
C
x
−
N
B
C
x
0
N
A
C
y
−
N
B
C
y
0
N
A
C
z
−
N
B
C
z
N
A
B
x
0
−
N
B
C
x
N
A
B
y
0
−
N
B
C
y
N
A
B
z
0
−
N
B
C
z
|
g
=
|
M
A
C
x
−
M
A
B
x
M
A
C
y
−
M
A
B
y
M
A
C
z
−
M
A
B
z
M
B
C
x
−
M
A
C
x
M
B
C
y
−
M
A
C
y
M
B
C
z
−
M
A
C
z
M
B
C
x
−
M
A
B
x
M
B
C
y
−
M
A
B
y
M
B
C
z
−
M
A
B
z
|
{\displaystyle H={\begin{vmatrix}N_{ABx}&-N_{ACx}&0\\N_{ABy}&-N_{ACy}&0\\N_{ABz}&-N_{ACz}&0\\0&N_{ACx}&-N_{BCx}\\0&N_{ACy}&-N_{BCy}\\0&N_{ACz}&-N_{BCz}\\N_{ABx}&0&-N_{BCx}\\N_{ABy}&0&-N_{BCy}\\N_{ABz}&0&-N_{BCz}\end{vmatrix}}\qquad \mathbf {g} ={\begin{vmatrix}M_{ACx}-M_{ABx}\\M_{ACy}-M_{ABy}\\M_{ACz}-M_{ABz}\\M_{BCx}-M_{ACx}\\M_{BCy}-M_{ACy}\\M_{BCz}-M_{ACz}\\M_{BCx}-M_{ABx}\\M_{BCy}-M_{ABy}\\M_{BCz}-M_{ABz}\\\end{vmatrix}}}
mAB , mAC and mBC :
N
A
B
x
m
A
B
−
N
A
C
x
m
A
C
=
M
A
C
x
−
M
A
B
x
N
A
C
y
m
A
C
−
N
B
C
y
m
B
C
=
M
B
C
y
−
M
A
C
y
N
A
B
z
m
A
B
−
N
B
C
z
m
B
C
=
M
B
C
z
−
M
A
B
z
{\displaystyle {\begin{aligned}N_{ABx}m_{AB}-N_{ACx}m_{AC}&=M_{ACx}-M_{ABx}\\N_{ACy}m_{AC}-N_{BCy}m_{BC}&=M_{BCy}-M_{ACy}\\N_{ABz}m_{AB}-N_{BCz}m_{BC}&=M_{BCz}-M_{ABz}\\\end{aligned}}}
The unit normal to the plane of ABC is,
N
D
=
A
C
×
A
B
‖
A
C
×
A
B
‖
{\displaystyle \mathbf {N_{D}} ={\dfrac {\mathbf {AC} \times \mathbf {AB} }{\left\Vert {\mathbf {AC} \times \mathbf {AB} }\right\|}}}
Solution
D
=
{
M
A
B
+
m
A
B
N
A
B
+
(
M
A
B
D
)
2
−
m
A
B
2
N
D
M
A
C
+
m
A
C
N
A
C
+
(
M
A
C
D
)
2
−
m
A
C
2
N
D
M
B
C
+
m
B
C
N
B
C
+
(
M
B
C
D
)
2
−
m
B
C
2
N
D
{\displaystyle \mathbf {D} ={\begin{cases}\mathbf {M_{AB}} +m_{AB}\mathbf {N_{AB}} +{\sqrt {(M_{AB}D)^{2}-m_{AB}^{2}}}\mathbf {N_{D}} \\\mathbf {M_{AC}} +m_{AC}\mathbf {N_{AC}} +{\sqrt {(M_{AC}D)^{2}-m_{AC}^{2}}}\mathbf {N_{D}} \\\mathbf {M_{BC}} +m_{BC}\mathbf {N_{BC}} +{\sqrt {(M_{BC}D)^{2}-m_{BC}^{2}}}\mathbf {N_{D}} \end{cases}}}
D
′
=
{
M
A
B
+
m
A
B
N
A
B
−
(
M
A
B
D
)
2
−
m
A
B
2
N
D
M
A
C
+
m
A
C
N
A
C
−
(
M
A
C
D
)
2
−
m
A
C
2
N
D
M
B
C
+
m
B
C
N
B
C
−
(
M
B
C
D
)
2
−
m
B
C
2
N
D
{\displaystyle \mathbf {D'} ={\begin{cases}\mathbf {M_{AB}} +m_{AB}\mathbf {N_{AB}} -{\sqrt {(M_{AB}D)^{2}-m_{AB}^{2}}}\mathbf {N_{D}} \\\mathbf {M_{AC}} +m_{AC}\mathbf {N_{AC}} -{\sqrt {(M_{AC}D)^{2}-m_{AC}^{2}}}\mathbf {N_{D}} \\\mathbf {M_{BC}} +m_{BC}\mathbf {N_{BC}} -{\sqrt {(M_{BC}D)^{2}-m_{BC}^{2}}}\mathbf {N_{D}} \end{cases}}}
where
M
A
B
D
=
A
D
1
−
[
(
A
D
)
2
+
(
A
B
)
2
−
(
B
D
)
2
(
A
B
)
(
A
D
)
]
2
{\displaystyle M_{AB}D=AD{\sqrt {1-\left[{\dfrac {(AD)^{2}+(AB)^{2}-(BD)^{2}}{(AB)(AD)}}\right]^{2}}}}
M
A
C
D
=
A
D
1
−
[
(
A
D
)
2
+
(
A
C
)
2
−
(
B
D
)
2
(
A
C
)
(
A
D
)
]
2
{\displaystyle M_{AC}D=AD{\sqrt {1-\left[{\dfrac {(AD)^{2}+(AC)^{2}-(BD)^{2}}{(AC)(AD)}}\right]^{2}}}}
M
B
C
D
=
B
D
1
−
[
(
B
D
)
2
+
(
B
C
)
2
−
(
C
D
)
2
(
B
C
)
(
B
D
)
]
2
{\displaystyle M_{BC}D=BD{\sqrt {1-\left[{\dfrac {(BD)^{2}+(BC)^{2}-(CD)^{2}}{(BC)(BD)}}\right]^{2}}}}
A
=
(
x
A
,
y
A
,
z
A
)
{\displaystyle \mathbf {A} =(x_{A},y_{A},z_{A})}
B
=
(
x
B
,
y
B
,
z
B
)
{\displaystyle \mathbf {B} =(x_{B},y_{B},z_{B})}
C
=
(
x
C
,
y
C
,
z
C
)
{\displaystyle \mathbf {C} =(x_{C},y_{C},z_{C})}
A
B
=
(
x
B
−
x
A
,
y
B
−
y
A
,
z
B
−
z
A
)
{\displaystyle \mathbf {AB} =(x_{B}-x_{A},y_{B}-y_{A},z_{B}-z_{A})}
A
C
=
(
x
C
−
x
A
,
y
C
−
y
A
,
z
C
−
z
A
)
{\displaystyle \mathbf {AC} =(x_{C}-x_{A},y_{C}-y_{A},z_{C}-z_{A})}
B
C
=
(
x
C
−
x
B
,
y
C
−
y
B
,
z
C
−
z
B
)
{\displaystyle \mathbf {BC} =(x_{C}-x_{B},y_{C}-y_{B},z_{C}-z_{B})}
A
C
∙
A
B
=
(
x
C
−
x
A
)
(
x
B
−
x
A
)
+
(
y
C
−
y
A
)
(
y
B
−
y
A
)
+
(
z
C
−
z
A
)
(
z
B
−
z
A
)
{\displaystyle \mathbf {AC} \bullet \mathbf {AB} =(x_{C}-x_{A})(x_{B}-x_{A})+(y_{C}-y_{A})(y_{B}-y_{A})+(z_{C}-z_{A})(z_{B}-z_{A})}
A
B
∙
A
C
=
(
x
C
−
x
A
)
(
x
B
−
x
A
)
+
(
y
C
−
y
A
)
(
y
B
−
y
A
)
+
(
z
C
−
z
A
)
(
z
B
−
z
A
)
{\displaystyle \mathbf {AB} \bullet \mathbf {AC} =(x_{C}-x_{A})(x_{B}-x_{A})+(y_{C}-y_{A})(y_{B}-y_{A})+(z_{C}-z_{A})(z_{B}-z_{A})}
A
B
∙
B
C
=
(
x
B
−
x
A
)
(
x
C
−
x
B
)
+
(
y
B
−
y
A
)
(
y
C
−
y
B
)
+
(
z
B
−
z
A
)
(
z
C
−
z
B
)
{\displaystyle \mathbf {AB} \bullet \mathbf {BC} =(x_{B}-x_{A})(x_{C}-x_{B})+(y_{B}-y_{A})(y_{C}-y_{B})+(z_{B}-z_{A})(z_{C}-z_{B})}
A
B
=
(
x
B
−
x
A
)
2
+
(
y
B
−
y
A
)
2
+
(
z
B
−
z
A
)
2
{\displaystyle AB={\sqrt {(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}+(z_{B}-z_{A})^{2}}}}
A
C
=
(
x
C
−
x
A
)
2
+
(
y
C
−
y
A
)
2
+
(
z
C
−
z
A
)
2
{\displaystyle AC={\sqrt {(x_{C}-x_{A})^{2}+(y_{C}-y_{A})^{2}+(z_{C}-z_{A})^{2}}}}
B
C
=
(
x
C
−
x
B
)
2
+
(
y
C
−
y
B
)
2
+
(
z
C
−
z
B
)
2
{\displaystyle BC={\sqrt {(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}+(z_{C}-z_{B})^{2}}}}
‖
A
C
−
A
C
∙
A
B
(
A
B
)
2
A
B
‖
=
(
A
C
)
2
−
(
A
C
∙
A
B
)
2
(
A
B
)
2
{\displaystyle \left\Vert {\mathbf {AC} -{\cfrac {\mathbf {AC} \bullet \mathbf {AB} }{(AB)^{2}}}\mathbf {AB} }\right\|={\sqrt {(AC)^{2}-{\cfrac {(\mathbf {AC} \bullet \mathbf {AB} )^{2}}{(AB)^{2}}}}}}
‖
A
B
−
A
B
∙
A
C
(
A
C
)
2
A
C
‖
=
(
A
B
)
2
−
(
A
B
∙
A
C
)
2
(
A
C
)
2
{\displaystyle \left\Vert {\mathbf {AB} -{\cfrac {\mathbf {AB} \bullet \mathbf {AC} }{(AC)^{2}}}\mathbf {AC} }\right\|={\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {AC} )^{2}}{(AC)^{2}}}}}}
‖
A
B
−
A
B
∙
B
C
(
B
C
)
2
B
C
‖
=
(
A
B
)
2
−
(
A
B
∙
B
C
)
2
(
B
C
)
2
{\displaystyle \left\Vert {\mathbf {AB} -{\cfrac {\mathbf {AB} \bullet \mathbf {BC} }{(BC)^{2}}}\mathbf {BC} }\right\|={\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {BC} )^{2}}{(BC)^{2}}}}}}
M
A
B
x
=
x
A
+
[
(
A
D
)
2
+
(
A
B
)
2
−
(
B
D
)
2
2
(
A
B
)
2
]
(
x
B
−
x
A
)
{\displaystyle {M_{ABx}}={x_{A}}+\left[{\dfrac {(AD)^{2}+(AB)^{2}-(BD)^{2}}{2(AB)^{2}}}\right](x_{B}-x_{A})}
M
A
B
y
=
y
A
+
[
(
A
D
)
2
+
(
A
B
)
2
−
(
B
D
)
2
2
(
A
B
)
2
]
(
y
B
−
y
A
)
{\displaystyle {M_{ABy}}={y_{A}}+\left[{\dfrac {(AD)^{2}+(AB)^{2}-(BD)^{2}}{2(AB)^{2}}}\right](y_{B}-y_{A})}
M
A
B
z
=
z
A
+
[
(
A
D
)
2
+
(
A
B
)
2
−
(
B
D
)
2
2
(
A
B
)
2
]
(
z
B
−
z
A
)
{\displaystyle {M_{ABz}}={z_{A}}+\left[{\dfrac {(AD)^{2}+(AB)^{2}-(BD)^{2}}{2(AB)^{2}}}\right](z_{B}-z_{A})}
M
A
C
x
=
x
A
+
[
(
A
D
)
2
+
(
A
C
)
2
−
(
C
D
)
2
2
(
A
C
)
2
]
(
x
C
−
x
A
)
{\displaystyle {M_{ACx}}={x_{A}}+\left[{\dfrac {(AD)^{2}+(AC)^{2}-(CD)^{2}}{2(AC)^{2}}}\right](x_{C}-x_{A})}
M
A
C
y
=
y
A
+
[
(
A
D
)
2
+
(
A
C
)
2
−
(
C
D
)
2
2
(
A
C
)
2
]
(
y
C
−
y
A
)
{\displaystyle {M_{ACy}}={y_{A}}+\left[{\dfrac {(AD)^{2}+(AC)^{2}-(CD)^{2}}{2(AC)^{2}}}\right](y_{C}-y_{A})}
M
A
C
z
=
z
A
+
[
(
A
D
)
2
+
(
A
C
)
2
−
(
C
D
)
2
2
(
A
C
)
2
]
(
z
C
−
z
A
)
{\displaystyle {M_{ACz}}={z_{A}}+\left[{\dfrac {(AD)^{2}+(AC)^{2}-(CD)^{2}}{2(AC)^{2}}}\right](z_{C}-z_{A})}
M
B
C
x
=
x
B
+
[
(
B
D
)
2
+
(
B
C
)
2
−
(
C
D
)
2
2
(
B
C
)
2
]
(
x
C
−
x
B
)
{\displaystyle {M_{BCx}}={x_{B}}+\left[{\dfrac {(BD)^{2}+(BC)^{2}-(CD)^{2}}{2(BC)^{2}}}\right](x_{C}-x_{B})}
M
B
C
y
=
y
B
+
[
(
B
D
)
2
+
(
B
C
)
2
−
(
C
D
)
2
2
(
B
C
)
2
]
(
y
C
−
y
B
)
{\displaystyle {M_{BCy}}={y_{B}}+\left[{\dfrac {(BD)^{2}+(BC)^{2}-(CD)^{2}}{2(BC)^{2}}}\right](y_{C}-y_{B})}
M
B
C
z
=
z
B
+
[
(
B
D
)
2
+
(
B
C
)
2
−
(
C
D
)
2
2
(
B
C
)
2
]
(
z
C
−
z
B
)
{\displaystyle {M_{BCz}}={z_{B}}+\left[{\dfrac {(BD)^{2}+(BC)^{2}-(CD)^{2}}{2(BC)^{2}}}\right](z_{C}-z_{B})}
N
A
B
x
=
(
x
C
−
x
A
)
−
A
C
∙
A
B
(
A
B
)
2
(
x
B
−
x
A
)
(
A
C
)
2
−
(
A
C
∙
A
B
)
2
(
A
B
)
2
{\displaystyle {N_{ABx}}={\cfrac {{(x_{C}-x_{A})}-{\cfrac {\mathbf {AC} \bullet \mathbf {AB} }{(AB)^{2}}}(x_{B}-x_{A})}{\sqrt {(AC)^{2}-{\cfrac {(\mathbf {AC} \bullet \mathbf {AB} )^{2}}{(AB)^{2}}}}}}}
N
A
B
y
=
(
y
C
−
y
A
)
−
A
C
∙
A
B
(
A
B
)
2
(
y
B
−
y
A
)
(
A
C
)
2
−
(
A
C
∙
A
B
)
2
(
A
B
)
2
{\displaystyle {N_{ABy}}={\cfrac {{(y_{C}-y_{A})}-{\cfrac {\mathbf {AC} \bullet \mathbf {AB} }{(AB)^{2}}}(y_{B}-y_{A})}{\sqrt {(AC)^{2}-{\cfrac {(\mathbf {AC} \bullet \mathbf {AB} )^{2}}{(AB)^{2}}}}}}}
N
A
B
z
=
(
z
C
−
z
A
)
−
A
C
∙
A
B
(
A
B
)
2
(
z
B
−
z
A
)
(
A
C
)
2
−
(
A
C
∙
A
B
)
2
(
A
B
)
2
{\displaystyle {N_{ABz}}={\cfrac {{(z_{C}-z_{A})}-{\cfrac {\mathbf {AC} \bullet \mathbf {AB} }{(AB)^{2}}}(z_{B}-z_{A})}{\sqrt {(AC)^{2}-{\cfrac {(\mathbf {AC} \bullet \mathbf {AB} )^{2}}{(AB)^{2}}}}}}}
N
A
C
x
=
(
x
B
−
x
A
)
−
A
B
∙
A
C
(
A
C
)
2
(
x
C
−
x
A
)
(
A
B
)
2
−
(
A
B
∙
A
C
)
2
(
A
C
)
2
{\displaystyle {N_{ACx}}={\cfrac {{(x_{B}-x_{A})}-{\cfrac {\mathbf {AB} \bullet \mathbf {AC} }{(AC)^{2}}}(x_{C}-x_{A})}{\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {AC} )^{2}}{(AC)^{2}}}}}}}
N
A
C
y
=
(
y
B
−
y
A
)
−
A
B
∙
A
C
(
A
C
)
2
(
y
C
−
y
A
)
(
A
B
)
2
−
(
A
B
∙
A
C
)
2
(
A
C
)
2
{\displaystyle {N_{ACy}}={\cfrac {{(y_{B}-y_{A})}-{\cfrac {\mathbf {AB} \bullet \mathbf {AC} }{(AC)^{2}}}(y_{C}-y_{A})}{\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {AC} )^{2}}{(AC)^{2}}}}}}}
N
A
C
z
=
(
z
B
−
z
A
)
−
A
B
∙
A
C
(
A
C
)
2
(
z
C
−
z
A
)
(
A
B
)
2
−
(
A
B
∙
A
C
)
2
(
A
C
)
2
{\displaystyle {N_{ACz}}={\cfrac {{(z_{B}-z_{A})}-{\cfrac {\mathbf {AB} \bullet \mathbf {AC} }{(AC)^{2}}}(z_{C}-z_{A})}{\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {AC} )^{2}}{(AC)^{2}}}}}}}
N
B
C
x
=
(
x
B
−
x
A
)
−
A
B
∙
B
C
(
B
C
)
2
(
x
C
−
x
B
)
(
A
B
)
2
−
(
A
B
∙
B
C
)
2
(
B
C
)
2
{\displaystyle {N_{BCx}}={\cfrac {{(x_{B}-x_{A})}-{\cfrac {\mathbf {AB} \bullet \mathbf {BC} }{(BC)^{2}}}(x_{C}-x_{B})}{\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {BC} )^{2}}{(BC)^{2}}}}}}}
N
B
C
y
=
(
y
B
−
y
A
)
−
A
B
∙
B
C
(
B
C
)
2
(
y
C
−
y
B
)
(
A
B
)
2
−
(
A
B
∙
B
C
)
2
(
B
C
)
2
{\displaystyle {N_{BCy}}={\cfrac {{(y_{B}-y_{A})}-{\cfrac {\mathbf {AB} \bullet \mathbf {BC} }{(BC)^{2}}}(y_{C}-y_{B})}{\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {BC} )^{2}}{(BC)^{2}}}}}}}
N
B
C
z
=
(
z
B
−
z
A
)
−
A
B
∙
B
C
(
B
C
)
2
(
z
C
−
z
B
)
(
A
B
)
2
−
(
A
B
∙
B
C
)
2
(
B
C
)
2
{\displaystyle {N_{BCz}}={\cfrac {{(z_{B}-z_{A})}-{\cfrac {\mathbf {AB} \bullet \mathbf {BC} }{(BC)^{2}}}(z_{C}-z_{B})}{\sqrt {(AB)^{2}-{\cfrac {(\mathbf {AB} \bullet \mathbf {BC} )^{2}}{(BC)^{2}}}}}}}
x
−
(
M
A
B
x
+
m
A
B
N
A
B
x
)
(
y
C
−
y
A
)
(
z
B
−
z
A
)
−
(
y
B
−
y
A
)
(
z
C
−
z
A
)
=
y
−
(
M
A
B
y
+
m
A
B
N
A
B
y
)
(
x
B
−
x
A
)
(
z
C
−
z
A
)
−
(
x
C
−
x
A
)
(
z
B
−
z
A
)
=
z
−
(
M
A
B
z
+
m
A
B
N
A
B
z
)
(
x
C
−
x
A
)
(
y
B
−
y
A
)
−
(
x
B
−
x
A
)
(
y
C
−
y
A
)
{\displaystyle {\cfrac {x-(M_{ABx}+m_{AB}N_{ABx})}{(y_{C}-y_{A})(z_{B}-z_{A})-(y_{B}-y_{A})(z_{C}-z_{A})}}={\cfrac {y-(M_{ABy}+m_{AB}N_{ABy})}{(x_{B}-x_{A})(z_{C}-z_{A})-(x_{C}-x_{A})(z_{B}-z_{A})}}={\cfrac {z-(M_{ABz}+m_{AB}N_{ABz})}{(x_{C}-x_{A})(y_{B}-y_{A})-(x_{B}-x_{A})(y_{C}-y_{A})}}}
A
C
×
A
B
=
|
i
j
k
x
C
−
x
A
y
C
−
y
A
z
C
−
z
A
x
B
−
x
A
y
B
−
y
A
z
B
−
z
A
|
{\displaystyle \mathbf {AC} \times \mathbf {AB} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\x_{C}-x_{A}&y_{C}-y_{A}&z_{C}-z_{A}\\x_{B}-x_{A}&y_{B}-y_{A}&z_{B}-z_{A}\\\end{vmatrix}}}
=
(
(
y
C
−
y
A
)
(
z
B
−
z
A
)
−
(
y
B
−
y
A
)
(
z
C
−
z
A
)
,
(
x
B
−
x
A
)
(
z
C
−
z
A
)
−
(
x
C
−
x
A
)
(
z
B
−
z
A
)
,
(
x
C
−
x
A
)
(
y
B
−
y
A
)
−
(
x
B
−
x
A
)
(
y
C
−
y
A
)
)
{\displaystyle ={\Big (}(y_{C}-y_{A})(z_{B}-z_{A})-(y_{B}-y_{A})(z_{C}-z_{A}),\ (x_{B}-x_{A})(z_{C}-z_{A})-(x_{C}-x_{A})(z_{B}-z_{A}),\ (x_{C}-x_{A})(y_{B}-y_{A})-(x_{B}-x_{A})(y_{C}-y_{A}){\Big )}}
‖
A
C
×
A
B
‖
=
{
(
(
y
C
−
y
A
)
(
z
B
−
z
A
)
−
(
y
B
−
y
A
)
(
z
C
−
z
A
)
)
2
+
(
(
x
B
−
x
A
)
(
z
C
−
z
A
)
−
(
x
C
−
x
A
)
(
z
B
−
z
A
)
)
2
+
(
(
x
C
−
x
A
)
(
y
B
−
y
A
)
−
(
x
B
−
x
A
)
(
y
C
−
y
A
)
)
2
}
1
2
{\displaystyle \left\Vert {\mathbf {AC} \times \mathbf {AB} }\right\|={\begin{Bmatrix}{\Big (}(y_{C}-y_{A})(z_{B}-z_{A})-(y_{B}-y_{A})(z_{C}-z_{A}){\Big )}^{2}+\\{\Big (}(x_{B}-x_{A})(z_{C}-z_{A})-(x_{C}-x_{A})(z_{B}-z_{A}){\Big )}^{2}+\\{\Big (}(x_{C}-x_{A})(y_{B}-y_{A})-(x_{B}-x_{A})(y_{C}-y_{A}){\Big )}^{2}\end{Bmatrix}}^{\frac {1}{2}}}
Discussion
While the same result can be obtained using only a few of these many equations, all possibilities are treated here. If the three spheres do not intersect, the least squares solution finds the axis of symmetry of the three spheres, or the closest solution. If the three spheres do intersect, the solution requires only a few of these equations.
See also
References
^ Borisenko, A. I. and Tarapov, I. E., (1968) "Vector and Tensor Analysis", General Publishing Company, p. 6. ISBN 0-486-63833-2
^ Leon, Steven J. (1980) "Linear Algebra", Macmillan Publishing Co., Inc., p. 152. ISBN 0-02-369870-5
External links 
![{\displaystyle \mathbf {M_{AB}} =\mathbf {A} +AD\cos(\angle {BAD}){\dfrac {\mathbf {AB} }{\left\Vert \mathbf {AB} \right\|}}=\mathbf {A} +\left[{\dfrac {(AD)^{2}+(AB)^{2}-(BD)^{2}}{2(AB)^{2}}}\right]\mathbf {AB} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/96d3082698db1f0f34629fa8a93aeab34eedcbc2)
![{\displaystyle \mathbf {M_{AC}} =\mathbf {A} +AD\cos(\angle {CAD}){\dfrac {\mathbf {AC} }{\left\Vert \mathbf {AC} \right\|}}=\mathbf {A} +\left[{\dfrac {(AD)^{2}+(AC)^{2}-(CD)^{2}}{2(AC)^{2}}}\right]\mathbf {AC} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/441d27a21335cb1c074b4260ab536abbe1cf813f)
![{\displaystyle \mathbf {M_{BC}} =\mathbf {B} +BD\cos(\angle {CBD}){\dfrac {\mathbf {BC} }{\left\Vert \mathbf {BC} \right\|}}=\mathbf {B} +\left[{\dfrac {(BD)^{2}+(BC)^{2}-(CD)^{2}}{2(BC)^{2}}}\right]\mathbf {BC} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f615e0dadd7bb7242f30109894b0d9d315d3762d)
![{\displaystyle M_{AB}D=AD{\sqrt {1-\left[{\dfrac {(AD)^{2}+(AB)^{2}-(BD)^{2}}{(AB)(AD)}}\right]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3457db7cb91c4c42df7f2f65256298554b9e131)
![{\displaystyle M_{AC}D=AD{\sqrt {1-\left[{\dfrac {(AD)^{2}+(AC)^{2}-(BD)^{2}}{(AC)(AD)}}\right]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/881f83cbf76e67dea714738c52c7b75596e09aa3)
![{\displaystyle M_{BC}D=BD{\sqrt {1-\left[{\dfrac {(BD)^{2}+(BC)^{2}-(CD)^{2}}{(BC)(BD)}}\right]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd18d160223ac09ff339951f42fee56ec0e44318)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cfda9bd50e26725fcb40ba45ce5cafda1b0bae6)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b2f40346675b4efcd08efc0c6cb12ae55958c35)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af2eeea234e9aa404500c344161dc7edc8559372)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6249fdd095b07b8b938a011fe1117ab6d7bfbedd)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90a2c0e1c6c4e2a8639d6ff5ef65171da9fd3243)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1625499fcee3d2954d001ba30d9f9b86a580a1ac)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/782d9bd2849702e0c2e61e81f8d673aa715885c6)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b84e46d5a6e2ccd7f2ea00befbba93a916a4ddab)
}](https://wikimedia.org/api/rest_v1/media/math/render/svg/194b6d671cb8ed5729bf3912d259c53cd8a92cba)
