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Triangulation in three dimensions is a method of finding the ___location of a point in three dimensions based on other known coordinates and distances, it is commonly used in surveying and astronomy .
Triangulation is also used in 2 dimensions to find the ___location of a point on a plane , this is commonly used in navigation to plot positions on a map.
One method to triangulate a ___location in 3D
This method uses vector analysis to determine the coordinates of the point where three lines meet given the scalar lengths of the lines and the coordinates of their bases. First treat these three lines as if they are the radii of three spheres of known centers (these known centres being the coordinates of the known end of each line), this method can then be used to calculate the intersection of the three spheres if they intersect . In the event that the three spheres don't intersect, this method obtains the closest solution to the axis of symmetry between three spheres.
Development
File:Triangulation illust 02.gif Figure 1. The apex and its mirror reflection about the plane of ABC precipitate D and D'. Three sticks of known lengths AD, BD, CD are anchored in the ground at known coordinates A, B, C . This development calculates the coordinates of the apex where the other ends of the three sticks will meet. These coordinates are given by the vector D . In the mirror case, D' is sub-apex where the three sticks would meet below the plane of A, B, C as well.
law of cosines ,
(
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})}
File:FacesABD ACD BCD 2.gif Figure 2. The normals are dropped on the sides from the apex and their intersections with AB, AC and BC are determined. [ 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} }
Figure 3. The red normals intersect at a common point.
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}}}}
Condition for intersection
If AD, BD, CD are assigned according to the arrangement,
A
D
≤
B
D
≤
C
D
{\displaystyle AD\leq BD\leq CD}
Then AD, BD, CD intersect if and only if,
A
D
+
B
D
≥
A
B
≥
B
D
−
A
D
{\displaystyle AD+BD\geq AB\geq BD-AD}
A
D
+
C
D
≥
A
C
≥
C
D
−
A
D
{\displaystyle AD+CD\geq AC\geq CD-AD}
B
D
+
C
D
≥
B
C
≥
C
D
−
B
D
{\displaystyle BD+CD\geq BC\geq CD-BD}
AD =rA =radius of sphere centered at A ,BD =rB =radius of sphere centered at B , and CD =rC =radius of sphere centered at C ,
r
A
≤
r
B
≤
r
C
{\displaystyle r_{A}\leq r_{B}\leq r_{C}}
then the three spheres intersect if and only if,
r
A
+
r
B
≥
A
B
≥
r
B
−
r
A
{\displaystyle r_{A}+r_{B}\geq AB\geq r_{B}-r_{A}}
r
A
+
r
C
≥
A
C
≥
r
C
−
r
A
{\displaystyle r_{A}+r_{C}\geq AC\geq r_{C}-r_{A}}
r
B
+
r
C
≥
B
C
≥
r
C
−
r
B
{\displaystyle r_{B}+r_{C}\geq BC\geq r_{C}-r_{B}}
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
=
(
M
A
B
x
,
M
A
B
y
,
M
A
B
z
)
{\displaystyle \mathbf {M_{AB}} =(M_{ABx},M_{ABy},M_{ABz})}
M
A
C
=
(
M
A
C
x
,
M
A
C
y
,
M
A
C
z
)
{\displaystyle \mathbf {M_{AC}} =(M_{ACx},M_{ACy},M_{ACz})}
M
B
C
=
(
M
B
C
x
,
M
B
C
y
,
M
B
C
z
)
{\displaystyle \mathbf {M_{BC}} =(M_{BCx},M_{BCy},M_{BCz})}
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
=
(
N
A
B
x
,
N
A
B
y
,
N
A
B
z
)
{\displaystyle \mathbf {N_{AB}} =(N_{ABx},N_{ABy},N_{ABz})}
N
A
C
=
(
N
A
C
x
,
N
A
C
y
,
N
A
C
z
)
{\displaystyle \mathbf {N_{AC}} =(N_{ACx},N_{ACy},N_{ACz})}
N
B
C
=
(
N
B
C
x
,
N
B
C
y
,
N
B
C
z
)
{\displaystyle \mathbf {N_{BC}} =(N_{BCx},N_{BCy},N_{BCz})}
N
A
B
x
=
(
x
C
−
x
A
)
−
A
C
∙
A
B
(
A
B
)
2
(
x
B
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x
A
)
(
A
C
)
2
−
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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
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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
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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
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2
(
z
C
−
z
A
)
(
A
B
)
2
−
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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
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−
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
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B
−
y
A
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−
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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}}}
Example
File:3 spheres2abcplane.gif Figure 4. The data is input into the equations to obtain the solution. The shaded area is the plane of ABC. A, B and C are the centers of each of the three spheres.
File:3 spheres2ABCcenters.gif Figure 5. Showing points D and D' as the result of 3 intersecting spheres at centers A, B, C from data given in Figure 4. The line adjoining D and D' imbedded in the interiors of all three spheres is the axis of symmetry of the three spheres.
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)
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