This article
needs attention from an expert on the subject . Please add a
reason or a
talk parameter to this template to explain the issue with the article.
When placing this tag, consider associating this request with a WikiProject . (April 2011 )
Development
Suppose three sticks of known length are anchored in the ground at known coordinates. 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 . D' is the vector to the coordinates of the apex where the three sticks would meet below the plane of A, B, C as well. This problem also calculates the axis of symmetry of three intersecting spheres. See the article Trilateration for the spherical case. Even if the three spheres don't intersect, this problem still calculates their axis of symmetry. This axis is the line drawn between D and D' . To proceed, 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 lengths AD, BD, CD are planted in the ground at known coordinates A, B, C . 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} }
File:FacesABD ACD BCD.gif
(
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}} }
File:Intersect normals ABC.gif 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
=
(
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
−
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
This is a robust solution incorporating all possibilities and accommodating non-intersecting cases with the closest answer using the least squares calculation of matrix H and b .
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)
