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In vector analysis , triangulation is a method of finding points in three dimensional spaces using distances, angles and vector functions such as magnitude , dot product and cross product . Among its uses are in surveying , navigation , and astronomy .
This article describes a method for determining the coordinates of the point where three lines meet, given the scalar lengths of the lines and the coordinates of their bases. If these three lines are the radii of three spheres of known centers, this method can 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
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.
File:Triangulation illust 02.gif 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})}
[ 1] AD onto AB and AC , and the projection of BD onto BC results in,
File:FacesABD ACD BCD 2.gif
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} }
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 arranged as,
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
−
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}}}
Example
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)
