VECTOR TRIPLE PRODUCT
F. Mujika
Department of Mechanical Engineering
Polytechnic School
University of the Basque Country
Plaza de Europa, 20018 San Sebastián/Donostia, Spain
e-mail: faustino.mujika@ehu.es
1
VECTOR TRIPLE PRODUCT
The vector triple product is given by:
r
( r r)
r
p = a × b ×c
(1)
(
)
r
r r
r
Assuming that b and c are in the plane π, the vector b × c is perpendicular with
r
respect to this plane, as shown in Figure 1. The vector a can be decompossed in a
component contained in the plane π and a component perpendicular to it, being:
r
r
r
a = aπ + an
(2)
According to the distributive property of the cross product it results:
r
( r r)
r
( r r)
r
p = a × b × c = aπ × b × c
(3)
(br × cr )
r
a
r
an
r
aπ
r
( r r)
aπ × b × c
r
c
r
b
π
Figure 1. Vector triple product. Components of vectors contained in the plane π and perpendicular with
respect to it.
Thus, the analysis can be carried out in the plane π, as shown in Figure 2. As the
double product is contained in the plane π, it can be decomposed according to the
r
r
directions of b and c :
r
r
r
p = λbu b + λcuc
(4)
r
r
Where u b and uc are the following unit vectors:
r
b
ub =
b
r
r
c
ub =
c
r
(5)
1
r
aπ
θab
θac
(
r r
r r
p = aπ × b × c
(b × c )
r
r
)
θpc
θ
r
c
θpb
r
b
Figure 2. Components of vectors contained in the plane π definition of angles.
After multiplying scalarly Equation (4) by the unit vectors of Equation (5) it results:
r r
r
r
p ⋅ u b = λb + λc (uc ⋅ u b )
r r
r
r
p ⋅ uc = λb (u b ⋅ uc ) + λc
(6)
(7)
According to the angles and vectors defined in Figure 2, Equations (6) and (7) can
be written as
p cos θ pb = λb + λc cos θ
r
(8)
r
(9)
p cos θ pc = λb cos θ + λc
By multiplying Equation (9) by cosθ and substracting it from Equation (8):
p (cos θ pb − cos θ pc cos θ ) = λb sin2 θ
r
(
(10)
)
r
Taking into account that θ pb = θ pc + θ and p = aπ bc sin θ , Equation (10) is:
λb = −aπ bc sin θ pc
(11)
π⎞
⎛
According to Figure 2, θ pc = ⎜θac − ⎟ . Replacing int Equation (11) it results that:
2⎠
⎝
λb = baπ c cos θac
(12)
Otherwise, according to the distributive property of the scalar product:
r r
r
r
r
r
r
(13)
r r
(14)
a ⋅ c = (aπ + an ) ⋅ c = aπ ⋅ c = aπ c cos θac
Combining Equations (12) and (13):
λb = b (a ⋅ c )
2
In order to obtain λc, by multiplying Equation (8) by cosθ and substracting it from
Equation (9):
p (cos θ pc − cos θ pb cos θ ) = λc sin2 θ
r
(
(15)
)
Taking into account that θ pc = θ pb − θ , Equation (15) becomes:
λc = aπ bc sin θ pb
(16)
π⎞
⎛
According to Figure 2, θ pb = ⎜θab − ⎟ . Replacing in Equation (16) it results:
2⎠
⎝
λc = −caπ b cos θab
(17)
In a similar manner than in Equation (13) it results that:
r r
a ⋅ b = aπ b cos θab
(18)
By comparing Equations (17) and (18):
(r r )
λc = −c a ⋅ b
(19)
Combining Equations (4), (5), (14) and (20) it is obtained:
r
r
( r r)
( )
r r r
r r r
p = a × b × c = b (a ⋅ c ) − c a ⋅ b
(20)
The e-δ IDENTITY
r
r
r
Being e1 , e2 , e3 the unit vectors of a orthonormal basis, according to the definition
of scalar and vector product it results:
r
r
e i ⋅ e j = δ ij
r
r
(21)
r
e i × e j = e ijk e k
Where δij is the Kronecker delta and eijk the e-permutation symbol. The vector triple
product of the unit vectos according to Equation (20) is:
e n × (e i × e j ) = e i (e n ⋅ e j ) − e j (e n ⋅ e i )
r
r
r
r r
r
r
r
r
(22)
According to Equations (21), Equation (22) becomes:
r
r
r
r
e ijk (e n × e k ) = e i δ nj − e j δ ni
r
r
r
e ijk e nkpe p = e i δ nj − e j δ ni
(23)
r
By multiplying scalarly Equation (23) by e m :
3
e ijk e nkp δ mp = δ im δ nj − δ jm δ ni
(24)
Operating in the left member of Equation (24) the e-δ identity is:
e ijk e mnk = δ im δ jn − δ in δ jm
(25)
4