Page 53 - Dynamics of Mechanical Systems

0593_C02_fm Page 34 Monday, May 6, 2002 1:46 PM

34 Dynamics of Mechanical Systems

Given three vectors A, B, and C, the scalar triple product has the form A · (B × C). The
result is a scalar. The scalar triple product is seen to be a combination of a vector product
and a scalar product.
Recall from Eqs. (2.6.21) and (2.7.23) that if n , n , and n are mutually perpendicular
2
1
3
unit vectors, the algorithms for evaluating the scalar and vector products are:
⋅
AB = A B + AB + AB (2.8.1)
1 1 2 2 3 3
and

n n n
1 2 3
AB = A A A (2.8.2)
×
1 2 3
B B B
1 2 3
where the A and B (i = 1, 2, 3) are the n components of A and B.
i
i
i
By comparing Eqs. (2.8.1) and (2.8.2), we see that the scalar triple product A · (B × C)
may be obtained by replacing n , n , and n in:
1
2
3
A 1 A 2 A 3
⋅(
×
AB C) = B B B (2.8.3)
1 2 3
C C C
1 2 3
Recall from the elementary rules of evaluating determinants that the rows and columns
may be interchanged without changing the value of the determinant. Also, the rows and
columns may be cyclically permuted without changing the determinant value. If the rows
or columns are anticyclically permuted, the value of the determinant changes sign. Hence,
we can rewrite Eq. (2.8.3) in the forms:

A 1 A 2 A 3 B 1 B 2 B 3 C 1 C 2 C 3
⋅(
×
AB C) = B 1 B 2 B 3 = C 1 C 2 C 3 = A 1 A 2 A mp
3
C 1 C 2 C 3 A 1 A 2 A 3 B 1 B 2 B 3
(2.8.4)
B 1 B 2 B 3 A 1 A 2 A 3 C 1 C 2 C 3
=− A 1 A 2 A 3 =− C 1 C 2 C 3 =− B 1 B 2 B B 3
C 1 C 2 C 3 B 1 B 2 B 3 A 1 A 2 A 3

By examining Eq. (2.8.4), we see that the dot and the cross may be interchanged in the
product. Also, the parentheses are unnecessary. Finally, the vectors may be cyclically
permuted without changing the value of the product. An anticyclic permutation changes
the sign of the result. Specifically,

⋅(
×
⋅
×
×
A B C) = A B C = AB C = C AB
×
⋅
⋅
(2.8.5)
⋅
×
×
×
⋅
= BC A = − B A C = − C B A = − A C B
⋅
×
⋅