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0593_C02_fm  Page 36  Monday, May 6, 2002  1:46 PM





                       36                                                  Dynamics of Mechanical Systems


                       the algorithms of Eqs. (2.8.1) and (2.8.2), we see that the vector triple products may be
                       expressed as:

                                                 A ×( B C) = ( A C B ) −( A B C )              (2.8.12)
                                                                       ⋅
                                                              ⋅
                                                       ⋅
                       and
                                                 ( AB) × C = ( A C B ) −( B C A )              (2.8.13)
                                                               ⋅
                                                   ×
                                                                       ⋅
                       Observe that the last terms in these expressions are different.


                       Example 2.8.2: Validity of Eqs. (2.8.6) and (2.8.7) and the Necessity of
                       Parentheses
                       Verify Eqs. (2.8.6) and (2.8.7) using the vectors of Eq. (2.8.7).
                        Solution: From Eqs. (2.8.2) and (2.8.9), A × (B × C) is:

                                                    n 1  n  2  n  3
                                         A ×( B C) = 3   −1    1  = −27 n − 29 n + 52 n 3      (2.8.14)
                                               ×
                                                                       1
                                                                             2
                                                     1   17   10
                        From Eq. (2.6.22), (A · C)B – (A · B)C is:
                                   ( AC B ) −( A C C ) = () − ( ) +− ( )( ) + () − ( )](2 n + 4 n − 7 n )
                                              ⋅
                                     ⋅
                                                                        5
                                                    [ 3
                                                                     1
                                                         1
                                                              1 3
                                                                             1
                                                                                  2
                                                                                       3
                                                    − ()() +− ( )( ) + () − ( )] − (  n + 3 n − 5 n )
                                                      [ 32
                                                               1 4
                                                                     1
                                                                         7
                                                                                        3
                                                                                   2
                                                                              1
                                                        )(           n )
                                                   =− ( 11 2 n + 4 n − 7                       (2.8.15)
                                                            1    2    3
                                                    −− ( ) − (  n + 3n − 5n )
                                                        5
                                                               3
                                                             1    2    3
                                                   =− 27n −  29n +  52n
                                                         1     2     3
                       Similarly, (A × B) × C and (A · C)B – (B · C)A are:
                                                     n 1  n 2  n 3
                                         ( AB) ×                       n +
                                            ×
                                                               14
                                                 C = 3
                                                          23
                                                                 =−157
                                                                            2
                                                                         1  n + 32 n 3         (2.8.16)
                                                     −1   3    −5
                       and
                                   ( AC B ) −( B C A )  = () − ( ) +− ( )( ) + () − ( )](2 n + 4 n − 7 n )
                                              ⋅
                                      ⋅
                                                                        5
                                                                     1
                                                         1
                                                    [ 3
                                                              1 3
                                                                                        3
                                                                             1
                                                                                  2
                                                                           5
                                                       − () − ( ) + ()( ) +− ( ) − ( )](3 n − n + )
                                                                                       n
                                                      [ 2
                                                           1
                                                                       7
                                                               4 3
                                                                                        3
                                                                                1
                                                                                    2
                                                        )(           n )
                                                   =− ( 11 2 n + 4 n − 7                       (2.8.17)
                                                            1    2    3
                                                       45
                                                       −( ) 3 (  n − n + )
                                                                    n
                                                                n
                                                             1   2   3
                                                   =− 157 n + n +  32 n
                                                          1   2     3
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