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0593_C02_fm  Page 28  Monday, May 6, 2002  1:46 PM





                       28                                                  Dynamics of Mechanical Systems


                       Hence,

                                                           A = 45                              (2.6.29)


                       Example 2.6.3: Angle Between Two Vectors
                       Let vectors A and B be expressed in terms of unit vectors n , n , and n  as:
                                                                               2
                                                                                      3
                                                                            1
                                                 −
                                           A = 3 n n + 4 n     and    B = −5 n + 2 n + 7 n     (2.6.30)
                                                    2    3             1    2    3
                       Determine the angle between A and B.
                        Solution: From Eq. (2.6.25), cosθ is:
                                                   ⋅
                                                        3
                                                           5
                                                                       4
                                                                 1 2
                                          cosθ=  AB   =  () − ( ) +− ( )( ) + ()( ) 7  =  . 0 244  (2.6.31)
                                                 AB            26 78
                       Then, θ is:
                                                   θ= cos −1 ( .0 244 ) = 75 .86 deg           (2.6.32)


                       Example 2.6.4: Projection of a Vector Along a Line
                       Consider a velocity vector V and points P and Q of a Cartesian coordinate system R as
                       shown in Figure 2.6.6. Let V be expressed in terms of mutually perpendicular unit vectors
                       i, j, and k as:

                                                     V = 3 i + 4j + 4 kft sec                  (2.6.33)
                       Let the coordinates of P and Q be as shown and let L be a line passing through P and Q.
                       Determine the projection of V along L.
                        Solution: From Eq. (2.6.9), the projection of V along L is simply where n is a unit vector
                       along L. Because L passes through P and Q, n is:

                                                               i (
                                                n =  PQ PQ = − + 5 j− 2 k)  29                 (2.6.34)
                       Therefore, the projection of V along (or onto) L is:

                                                             i (
                                           ⋅
                                          vn = (3 i + 4 j + 4 k) ⋅− + 5 j − 2 k)  29
                                                                                               (2.6.35)
                                              = () − ( ) + ()( ) + () − ( )]  29  = 1 67 ft sec
                                                                  2
                                                                            .
                                                               4
                                                    1
                                               [ 3
                                                        4 5
                       2.7  Vector Multiplication: Vector Product

                       Next, consider the “vector” product, so called because the result is a vector. Given any
                       vectors A and B, the vector product, written A × B, is defined as:

                                                      AB =   A B sin   n                        (2.7.1)
                                                        ×
                                                                   θ
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