Tham khảo tài liệu 'dynamics of mechanical systems 2009 part 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 32 Dynamics of Mechanical Systems FIGURE Vectors A D nA DII and D . Returning to the product A X D let A and D be depicted as in Figure where 0 is the angle between the vectors. As before let nA be a unit vector parallel to A. Then A nA X D is a vector perpendicular to nA and with magnitude I D II sin0 . From Figure we see that Dll sin 0 DJ By similar reasoning we have A X B A X B and A X C A X C Therefore by comparing Eqs. and we have A X D A x B C A X B A X C This establishes the distributive law. Finally suppose that n1 n2 and n3 are mutually perpendicular unit vectors and suppose that vectors A and B are expressed in the forms A A1n1 A2n2 A3n3 and B B1n1 B2n2 B3n 3 Then by repeated use of Eqs. and we see that A X B may be expressed as A X B A2B3 - A3B2 n1 A3B1 - A3B1 n2 AB2 - A2B1 n3 333 -ÍÍÊ i 1 j 1 k 1 By recalling the elementary rules for expanding determinants we see that Eq. may be written as n1 n2 n3 A X B Ai A2 A3 B1 B2 B3 This is a useful algorithm for computing the vector product. Review of Vector Algebra 33 Example Vector Product Computation and Geometric Properties of the Vector Product Let vectors A and B be expressed in terms of mutually perpendicular dextral unit vectors n1 n2 and n3 as A 7n1 - 2n2 4n3 and B n1 3n2 - 8n3 Let C be the vector product A X B. a. Find C. b. Show that C is perpendicular to both A and B. c. Show that B X A -C. Solution a. From Eq. C is C n1 n2 n3 7 -2 4 1 3 -8 4n1 60n2 23n3 b. If C is perpendicular to A with the angle 0 between C and A being 90 C A is zero because cos0 is zero. Conversely if C A is zero and neither C nor A is zero then cos0 is zero making C perpendicular to A. From Eq. C A is C A 4 7 60 -2 23 4 0 Similarly C B is C B 4 1 60 3 23 -8 0 c. From Eq. B X A is n1 n2 n3 B X A 1 3 -8 -4n1 - 60n2 - 23n3 7 -2 4 which is seen to be from Eq. .
