6.4. THE CROSS PRODUCT 87For @ a scalar,(aa) xb= a(ax b) = ax (ab), (6.23)For a,b, and c vectors, one obtains the distributive laws,ax (b+c)=axb+axe, (6.24)(b+c)xa=bxa+cxa. (6.25)Formula 6.22 follows immediately from the definition. The vectors a x b and bxahave the same magnitude, |a| |b| sin 8, and an application of the right hand rule shows theyhave opposite direction. Formula 6.23 is also fairly clear. If a is a nonnegative scalar, thedirection of (@a) xb is the same as the direction of a x b,a(a@ x b) and ax (ab) whilethe magnitude is just @ times the magnitude of a x b which is the same as the magnitudeof a(a x b) and ax (ab). Using this yields equality in 6.23. In the case where a < 0,everything works the same way except the vectors are all pointing in the opposite directionand you must multiply by |a@| when comparing their magnitudes. The distributive laws aremuch harder to establish but the second follows from the first quite easily. Thus, assumingthe first, and using 6.22,(b+c) xa=-—ax(b+c)=-(axb+axc)=bxa+cxa.A proof of the distributive law is given in a later section for those who are interested.Now from the definition of the cross product,axjo=k, jxi=-kkxt=j, txk=-Jjxk=i, kxj=-iWith this information, the following gives the coordinate description of the cross product.Proposition 6.4.3 Let a = ajt+a2j +. a3k and b= bit +b27 + b3k be two vectors. Thena x b= (agb3 — a3b2) t+ (a3b, — a, b3) 7+ (ayb2 — anb)) k. (6.26)Proof: From the above table and the properties of the cross product listed,(a\t+arj +a3k) x (bit +boj + b3k) =abot xX j +ayb3t x k+azgbyj x i+agb3j x k++a3b\k x t+a3bok xj= aj byk — a, b39 — anb\ k + anb3t + a3b,j — azb2t= (aab3 —a3b7) t+ (a3b) — ayb3) j+ (aib2 — azb,) k (6.27)|It is probably impossible for most people to remember 6.26. Fortunately, there is asomewhat easier way to remember it.i jkaxb=|)a @ (6.28)by by by