Kenneth Kuttler

294 CHAPTER 14. VECTOR PRODUCTS

For a,b, and c vectors, one obtains the distributive laws,

a×(b+c) = a×b+a×c, (14.17)

(b+c)×a= b×a+c×a. (14.18)

Formula 14.15 follows immediately from the definition. The vectors a×b and b×ahave the same magnitude, |a| |b|sinθ , and an application of the right hand rule shows theyhave opposite direction. Formula 14.16 is also fairly clear. If α is a nonnegative scalar, thedirection of (αa)×b is the same as the direction of a×b,α (a×b) and a×(αb) whilethe magnitude is just α times the magnitude of a×b which is the same as the magnitudeof α (a×b) and a×(αb) . Using this yields equality in 14.16. In the case where α < 0,everything works the same way except the vectors are all pointing in the opposite directionand you must multiply by |α| 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 14.15,

(b+c)×a=−a×(b+c) =−(a×b+a×c) = b×a+c×a.

A proof of the distributive law is given later.Now from the definition of the cross product,

i×j = k, j× i=−kk× i= j, i×k=−jj×k= i, k×j =−i

With this information, the following gives the coordinate description of the cross product.

Proposition 14.4.3 Let a= a1i+a2j+a3k and b= b1i+b2j+b3k be two vectors.Then

a×b= (a2b3 −a3b2) i+ (a3b1 −a1b3)j+ (a1b2 −a2b1)k. (14.19)

Proof: From the above table and the properties of the cross product listed,

(a1i+a2j+a3k)× (b1i+b2j+b3k) =

a1b2i×j+a1b3i×k+a2b1j× i+a2b3j×k+a3b1k× i+a3b2k×j

= a1b2k−a1b3j−a2b1k+a2b3i+a3b1j−a3b2i

= (a2b3 −a3b2) i+ (a3b1 −a1b3)j+ (a1b2 −a2b1)k (14.20)

It is probably impossible for most people to remember 14.19. Fortunately, there is asomewhat easier way to remember it.

a×b=

∣∣∣∣∣∣i j k

a1 a2 a3b1 b2 b3

∣∣∣∣∣∣ (14.21)

where you formally expand the determinant along the top row. For those who have not seendeterminants, here is a short description. All you need here is how to evaluate 2× 2 and3×3 determinants. ∣∣∣∣ x y

z w

∣∣∣∣= xw− yz

294 CHAPTER 14. VECTOR PRODUCTSFor a, b, and c vectors, one obtains the distributive laws,ax (b+c)=axb+axe, (14.17)(b+c)xa=bxa+cxa. (14.18)Formula 14.15 follows immediately from the definition. The vectors a x b and bx ahave the same magnitude, |a| |b| sin 8, and an application of the right hand rule shows theyhave opposite direction. Formula 14.16 is also fairly clear. If & 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 14.16. 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 14.15,(b+c)xa=-—ax(b+c)=-—(axb+axc)=bxa+cxa.A proof of the distributive law is given later.Now from the definition of the cross product,ixj=k, jxi=—kkxi=j, ixk=—-jjxk=i, kxj=-iWith this information, the following gives the coordinate description of the cross product.Proposition 14.4.3 Let a = ayi+arj +.a3k and b = bi +b2j +.b3k be two vectors.Thena x b= (axb3 — a3bz) t+ (a3b1 — a,b3) J+ (aib2 — azb1) k. (14.19)Proof: From the above table and the properties of the cross product listed,(a\t+ aj +a3k) x (bit +boj + b3k) =abot X 7 +a,b3t xX k+anbyjg Xt+arb3j Xk +a3b,k xt+azbok x9= abyk — a,b39 — anbik +. azb3i + a3b19 — agbot= (anb3 — a3b2) t+ (a3b1 — ayb3) J+ (ayb2 — anh, )k (14.20)|It is probably impossible for most people to remember 14.19. Fortunately, there is asomewhat easier way to remember it.ig ikaxb=|aq, @ @ (14.21)bh bo bywhere you formally expand the determinant along the top row. For those who have not seendeterminants, here is a short description. All you need here is how to evaluate 2 x 2 and3 x 3 determinants.x y=xXw— yzZ WwW | y