Kenneth Kuttler

3.4. THE CROSS PRODUCT 43

have opposite direction. Formula 3.20 is also fairly clear. If α is a nonnegative scalar,the direction 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 3.20. 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 3.19,

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

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,

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 3.4.3 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. (3.23)

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 (3.24)

■It is probably impossible for most people to remember 3.23. Fortunately, there is a

somewhat easier way to remember it. Define the determinant of a 2×2 matrix as follows∣∣∣∣∣ a bc d

∣∣∣∣∣≡ ad−bc

Then

a×b =

∣∣∣∣∣∣∣i j k

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣∣ (3.25)

3.4. THE CROSS PRODUCT 43have opposite direction. Formula 3.20 is also fairly clear. If @ is a nonnegative scalar,the direction 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(axb) and ax (ab). Using this yields equality in 3.20. 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 3.19,(b+c) xa=-—ax (b+c) =—(axb+axc)=bxa+cexa.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,ixj=k jxi=-—-kkxi=j ixk=-jjxk=i kxj=—iWith this information, the following gives the coordinate description of the cross product.Proposition 3.4.3 Let a = aji+ aj +a3k and b = bji+ boj +.b3k be two vectors. Thenaxb= (azb3 —a3zbz)i+ (a3by —a,b3)j++ (a,b2 — anb,)k. (3.23)Proof: From the above table and the properties of the cross product listed,(aji+ aoj+a3k) x (bi + boj +b3k) =ayboi x j+a,b3i x kK+ anbij x i+azb3j x k++ta3bik x i+a3bok xj= ay bok — ay b3j — ab, k + anb3i + a3b1j — agbri= (ab3 — a3b2) i+ (a3b1 — a1 b3) j+ (ab — abi) k (3.24)aIt is probably impossible for most people to remember 3.23. Fortunately, there is asomewhat easier way to remember it. Define the determinant of a 2 x 2 matrix as followsb| =ad—bcc dTheni j ikaxb= ay ag a3 (3.25)by bo b3