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Derivation of Vector Product 
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#1
Oct1705, 09:15 PM

P: 8

In my calc book the derivation of the vector cross product is not derrived but rather just given. I've read in another book that William Rowan Hamilton, after years of work, came up with the basic form we momorize today and symbolize with determinants. Does anybody know how this vector cross product is derrived. Some might say "assume the determinat definition and the results follow". The question is how was this definition derrived. Thanks



#2
Oct1805, 04:15 AM

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P: 661

I personally believe that the cross product is a definition rather than any kind of result. On the same level as the rules of multiplication and division. We've defined a new operation we can do on vectors.



#3
Oct1805, 04:36 AM

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P: 11,896

Just follow some exterior algebra and some geometry and you'll find that the "vector product" is a particular (up to Hodge duality) example of exterior product.
Daniel. 


#4
Oct1805, 06:19 PM

HW Helper
P: 2,263

Derivation of Vector Product
Definitions are not really derived, they are not chosen at random either. It is a matter of seeing what is nice about the object in question and seeing how that leads to a definition. Often different rotes can be followed. Here is one for this case
we desire a function z=f(x,y) f:R^3xR^3>R^3 That is we want to map two 3space vectors into one. We also would like the function to linearly map products of componets of x and y to componets the function to not depend upon the coordinate system used for x and y that is it should be invariant So one (tedious) way to procede would be to write down a function with the linear componet map property that would have 81 constants. Then the invarence property would give 80 independent equations. Thus the cross product would be determined up to a multiplicative constant. 


#5
Oct1805, 10:39 PM

P: 8

lurflurf you gave me enough to know that I will wait till my calc 3 undegraduate level of math is completed before I really follow your reply. For now I think I know that the need for the properties of a cross product drove a legit derrivation  originally maybe the one by Irish Mathematician Hamiliton, and more recently the process you so kindly mentioned. Thanks!



#6
Oct2105, 11:33 PM

HW Helper
P: 2,263

I will try again.
We want to define a (pseudo)vector axb where a and b are vectors and a,b,axb are in 3space so that 1) it is bilinear that is (ka)xb=k(axb) ax(kb)=k(axb) (a+c)xb=axb+cxb ax(b+c)=axb+axc a,b,c vectors k a scalar 2) it is invariant under rotation that is is v' is the vector v after a (proper) rotation (axb)'=a'xb' A proper rotation can be written with a (special orthogonal) matrix A v'=Av where det(A)=1 and inverse(A)=transpose(A) or thought of as a rotation geometrically The point of this is if two coordinate systems are used the results should be the same. The approuch define the general product reduce the possibilities by enforcing rotation invariance reduce 27=3^3 variables to 1 (I said 81 before oops) only rightangle rotations will be needed we can visuallize right angle rotations as the 24 ways a cube can be possitioned so since we have bilinearity we will have defined the cross produce when we have defined the 9 quantities ixi,ixj,ixk,jxi,jxj,jxk,kxi,kxj,kxk each of which will have the form ixj=Ai+Bj+Ck thus the 27 values we need to specify consider the rotation i'=i j'=j k'=k so ixj=Ai+Bj+Ck Invarience requres the equation to hold when i'xj'=Ai'+Bj'+Ck' which becomes (i)x(j)=AiBj+Ck but by bilinearity (i)x(j)=ixj so Ai+Bj+Ck=AiBj+Ck hence A=A>2A=0>A=0 B=B>2B=0>B=0 so ixj=CK this is a big start we have directly eliminated 2 of our 27 and using this fact later we have went from 27 to 9 as similar resoning applies to all nine products ixi,ixj,ixk,jxi,jxj,jxk,kxi,kxj,kxk in particular if kxk=Di+Ej+Fk then k'xk'=Di'+Ej'+Fk' so kxk=DiEj+Fk so d=E=0 and kxk=Fk now consider the rotation i'=j j'=i k'=k i'xj'=Ck' (recall A=B=0) jxi=Ck also consider kxk=Fk k'xk'=Fk' (k)x(k)=Fk so F=0 kxk=0 The full effect of this line of reasoning reduces the 9 variables to 3 so far we have ixj=Ck jxi=Ck kxk=0 now consider the rotation i'=k j'=i k'=j i'xj'=Ck' so kxi=Cj j'xi'=Ck' so ixk=Cj kxk=0 so jxj=0 now the rotation i'=j j'=k k'=i i'xj'=Ck' so jxk=Ci j'xi'=Ck' kxj=Ci k'xk'=0 ixi=0 collecting things up again ixi=0 ixj=Ck ixk=Cj jxi=Ck jxj=0 jxk=Ci kxi=Cj kxj=Ci kxk=0 So any such product is a multiple of the standard one in which C is chosen C=1 giving ixi=0 ixj=k ixk=j jxi=k jxj=0 jxk=i kxi=j kxj=i kxk=0 Similar reasoning applies to higher products a.bxc=axb.c is the only scalar triple product all vector triple products are of the form u(a.b)c+v(b.c)a+w(c.a)b and so on 


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