Understanding Triple Product Rules in Vector Calculus

[quietrain]

triple product is given by

A.(B x C) = B. (C x A) = C. (A x B)


so why is ∇.(F x G) = [∇ x F].G - F.[∇ x G] ?




if i let A = ∇, B = F , C = G,

then ∇.(F x G) = - F .( ∇ x G) = G.(∇ x F)
its as though they carried the 3rd term over to add to the 2nd term ?

 

[lurflurf]

wow.
That rule does not apply to ∇, because ∇ acts to the right.
It is very much like the single variable rule
D(uv)=u(Dv)+(Du)v
Where is D were bidirectional we would have
Duv=uDv=uvD

 

[quietrain]

? you mean the triple product rules is not universal?

so the triple cross product A x (BxC) is also not applicable for the del?

so i won't get B(A.C)-C(A.B)?

so how do i know that the del will give me Duv=uDv=uvD ?

is there an identity or i must proof from scratch?

 

[lurflurf]

In general an algebric vector identity will not be an identity for ∇.
A.(B x C) = B. (C x A) = C. (A x B)
A x (BxC)=B(A.C)-C(A.B)
are not always true for ∇
In most cases this is because ∇ is not bidirectional.
Thus a vector identity where the vectors change places tends not to hold for ∇
For example
A.B=B.A
but
∇.A!=A.∇

For
∇.[F x G] = G.[∇ x F] - F.[∇ x G]
Suppose F is constant then we have
∇.[F x G] = - F.[∇ x G]
F can move through ∇ because it is constant
Suppose G is constant then we have
∇.[F x G] =G.[∇ x F]
G can move through ∇ because it is constant
The case where neither F nor G need be constant is the sum of these
∇.[F x G] =G.[∇ x F] - F.[∇ x G]

 

[quietrain]

wow... i have more to learn

thanks!