reverse of cross product ?
RichardPuchmayer
Hi,
Is there such a thing as the inverse of a cross product.
ie. if I have C = AxB, and I know C and B.
can I find A?
something along the lines of
F(C,B) = A
where A,B and C are vectors, and F() is a vector function
of C and B.
Thanks,
Richard.
--
Richard Puchmayer == pu...@cutmcvax.cs.curtin.edu.au | Some of us are poets,
Masters Student at Curtin University of Technology, WA.| some of us are not!
-------------------------------------------------------+------------------------
I know nothing, so can hold no opinions for myself or others...
Rouben Rostamian
> Is there such a thing as the inverse of a cross product.
Case 1: B=0
If C = 0, then A can be any arbitrary vector.
If C != 0, then there is no such A.
Case 2: B != 0
Then A = BxC/|B|^2 + aB, where a is an arbitrary scalar multiplier.
Proof is left for an exercise :-)
--
Rouben Rostamian Telephone: (301) 455-2458
Department of Mathematics and Statistics e-mail:
University of Maryland Baltimore County bitnet: rost...@umbc.bitnet
Baltimore, MD 21228, U.S.A. internet: rou...@math9.math.umbc.edu
jo...@ohstpy.mps.ohio-state.edu
> Is there such a thing as the inverse of a cross product.
> ie. if I have C = AxB, and I know C and B.
> can I find A?
>
> something along the lines of
>
> F(C,B) = A
>
> where A,B and C are vectors, and F() is a vector function
> of C and B.
of thought may indicate that the inverse will not be unique. That is, if
A and C are specified, then B in C=AXB will not be unique; that is, many
B's may be found which will satisfy the equation. (It is tacitly assumed
of course that neither A nor C vanish and that A and C are not colinear.)
Indeed, write the components of the cross product as:
C[i] = A[j] * B[k] - A[k] * B[j] (i,j,k)=cyclic(1,2,3)
where A[i] is the i'th component of the vector A with i = 1,2,3; and similar
for B and C. We can then rewrite C=AXB to look like a matrix equation:
( ) ( ) ( )
( C[1] ) ( 0 -A[3] A[2] ) ( B[1] )
( ) ( ) ( )
( C[2] ) = ( A[3] 0 -A[1] ) ( B[2] )
( ) ( ) ( )
( C[3] ) ( -A[2] A[1] 0 ) ( B[3] )
( ) ( ) ( )
It is a simple matter to show that the determinant of the matrix is 0
meaning of course that the matrix cannot be inverted and so B can not
be expressed in terms of A and C.
--
bob jones; Disclaimer: "I just said what?"; jo...@ohstpy.mps.ohio-state.edu
pfen...@uni2a.unige.ch
>...
> ie. if I have C = AxB, and I know C and B.
> can I find A?
This is discussed by:
Frawley W.J., 1985, in "Applications of Computer Algebra" ed. R. Pavelle
(Kluwer: Boston) p. 415-426
If AxB is considered as a matrix multiplication:
[ 0 -a3 a2] [b1]
AxB = [ a3 0 -a1] [b2]
[-a2 a1 0] [b3]
then A has a rank 2 so is not uniquely invertible. However the Moore-Penrose
generalized pseudo-inverse always exists and in this example is:
1 [ 0 a3 -a2]
A' = -------------- [-a3 0 a1]
a1^2+a2^2+a3^2 [ a2 -a1 0]
This idea applies to higher dimensional anti-symmetric products.
Daniel Pfenniger
University of Geneva
Trent Tobler
>
> Hi,
>
> Is there such a thing as the inverse of a cross product.
> ie. if I have C = AxB, and I know C and B.
> can I find A?
Yes, A can be found. But unless you have a very specific case, it will
not be unique. For example, suppose you have A = (1,0,0), C = (0,0,1),
then B = (1,y,0) for y real - (1,0,0) works, but so does (1,5,0) and (1,100,0)
and (1,pi,0).
>
> something along the lines of
>
> F(C,B) = A
>
> where A,B and C are vectors, and F() is a vector function
> of C and B.
No. F will not be a vector. It will be a vector space. (it will be
a vector with one independant variable.); You could define it to
be such that all angles are 90 degrees (This restricts it to a single
vector, and is found by:
|C|*(AxC)
B = ----------
|A|*|AxC|
Just remember that this is just one solution out of infinitely many that fill
the AxB=C solved for B criteria.
*note, |A| is notation to mean the length of A
--
Trent Tobler - tto...@csulx.weber.edu