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Physics 14 Online
OpenStudy (anonymous):

: if (vector)a X (vector)b = (vector)c X (vector)b , is it possible to prove that (vector)a = (vector)c

OpenStudy (amistre64):

is cross product commutable?

OpenStudy (amistre64):

AB not= BA

OpenStudy (amistre64):

But, its prolly not pertinent is it

OpenStudy (amistre64):

there are more comprehensive proofs online then I can think up at the moment

OpenStudy (anonymous):

why is cross product's commutability entering the matter ...it's already been stated the AB=CB

OpenStudy (amistre64):

becasue that is what i thought about when I read your post, but after posting it i realized that it had nothing to do with the matter.

OpenStudy (anonymous):

okay

OpenStudy (amistre64):

Suppose a and c are not the same; and try to prove otherwise. <a,b>x<m,n> = <c,d>x<m,n> that might be something worth doing

OpenStudy (anonymous):

no... we cant say that because if (vector)a X (vector)b = (vector)c X (vector)b, it implies that absin(angle bet a &b)= bcsin(angle bet b &c), so even though it is stated in that way, u cant say vector a= vector c unless angles between a&b and b&c are equal... I think this gives u ur required answer..

OpenStudy (amistre64):

\[<a_1,a_2>x<b_1,b_2>=<c_1,c_2>x<b_1,b_2>\] \[a_2b_1 -a_1b_2=c_2b_1-c_1b_2\] \[a_2b_1-c_2b_1= a_1b_2-c_1b_2\] \[b_1(a_2-c_2)= b_2(a_1-c_1)\] anushas looks viable :)

OpenStudy (anonymous):

thanks for the medals @amistre64 & siddarth95...

OpenStudy (anonymous):

@anusha : .... thanks ...i think i understand it now :)

OpenStudy (anonymous):

:)

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