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NoliecBest ResponseYou've already chosen the best response.0
Initially I set u=(1)(v+w), v=(1)(u+w), w=(1)(u+v) and then try to just insert those vectors into the thing I'm supposed to show; I don't know how to go from where I am though.
 6 months ago

NoliecBest ResponseYou've already chosen the best response.0
Currently at vxu+vxw+wxu
 6 months ago

NoliecBest ResponseYou've already chosen the best response.0
Quite certain you're supposed to use the fact that some of the vector products are orthogonal to each other and therefore 0
 6 months ago

NoliecBest ResponseYou've already chosen the best response.0
Just arrived at wxu; still haven't shown for vxw though, it seems very chaotic just inserting relations over and over, is there a better way?
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
So say we have u=(u1,u2,u3) v=(v1,v2,v3) w=(w1,w2,w3) ,then u1+v1+w1=0 u2+v2+w2=0 u3+v3+w3=0. uxv= dw:1380907086373:dw =i(u2v3u3v2)j(u1v3v1u3)+k(u1v2u2v2) SO I guess you could use this to show wxu by writing what I just wrote but your ui's as wi's and your vi's as ui's. (Then do the same for vxw) But this seams pretty long to me. I will have to see if I can find a shorter way.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
You would write your ui's as wi's by using ui+vi+wi=0 as ui=(wi+vi)
 6 months ago

NoliecBest ResponseYou've already chosen the best response.0
Indeed, that's what I went for, managed to show that the left term was the right term; never got to the middle one though. It seems like a mess using the laws though.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
Hey! We use some cool properties!? :)
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
Like can we use ax(b+c)=(axb)+(axc) This would cut out a lot of time. This would make the proof like 100 times easier.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.1
Because since w=(u+v), then w=u+(v) So vxw=vx(u+(v)) =vx(u) + vx(v) Then we can also use bxa=axb and use axa=0
 6 months ago
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