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Noliec
 one year ago
Best ResponseYou've already chosen the best response.0Initially 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.

Noliec
 one year ago
Best ResponseYou've already chosen the best response.0Currently at vxu+vxw+wxu

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

Noliec
 one year ago
Best ResponseYou've already chosen the best response.0Just 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?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1So 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1You would write your ui's as wi's by using ui+vi+wi=0 as ui=(wi+vi)

Noliec
 one year ago
Best ResponseYou've already chosen the best response.0Indeed, 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Hey! We use some cool properties!? :)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Like 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Because 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
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