At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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.
Currently at vxu+vxw+wxu
Quite certain you're supposed to use the fact that some of the vector products are orthogonal to each other and therefore 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?
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(u2v3-u3v2)-j(u1v3-v1u3)+k(u1v2-u2v2) 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.
You would write your ui's as wi's by using ui+vi+wi=0 as ui=-(wi+vi)
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.
Hey! We use some cool properties!? :)
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.
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