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Noliec

  • 2 years ago

If u+v+w=0, show that u x v = v x w = w x u.

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  1. Noliec
    • 2 years ago
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    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.

  2. Noliec
    • 2 years ago
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    Currently at vxu+vxw+wxu

  3. Noliec
    • 2 years ago
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    Quite certain you're supposed to use the fact that some of the vector products are orthogonal to each other and therefore 0

  4. Noliec
    • 2 years ago
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    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?

  5. myininaya
    • 2 years ago
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    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.

  6. myininaya
    • 2 years ago
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    You would write your ui's as wi's by using ui+vi+wi=0 as ui=-(wi+vi)

  7. Noliec
    • 2 years ago
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    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.

  8. myininaya
    • 2 years ago
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    Hey! We use some cool properties!? :)

  9. myininaya
    • 2 years ago
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    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.

  10. myininaya
    • 2 years ago
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    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

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