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Help? Linear algebra proof

  • Thread starter mitch987
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  • #1
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Help!? Linear algebra proof

Homework Statement


Suppose that u,v,w are geometric vectors such that u[tex]\neq[/tex]0,
u[tex]\cdot[/tex]v=u[tex]\cdot[/tex]w and uxv=uxw

Prove that v=w


Homework Equations





The Attempt at a Solution


So far, I'm not sure if this is correct
u[tex]\cdot[/tex]v=u[tex]\cdot[/tex]w
|u||v|cos[tex]\theta[/tex]=|u||w|cos[tex]\theta[/tex]
|v|=|u|


uxv=uxw
|u||v|sin[tex]\theta[/tex][tex]\hat{(u\times v)}[/tex]=|u||w|sin[tex]\theta[/tex][tex]\hat{(u\times w)}[/tex]
|w|sin[tex]\theta[/tex][tex]\hat{(u\times v)}[/tex]=|w|sin[tex]\theta[/tex][tex]\hat{(u\times w)}[/tex]
[tex]\hat{(u\times v)}[/tex]=[tex]\hat{(u\times w)}[/tex]
therefore, v=w
 

Answers and Replies

  • #2
150
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I think you're on the right track. One thing to keep in mind is that we can't assume a priori that the two angles are equal, so we're looking at

[tex]|u||v|\cos\theta_1 = |u||w|\cos\theta_2

|v|\cos\theta_1 = |w|\cos\theta_2[/tex]

Similarly,

[tex]|v|\sin\theta_1 = |w|\sin\theta_2[/tex]

See what you can do from there
 
  • #3
8
0


Thanks, i completely forgot bout the angles not being equal.
however, going with that i can only simplify it down to

[tex]\upsilon[/tex] [tex]\cdot[/tex] [tex]\upsilon[/tex] = [tex]\omega[/tex] [tex]\cdot[/tex] [tex]\omega[/tex]
 
  • #4
lanedance
Homework Helper
3,304
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so that shows the magnitudes are the same...
 
  • #5
8
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yeah, but that only encompasses magnitude not direction.
anyway i figured it out by using the various laws.

if, u · v = u · w
then, u · (v-w) = 0

if, u x v = u x w
then, u x (v-w) = 0

therefore, v-w is both orthogonal and parallel to the non zero vector u, hence v-w = 0
therefore v=w
 
  • #6
lanedance
Homework Helper
3,304
2


once you have the magnitudes, it follows from either

but yeah thats heaps nicer
 

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