Is This Approach to Linear Algebra Proof Correct?

[mitch987]

Help!? Linear algebra proof

Homework Statement

Suppose that u,v,w are geometric vectors such that u$$\neq$$0,
u$$\cdot$$v=u$$\cdot$$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$$\cdot$$v=u$$\cdot$$w
|u||v|cos$$\theta$$=|u||w|cos$$\theta$$
|v|=|u|uxv=uxw
|u||v|sin$$\theta$$$$\hat{(u\times v)}$$=|u||w|sin$$\theta$$$$\hat{(u\times w)}$$
|w|sin$$\theta$$$$\hat{(u\times v)}$$=|w|sin$$\theta$$$$\hat{(u\times w)}$$
$$\hat{(u\times v)}$$=$$\hat{(u\times w)}$$
therefore, v=w

 

[tjackson3]

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

$$|u||v|\cos\theta_1 = |u||w|\cos\theta_2<br /> <br /> |v|\cos\theta_1 = |w|\cos\theta_2$$

Similarly,

$$|v|\sin\theta_1 = |w|\sin\theta_2$$

See what you can do from there

 

[mitch987]

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

$$\upsilon$$ $$\cdot$$ $$\upsilon$$ = $$\omega$$ $$\cdot$$ $$\omega$$

 

[lanedance]

so that shows the magnitudes are the same...

 

[mitch987]

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

 

[lanedance]

once you have the magnitudes, it follows from either

but yeah that's heaps nicer