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OK, i am trying to prove that if $\vec a\times \vec b = \vec a \times \vec c$ and also $\vec a\cdot \vec b = \vec a \cdot \vec c$ then $\vec b = \vec c$.

so far i got to $\vec n \tan \alpha = \vec m \tan \beta$ and do not know how to continue to get the result

this seems too easy :)

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we can assume that $\vec a$ cannot be 0 –  ASROMA Oct 14 '12 at 4:32
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There are formulas for the (magnitude of) the cross and dot products in terms of the magnitudes of the vectors and the angles between them - that might be a good place to start. –  Gerry Myerson Oct 14 '12 at 4:35
    
that is how i got to a place where i have this: $\vec n \tan \alpha = \vec m \tan \beta$ , n and m are unit vectors –  ASROMA Oct 14 '12 at 4:41
    
@Vahe Just out of curiosity, how can it seem too easy if you don't know how to continue? –  EuYu Oct 14 '12 at 4:45
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EuYu, I took it to mean, "it seems like it should be easy". –  Gerry Myerson Oct 14 '12 at 4:46

2 Answers 2

up vote 3 down vote accepted

This is basically a fleshing out of the hint by @GerryMyerson. We assume that ${\bf a}\ne {\bf 0}$.

Let ${\bf d} = {\bf b}-{\bf c}$. Then ${\bf a}\times{\bf d} = {\bf 0}$ and ${\bf a}\cdot{\bf d} = 0$, and so $\|{\bf a}\| \|{\bf d}\|\sin\theta = 0$ and $\|{\bf a}\| \|{\bf d}\|\cos\theta = 0$, where $\theta$ is the angle between ${\bf a}$ and ${\bf d}$. Since there is no $\theta$ for which $\sin\theta = \cos\theta = 0$, we can conclude that ${\bf d} = {\bf 0}$, and so ${\bf b} = {\bf c}$.

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is $a \times (b-c) = a \times b - a \times c$ ?? I thought the equation is correct for addition... i guess you can write b+(-c) –  ASROMA Oct 14 '12 at 5:23

Hint: Consider the identity $$\|\mathbf{x} \times \mathbf{y}\|^2 + (\mathbf{x}\cdot \mathbf{y})^2 = 2\|\mathbf{x}\|^2\|\mathbf{y}\|^2$$ In particular, this is a special case of the angle identities that Gerry was mentioning.

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but how does this help? i have to prove that vectors are equal, not their scalars –  ASROMA Oct 14 '12 at 4:50
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Can you think of some way to apply this identity to your problem? Try proving that $\mathbf{b - c} = \mathbf{0}$. –  EuYu Oct 14 '12 at 4:51
    
there is a gray area when it comes to angles, i have to use different angles with A, B and A, C. so basically i have to show that these angles are equal –  ASROMA Oct 14 '12 at 4:53
    
I'm not sure what you mean by the grey area. You don't actually need to invoke angles explicitly in this problem. Try applying the above identity to $\mathbf{x} = \mathbf{a}$ and $\mathbf{y} = \mathbf{b-c}$. –  EuYu Oct 14 '12 at 4:56

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