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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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    @Eu Yu, because when you look at it, it is like the answer is right before your eyes, that is why i said so. But grammatically, you are correct ;)2012-10-14

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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)2012-10-14
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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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    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 $\mathbf{y} = \mathbf{b-c}$.2012-10-14