How to prove the equality of two vectors?

Question

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 :)

Answer 1

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 ${\bf b} = {\bf c}$.

Answer 2

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.