<s>[INST] For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal? For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal? I tried looking for counter-examples or using coordinate-by-coordinate proofs, but that didn't get me anywhere.
 [/INST] We can conclude that $\mathbf b= \mathbf c$ provided $\mathbf a \neq \mathbf 0$.
Write $\mathbf d = \mathbf b - \mathbf c$. The given equations can be written as 
$$
\mathbf a \cdot \mathbf d = 0, \ \ \ 
\mathbf a \times \mathbf d = \mathbf 0.
$$
Now, suppose $\theta$ is the angle between $\mathbf a$ and $\mathbf d$. 
$$
|\mathbf a \cdot \mathbf d| = |\mathbf a| |\mathbf d| |\cos \theta|, \ \ \ 
|\mathbf a \times \mathbf d| = |\mathbf a| |\mathbf d| |\sin \theta|, \ \ \ 
$$
Then, squaring and adding,
$$
0 = 0 + 0 = |\mathbf a \cdot \mathbf d|^2 + |\mathbf a \times \mathbf d|^2 = |\mathbf a|^2 |\mathbf d|^2 (\cos^2 \theta + \sin^2 \theta) = |\mathbf a|^2 |\mathbf d|^2,$$
which implies that either $\mathbf a = \mathbf 0$ or $\mathbf d = \mathbf 0$. Finally, $\mathbf d = \mathbf 0 \iff \mathbf b = \mathbf c$.
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