= Solution
For the <product order>, <reflexivity> follows from $a_i\le a_i$ in every coordinate. If $\mathbf a\preceq\mathbf b$ and $\mathbf b\preceq\mathbf a$, the <antisymmetric relation> axiom for the usual real order gives $a_i=b_i$ for every $i$, hence $\mathbf a=\mathbf b$. If $\mathbf a\preceq\mathbf b\preceq\mathbf c$, <transitivity> in each coordinate gives $a_i\le c_i$, hence $\mathbf a\preceq\mathbf c$. Thus this is a <partial order>.
For $n\ge2$, the <vectors> $(1,0,0,\ldots)$ and $(0,1,0,\ldots)$ are incomparable: each has a coordinate larger than the corresponding coordinate of the other. \b[It is not a <total order> for $n\ge2$.] The dimension qualification matters: for $n=1$ this is the usual <total order> on $\mathbb R$, so the printed non-totality claim needs the implicit assumption $n\ge2$.
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