Let S be the set of all vectors in a two-dimensional space ℝ^2 . A...

For reflexivity, consider a vector $\vec{a} \in S$.
$\vec{a} \cdot \vec{a} = |\vec{a}|^2 = |\vec{a}||\vec{a}|$
Thus, $\vec{a} R \vec{a}$ is true for all $\vec{a} \in S$. The relation is reflexive.

For symmetry, let $\vec{a} R \vec{b}$.
$\Rightarrow \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|$
$\Rightarrow \vec{b} \cdot \vec{a} = |\vec{b}||\vec{a}|$
Thus, $\vec{b} R \vec{a}$ is true. The relation is symmetric.

For transitivity, let $\vec{a} R \vec{b}$ and $\vec{b} R \vec{c}$.
Consider the case where $\vec{b} = \vec{0}$ (the zero vector).
For any vector $\vec{a}$, $\vec{a} \cdot \vec{0} = 0$ and $|\vec{a}||\vec{0}| = 0$. Thus, $\vec{a} R \vec{0}$ is true for all $\vec{a} \in S$.
Let $\vec{a} = \hat{i}$ and $\vec{c} = \hat{j}$.
We have $\hat{i} R \vec{0}$ and $\vec{0} R \hat{j}$.
However, $\hat{i} \cdot \hat{j} = 0$, while $|\hat{i}||\hat{j}| = 1 \times 1 = 1$.
Since $0 \neq 1$, $\hat{i}$ is not related to $\hat{j}$.
Therefore, the relation is not transitive.

Hence, $R$ is reflexive and symmetric but not transitive.

Answer: Reflexive and symmetric but not transitive