Let $L$ be the set of all lines in a plane and $R$ be the relation in $L$ defined as $R = \{(L_{1}, L_{2}) : L_{1} \text{ is perpendicular to } L_{2}\}$. Show that $R$ is symmetric but neither reflexive nor transitive.

(N/A) $R$ is not reflexive,as a line $L_{1}$ cannot be perpendicular to itself,i.e.,$(L_{1}, L_{1}) \notin R$.
$R$ is symmetric as $(L_{1}, L_{2}) \in R$
$\Rightarrow L_{1} \text{ is perpendicular to } L_{2}$
$\Rightarrow L_{2} \text{ is perpendicular to } L_{1}$
$\Rightarrow (L_{2}, L_{1}) \in R$
$R$ is not transitive. Indeed,if $L_{1}$ is perpendicular to $L_{2}$ and $L_{2}$ is perpendicular to $L_{3}$,then $L_{1}$ can never be perpendicular to $L_{3}$. In fact,$L_{1}$ is parallel to $L_{3}$,i.e.,$(L_{1}, L_{2}) \in R, (L_{2}, L_{3}) \in R$ but $(L_{1}, L_{3}) \notin R$.