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

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Solution

Given the relation $R$ is defined as $R = {(L_{1}, L_{2}) : L_{1} ⊥ L_{2}}$ .
Now this relation is not reflexive as $_{L_{1}} R_{L_{1}}$ does not hold as every line is not perpendicular to itself.
The relation is symmetric as $_{L_{1}} R_{L_{2}}$ gives $_{L_{2}} R_{L_{1}}$ . [ As if $L_{1}$ is perpendicular to $L_{2}$ then $L_{2}$ is also perpendicular to $L_{1}$ ].
The relation is not transitive as $_{L_{1}} R_{L_{2}},_{L_{2}} R_{L_{3}}$ does not gives $_{L_{1}} R_{L_{3}}$ . [ As, if $L_{1}$ is perpendicular to $L_{2}$ and $L_{2}$ is perpendicular to $L_{3}$ then $L_{1}$ may of may not be perpendicular to $L_{3}$ ]

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