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

Step 1: Check for Reflexivity

For a relation to be reflexive, every element must be related to itself. In this case, a line $L_1$ would need to be perpendicular to itself. However, a line cannot be perpendicular to itself, as the angle between a line and itself is $0^\circ$, not $90^\circ$. Therefore, the relation R is not reflexive.

Step 2: Check for Symmetry

For a relation to be symmetric, if $L_1$ is related to $L_2$, then $L_2$ must be related to $L_1$. If line $L_1$ is perpendicular to line $L_2$, it implies that the angle between them is $90^\circ$. This also means that line $L_2$ is perpendicular to line $L_1$. Therefore, the relation R is symmetric.

Step 3: Check for Transitivity

For a relation to be transitive, if $L_1$ is related to $L_2$ and $L_2$ is related to $L_3$, then $L_1$ must be related to $L_3$. If $L_1$ is perpendicular to $L_2$, and $L_2$ is perpendicular to $L_3$, then $L_1$ and $L_3$ must be parallel to each other (or coincident). Parallel lines are not perpendicular. Therefore, the relation R is not transitive.

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