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Question: Let L denote the set of all straight lines in a plane. Let a relation R be defined by \(\alpha R\bet...

Let L denote the set of all straight lines in a plane. Let a relation R be defined by αRβ⇔α⊥β,α,β∈L\alpha R\beta \Leftrightarrow \alpha \bot \beta ,\alpha ,\beta \in L. Then R is,
A. Reflexive
B. Symmetric
C. Transitive
D. None of these

Explanation

Solution

Hint: We will be using the concepts of functions and relations to solve the problem. We will be using the definitions of reflexive relation, symmetric relations and transitive relations to verify if each relation holds or not and hence deduce the answer.

Complete step-by-step solution -
Now, we have been given a relation and we have to find whether the relation is reflexive, symmetric, transitive or a combination of these.
Now, we know that reflexive relations are those in which every element is mapped to itself i.e. (a,a)∈R\left( a,a \right)\in R while symmetric relations are those for which if a R b then b R a. Also, holds and transitive are those relations in which if a R b and b R c then a R c must be held.
Now, we know different types of relations. We will check the given relation for these.
Now, we have been given a set L that denotes all straight lines in a plane. Now, we have been given a relation R defined by,
αRβ⇔α⊥β,α,β∈L\alpha R\beta \Leftrightarrow \alpha \bot \beta ,\alpha ,\beta \in L
Now, we will first test for reflexivity. We know that for reflexivity (α,α)\left( \alpha ,\alpha \right) must belong to R but since, the same line can’t be perpendicular to itself. Therefore, (α,α)∉R\left( \alpha ,\alpha \right)\notin R and the relation is not reflexive.
Now, for symmetric relations we have if αRβ⇒α⊥β\alpha R\beta \Rightarrow \alpha \bot \beta .
Or we write it as β⊥α\beta \bot \alpha since both the lines are perpendicular to each other.
Therefore, βRα as β⊥α\beta R\alpha \ as\ \beta \bot \alpha and hence, αRβ⇒βRα\alpha R\beta \Rightarrow \beta R\alpha . Therefore, the relation is transitive.
Now, for transitive we have if,
αRβ⇒α⊥β and βRγ⇒β⊥γ \begin{aligned} & \alpha R\beta \Rightarrow \alpha \bot \beta \ and \\\ & \beta R\gamma \Rightarrow \beta \bot \gamma \\\ \end{aligned}
Now, we know that if α⊥β and β⊥γ or α and γ are both ⊥ to β then α ∥ γ\alpha \bot \beta \ and\ \beta \bot \gamma \ or\ \alpha \ and\ \gamma \ are\ both\ \bot \ to\ \beta \ then\ \alpha \ \parallel \ \gamma lines perpendicular to same base are parallel. Therefore, (α,γ)∉R\left( \alpha ,\gamma \right)\notin R and hence, the relation is not transitive.
Therefore, the correct option is (B).

Note: To solve these types of questions it is important to note that a R b means that a is related to b by a relation R. Also these types of questions are solved easily by giving examples and counter examples. Also, we have to check the relation for reflexive, symmetric and transitive relation to check it for equivalence relation.