Check the following relations $ R $ and $ S $ for reflexivity, symmetry and transitivity:
(a) $ aRb $ iff $ b $ is divisible by $ a $ , $ a,b \in N $
(ii) $ {l_1}S\,{l_2} $ iff $ {l_1} \bot {l_2} $ , where $ {l_1} $ and $ {l_2} $ are straight lines in a plane.

Hint: To solve this problem we will have to check for the reflexivity, symmetry and transitivity separately. To check reflexivity of a relation $ R $ , we will check whether $ \left( {a,a} \right) \in R $ for every $ a \in R $ .To check symmetry, we will check if $ \left( {a,b} \right) \in R $ then $ \left( {b,a} \right) \in R $ . To check transitivity, we will check if $ \left( {a,b} \right) \in R $ and $ \left( {b,c} \right) \in R $ then $ \left( {a,c} \right) \in R $ .

Complete step-by-step answer:
(i)
Given:
Relation $ R\left( {a,b} \right) $ is such that $ b $ is divisible by $ a $ .
Now, we can write relation $ R $ as,
 $ R = \left\{ {\left( {a,b} \right):b\;{\rm{is divisible by }}a} \right\} $
In order to check the reflexivity of the relation, $ \left( {a,a} \right) $ must belong to the relation $ R $ such that $ a $ is divisible by $ a $ . Since $ \left( {a,a} \right) \in R $ .
Therefore $ R $ is reflexive.
In order to check the symmetry of the relation, we will have to check if $ \left( {a,b} \right) \in R $ then $ \left( {b,a} \right) \in R $ . To check for symmetry, we will have to take an example.
Suppose $ \left( {a,b} \right) $ are $ \left( {3,9} \right) $
Here $ \left( {3,9} \right) \in R $ , as 9 is divisible by 3, but $ \left( {9,3} \right) \in R $ 3 is not divisible by 9.
Therefore, from the example it is clear that $ R $ is not symmetric.
In order to check the transitivity, we will have to assume another value $ c \in R $ such that $ \left( {a,b} \right) \in R $ , $ b $ is divisible by a and $ \left( {b,c} \right) \in R $ , $ c $ is divisible by $ b $ .
If $ b $ is divisible by a and $ c $ is divisible by $ b $ then $ c $ is divisible by $ a $ .
Since, $ \left( {a,b} \right) \in R $ and $ \left( {b,c} \right) \in R $ therefore, $ \left( {a,c} \right) \in R $ .
Hence $ R $ is transitive.

(ii)
Given: Relation $ S\left( {{l_1},{l_2}} \right) $ such that $ {l_1} \bot {l_2} $ , where $ {l_1} $ and $ {l_2} $ are straight lines in a plane.
Relation $ S $ can be expressed as,
 $ S = \left\{ {\left( {{l_1},{l_2}} \right):{l_1}\;{\rm{is \text perpendicular to }}{l_2}} \right\} $
In order to check the reflexivity of the relation, $ \left( {{l_1},{l_1}} \right) $ must belong to the relation $ S $ but a line can never be perpendicular to itself. So, $ \left( {{l_1},{l_1}} \right) \notin S $ .
Therefore $ S $ is not reflexive.
In order to check the symmetry of the relation, we will have to check if $ \left( {{l_1},{l_2}} \right) \in S $ then $ \left( {{l_2},{l_1}} \right) \in S $ .
If $ {l_1} $ is perpendicular to $ {l_2} $ then $ {l_2} $ is also perpendicular to $ {l_1} $ . Hence $ \left( {{l_2},{l_1}} \right) \in S $ .
Therefore, $ S $ is symmetric.
In order to check the transitivity, we will have to assume another value $ {l_3} \in S $ such that $ \left( {{l_1},{l_2}} \right) \in S $ , $ {l_1} $ is perpendicular to $ {l_2} $ and $ \left( {{l_2},{l_3}} \right) \in S $ , $ {l_2} $ is perpendicular to $ {l_3} $ .
If $ {l_1} $ is perpendicular to $ {l_2} $ and $ {l_2} $ is perpendicular to $ {l_3} $ then $ {l_1} $ is not perpendicular to $ {l_3} $ .
Hence $ \left( {{l_1},{l_2}} \right) \notin S $ .
Hence $ S $ is not transitive.

Note: In this type of problem, all the values of the relation must satisfy the conditions of reflexivity, symmetry and transitivity. This can be done taking different examples of the values from the same relation.