PQRS is a parallelogram. If \(\overrightarrow{\text{PR}}=\vec{\text{a}}\) and \(\overrightarrow{\text{QS}}=\vec{\text{b}}\), then what is \(\overrightarrow{\text{PQ}}\) equal to ?

Concept:

The diagonals of the parallelogram bisect each other.

In a triangle PQR,  ​\(\overrightarrow{\text{PQ}} + \overrightarrow{\text{QR}}+ \overrightarrow{\text{RP}}=0\) 

For any vector \(\overrightarrow{\text{AB}}, \ \ \overrightarrow{\text{AB}}= -\overrightarrow{\text{BA}}\)

Calculation:

Given:  \(\overrightarrow{\text{PR}}=\vec{\text{a}}\) and \(\overrightarrow{\text{QS}}=\vec{\text{b}}\), 

Let the diagonals of the parallelogram intersect each other at O.

⇒ \(\overrightarrow{\text{PO}} = ​​\overrightarrow{\text{OR}}\ \text {and} \ \overrightarrow{\text{QO}}= \overrightarrow{\text{OS}}\)  __(1)

(∵ The diagonals of the parallelogram bisect each other.)

As,  \(\overrightarrow{\text{PR}}=\vec{\text{a}}\) and \(\overrightarrow{\text{QS}}=\vec{\text{b}}\), 

⇒  \(\overrightarrow{\text{PO}}+ \overrightarrow{\text{OR}}=\vec{\text{a}}\) and \(\overrightarrow{\text{QO}}+ \overrightarrow{\text{OS}}=\vec{\text{b}}\), __(2)

From (1) and (2),

⇒ \(\overrightarrow{\text{PO}} = ​​\overrightarrow{\text{OR}}\ = { \overrightarrow{\text{a}} \over 2}\ \text{and} \ \overrightarrow{\text{QO}}= \overrightarrow{\text{OS}} ={ \overrightarrow{\text{b}} \over 2}\)  __(3)

Now in triangle PQO,

\(\overrightarrow{\text{PQ}}\) + \(\overrightarrow{\text{QO}}\) + \(\overrightarrow{\text{OP}}\) = 0

⇒ \(\overrightarrow{\text{PQ}} = -\overrightarrow{\text{QO}}- \overrightarrow{\text{OP}}\)

⇒ \(\overrightarrow{\text{PQ}} = -\overrightarrow{\text{QO}}+ \overrightarrow{\text{PO}}\)

Putting the values from equation (3),

⇒ \(\overrightarrow{\text{PQ}} = -{ \overrightarrow{\text{b}} \over 2}+ { \overrightarrow{\text{a}} \over 2}\)

⇒ \(\overrightarrow{\text{PQ}}\) = \(\frac{\vec{\text{a}}−\vec{\text{b}}}{2}\)

∴ The correct option. is (4) .