Question: In the given figure, ABCD is a parallelogram. If $\vec{OA} = \vec{a}, \vec{OB} = \vec{b}, and \; \vec{OC} = \vec{c}, find \; \vec{OD}$.

Solution:

To find: value of $\vec{OD}$

Given, $\vec{OA} = \vec{a}, \vec{OB} = \vec{b}, and \; \vec{OC} = \vec{c}$

ABCD is a parallelogram. So, opposite sides are equal and opposite vector in same direction are equal. $or, \vec{AB} = \vec{DC}$



Now,

In ∆ OAB,
Using ∆ law of vector addition,

$\vec{AB} = \vec{AO} + \vec{OB}$

$or, \vec{AB} = \vec{OB} - \vec{OA}$

$\therefore \vec{AB} = \vec{b} - \vec{a}$

From above,
$\vec{DC} = \vec{AB} = \vec{b} - \vec{a}$

In ∆ ODC,
Using ∆ law of vector addition,

$\vec{OD} = \vec{OC} + \vec{CD}$

$or, \vec{OD} = \vec{c} - \vec{DC}$

$or, \vec{OD} = \vec{c} - ( \vec{b} - \vec{a})$

$or, \vec{OD} = \vec{c} - \vec{b} + \vec{a}$

$\therefore \vec{OD} = \vec{a} - \vec{b} + \vec{C}$

Hence, the required value of $\vec{OD} $ is $\vec{a} - \vec{b} + \vec{C}$.