Question: In the figure, $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \;and\; \vec{AB} = 2\vec{AP}$, find the value of $\vec{OP}$ in terms of $\vec{a}$ and $\vec{b}$.

Solution:

Given: $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \; and \; \vec{AB} = 2\vec{AP}$

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


In ∆ OAP,
Using ∆ law of vector addition,
$\vec{AP} = \vec{AO} + \vec{OP}$

$or, \vec{AP} = \vec{OP} - \vec{OA}$

$or, \vec{AP} = \vec{OP} - \vec{a}$ - (i)

In ∆ OAB,
Using ∆ law of vector addition,
$\vec{AB} = \vec{AO} + \vec{OB}$

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

$or, \vec{AB} = \vec{b} - \vec{a}$

We have,

$\vec{AB} = 2 \vec{AP}$ - (ii)

[ Put value of $\vec{AB}$ and $\vec{AP}$ from equation (ii) and equation (i) respectively. ]

$or, \vec{b} - \vec{a} = 2 (  \vec{OP} - \vec{a})$

$or, \vec{b} - \vec{a} = 2\vec{OP} - 2\vec{a}$

$or, 2\vec{a} - \vec{a} + \vec{b} = 2 \vec{OP}$

$or, 2\vec{OP} = \vec{a} + \vec{b}$

$or, \vec{OP} = \dfrac{\vec{a} + \vec{b}}{2}$

$\therefore \vec{OP} = \dfrac{1}{2} (\vec{a} + \vec{b})$

Therefore, the required value of $\vec{OP}$ in terms of $\vec{a} \; and \; \vec{b}$ is $ \frac{1}{2} (\vec{a} + \vec{b})$.