In the figure, vec.OA = vec.a, vec.OB = vec.b and vec.AB = 3 vec.AP. Find the value of vec.OP in terms of vec.a and vec.b. | Vector Geometry | SciPiPupil

Question: In the figure, $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \; and \; \vec{AB} = 3 \vec{AP}$. Find the value of $\vec{OP}$ in terms of $\vec{a}$ and $\vec{b}$.

Solution:

Given: $\vec{OA} = \vec{a}, \vec{OB} = \vec{b} \; and \; \vec{AB} = 3 \vec{AP}$

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

In ∆ OAB,

Using ∆ law of vector addition,

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

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

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

In ∆ OAP,

Using ∆ law of vector addition,

$\vec{AP} = \vec{AO} + \vec{OP}$

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

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

We have,

$\vec{AB} = 3 \vec{AP}$

[ Put the respective values from above]

$or, \vec{b} - \vec{a} = 3 ( \vec{OP} - \vec{a} )$

$or, \vec{b} - \vec{a} = 3 \vec{OP} - 3 \vec{a}$

$or, 3 \vec{a} - \vec{a} + vec{b} = 3 \vec{OP}$

$or, 2 \vec{a} + vec{b} = 3 \vec{OP}$

$\therefore \vec{OP} = \dfrac{1}{3} (2\vec{a} + \vec{b} )$

Therefore, the required value of $\vec{OP}$ is $\frac{1}{3} (2\vec{a} + \vec{b} )$