Question: In the diagram, $\vec{PQ} = \vec{p}  \; and \;  \vec{QS} = \vec{q}. If \frac{ \vec{QS}}{ \vec{SR}} = \frac{1}{3}$, express $\vec{PR}$ in terms of $\vec{p}$ and $\vec{q}$.

Solution:

Given: $\vec{PQ} = \vec{p}, \vec{QS} = \vec{q}  \; and \;   \frac{ \vec{QS}}{ \vec{SR}} = \frac{1}{3}$

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



In ∆ PQS, using ∆ law of vector addition,

$\vec{PS} = \vec{PQ} + \vec{QS}$

$or, \vec{PS} = \vec{p} + \vec{q}$

We have,

$\dfrac{QS}{SR} = \dfrac{1}{3}$

$or, \dfrac{ \vec{QS} }{ \vec{SR} } = \dfrac{1}{3}$

$or, 3 \vec{QS} = \vec{SR}$

$or, 3 \vec{q} = \vec{SP} + \vec{PR}$

$or, 3 \vec{q} = \vec{PR} - \vec{PS}$

$or, 3\vec{q} + \vec{PS} = \vec{PR}$

$or, 3\vec{q} + ( \vec{p} + \vec{q} ) = \vec{PR}$

$or, 3\vec{q} + \vec{p} + \vec{q} = \vec{PR}$

$\therefore \vec{PR} = \vec{p} + 4 \vec{q}$

Hence, the required value of $\vec{PR}$ is $\vec{p} + 4 \vec{q}$.