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}.