If P, Q, R are the mid-points of the sides AB | vedclass

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(B) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Since $P, Q, R$ are mid-points of $AB, BC, CA$,their po

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$PQ$

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(B) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Since $P, Q, R$ are mid-points of $AB, BC, CA$,their position vectors are $\vec{p} = \frac{\vec{a} + \vec{b}}{2}$,$\vec{q} = \frac{\vec{b} + \vec{c}}{2}$,and $\vec{r} = \frac{\vec{c} + \vec{a}}{2}$ respectively.Now,we calculate the vectors $PC$ and $BQ$:$\vec{PC} = \vec{c} - \vec{p} = \vec{c} - \frac{\vec{a} + \vec{b}}{2} = \frac{2\vec{c} - \vec{a} - \vec{b}}{2}$$\vec{BQ} = \vec{q} - \vec{b} = \frac{\vec{b} + \vec{c}}{2} - \vec{b} = \frac{\vec{c} - \vec{b}}{2}$Therefore,$\vec{PC} - \vec{BQ} = \frac{2\vec{c} - \vec{a} - \vec{b}}{2} - \frac{\vec{c} - \vec{b}}{2} = \frac{2\vec{c} - \vec{a} - \vec{b} - \vec{c} + \vec{b}}{2} = \frac{\vec{c} - \vec{a}}{2}$.Now,check the options:$PQ = \vec{q} - \vec{p} = \frac{\vec{b} + \vec{c}}{2} - \frac{\vec{a} + \vec{b}}{2} = \frac{\vec{c} - \vec{a}}{2}$.$AR = \vec{r} - \vec{a} = \frac{\vec{c} + \vec{a}}{2} - \vec{a} = \frac{\vec{c} - \vec{a}}{2}$.Thus,$PC - BQ = PQ = AR$.

If $P, Q, R$ are the mid-points of the sides $AB, BC$ and $CA$ of $\triangle ABC$ respectively,then $PC - BQ =$

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