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If $P, Q, R$ are the mid points of the sides $A B, B C, C A$, respectively of a triangle $A B C$ and if $\vec{a}, \vec{p}, \vec{q}$ are the position vector of $A, P, Q$ respectively, then what is the position vector of $R ?$
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Verified Answer
The correct answer is:
$\quad \vec{a}-(\vec{p}-\vec{q})$
Let the position vectors of $B, C$ and $R$ are $\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ and
$\overrightarrow{\mathbf{r}}$ respectively.
$\therefore \overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}}{2} \Rightarrow \overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}}$
$\overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{2}$
$\Rightarrow \vec{c}=2 \vec{q}-\vec{b}$
$\Rightarrow \overrightarrow{\mathrm{c}}=2 \overrightarrow{\mathrm{q}}-(2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}})=2 \overrightarrow{\mathrm{q}}-2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}$
and $\overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}}{2}$
$=\frac{2 \overrightarrow{\mathrm{q}}-2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}}}{2}=\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})$
$\overrightarrow{\mathbf{r}}$ respectively.
$\therefore \overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}}{2} \Rightarrow \overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}}$
$\overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{2}$
$\Rightarrow \vec{c}=2 \vec{q}-\vec{b}$
$\Rightarrow \overrightarrow{\mathrm{c}}=2 \overrightarrow{\mathrm{q}}-(2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}})=2 \overrightarrow{\mathrm{q}}-2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}$
and $\overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}}{2}$
$=\frac{2 \overrightarrow{\mathrm{q}}-2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}}}{2}=\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})$
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