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If the middle points of sides $\mathrm{BC}, \mathrm{CA} \& \mathrm{AB}$ of triangle $\mathrm{ABC}$ are respectively $\mathrm{D}, \mathrm{E}, \mathrm{F}$ then position vector of centre of triangle $\mathrm{DEF}$, when position vector of $A, \quad B, \quad C$ are respectively $\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{k}+\hat{i}$ is
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The correct answer is:
$\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$
The position vector of points $D, E, F$ are respectively $\hat{i}+\hat{j}+\hat{k}, \hat{i}+\frac{\hat{k}+\hat{j}}{2}$ and $\frac{\hat{i}+\hat{k}}{2}+\hat{j}$
So, position vector of centre of $\Delta \mathrm{DEF}$ $=\frac{1}{3}\left[\frac{\hat{i}+\hat{j}}{2}+\hat{k}+\hat{i} \frac{\hat{k}+\hat{j}}{2}+\frac{\hat{i}+\hat{k}}{2}+\hat{j}\right]$
$=\frac{2}{3}[\hat{i}+\hat{j}+\hat{k}]$
So, position vector of centre of $\Delta \mathrm{DEF}$ $=\frac{1}{3}\left[\frac{\hat{i}+\hat{j}}{2}+\hat{k}+\hat{i} \frac{\hat{k}+\hat{j}}{2}+\frac{\hat{i}+\hat{k}}{2}+\hat{j}\right]$
$=\frac{2}{3}[\hat{i}+\hat{j}+\hat{k}]$
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