If the middle points of sides $\mathrm{BC}, \mathrm{CA} ~\&~ \mathrm{AB}$ of triangle $A B C$ are respectively $D, E, F$ then position vector of centre of triangle $\mathrm{DEF}$, when position vector of $A, B, C$ are respectively $\hat{i}+\hat{j}$, $\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{k}}+\hat{\mathrm{i}}$ is -

Question

If the middle points of sides $\mathrm{BC}, \mathrm{CA} ~\&~ \mathrm{AB}$ of triangle $A B C$ are respectively $D, E, F$ then position vector of centre of triangle $\mathrm{DEF}$, when position vector of $A, B, C$ are respectively $\hat{i}+\hat{j}$, $\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{k}}+\hat{\mathrm{i}}$ is -, $(1 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$, $(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$, $2(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$, $(2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$

MARKS App · Accepted Answer

The position vector of points $D, E, F$ are respectively $\frac{\mathrm{i}+\mathrm{j}}{2}+\mathrm{k}, \mathrm{i}+\frac{\mathrm{k}+\mathrm{j}}{2} \text { and } \frac{\mathrm{i}+\mathrm{k}}{2}+\mathrm{j}$ So, position vector of centre of $\triangle \mathrm{DEF}$ $=\frac{1}{3}\left[\frac{\mathrm{i}+\mathrm{j}}{2}+\mathrm{k}+\mathrm{i} \frac{\mathrm{k}+\mathrm{j}}{2}+\frac{\mathrm{i}+\mathrm{k}}{2}+\mathrm{j}\right]=\frac{2}{3}[\mathrm{i}+\mathrm{j}+\mathrm{k}]$

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