Q76 / 125 · BITSAT 2010 · Mathematics

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 -

  • A
    \((1 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
  • B
    \((\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
  • C
    \(2(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
  • Correct answer
    D
    \((2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

Solution

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 \(\triangle \mathrm{DEF}\) \(=\frac{1}{3}\left[\frac{\mathrm{i}+\mathrm{j}}{2}+\mathrm{k}+\mathrm{i} \frac{\mathrm{k}+\mathrm{j}}{2}+\frac{\mathrm{i}+\mathrm{k}}{2}+\mathrm{j}\right]=\frac{2}{3}[\mathrm{i}+\mathrm{j}+\mathrm{k}]\)

Hence, the correct answer is option: 'D'.

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