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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2010 · Mathematics · Three Dimensional Geometry

If the middle points of sides (BC, CA ~\&~ AB) of triangle A B C...

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

Options

  1. A((1 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}))
  2. B((\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}))
  3. C(2(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}))
  4. D((2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}))

Correct answer

D. ((2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}))

Step-by-step solution

The position vector of points D,E,FD, 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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