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Question: Answered & Verified by Expert
If $\overline{\mathrm{a}}=2 \hat{\imath}+3 \hat{\jmath}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=4 \hat{\imath}+5 \hat{\jmath}+3 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=6 \hat{\imath}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ are the position vectors of the
vertices of a triangle ABC respectively, then the position vector of the intersection of the medians of the triangle $\mathrm{ABC}$ is
MathematicsVector AlgebraMHT CETMHT CET 2020 (20 Oct Shift 1)
Options:
  • A $4 \hat{\imath}+3 \hat{\jmath}+3 \hat{\mathrm{k}}$
  • B $2 \hat{\imath}+3 \hat{\jmath}+3 \hat{\mathrm{k}}$
  • C $5 \hat{\imath}+3 \hat{\jmath}+3 \hat{\mathrm{k}}$
  • D $3 \hat{\imath}+3 \hat{\jmath}+4 \hat{\mathrm{k}}$
Solution:
1280 Upvotes Verified Answer
The correct answer is: $4 \hat{\imath}+3 \hat{\jmath}+3 \hat{\mathrm{k}}$
Centroid $(\overline{\mathrm{g}})=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}}{3}=\frac{12 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}}{3}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

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