TS EAMCET · Maths · Three Dimensional Geometry
ABCD is a tetrahedron. \(\bar{i}-2 \bar{j}+3 \bar{k},-2 \bar{i}+\bar{j}+3 \bar{k}, 3 \bar{i}+2 \bar{j}-\bar{k}\) are the position vectors of the points \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) respectively. \(-\bar{i}+2 \bar{j}-3 \bar{k}\) is the position vector of the centroid of the triangular face BCD. If G is the centroid of the tetrahedron, then \(\mathrm{GD}=\)
- A \(\frac{\sqrt{13}}{\sqrt{2}}\)
- B \(\sqrt{23}\)
- C \(\frac{\sqrt{213}}{\sqrt{2}}\)
- D \(\sqrt{46}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{213}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\vec{d} = 3(-\bar{i}+2 \bar{j}-3 \bar{k}) - ((-2 \bar{i}+\bar{j}+3 \bar{k}) + (3 \bar{i}+2 \bar{j}-\bar{k})) = (-3\bar{i}+6 \bar{j}-9 \bar{k}) - (\bar{i}+3\bar{j}+2\bar{k}) = -4\bar{i}+3\bar{j}-11\bar{k}\)…
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