TS EAMCET · Maths · Vector Algebra
\(\bar{i}+\bar{j}+\bar{k}, a_1 \bar{i}+b_1 \bar{j}+c_1 \bar{k}, a_2 \bar{i}+b_2 \bar{j}+c_2 \bar{k}, a_3 \bar{i}+b_3 \bar{j}+c_3 \bar{k}\) are the position vectors of the points \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) respectively. \(\frac{2}{3}(\bar{i}+\bar{j}+\bar{k})\) is the position vector of the centroid of the triangular face BCD of the tetrahedron ABCD. If \(\alpha \bar{i}+\beta \bar{j}+\gamma \bar{k}\) is the position vector of the centroid of the tetrahedron, then \(2 \alpha+\beta+\gamma=\)
- A 3
- B 2
- C \(\frac{2}{3}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
Let \(\bar{a}, \bar{b}, \bar{c}, \bar{d}\) be the position vectors of A, B, C, D respectively. Given \(\bar{a} = \bar{i}+\bar{j}+\bar{k}\). Centroid of face BCD: \( \frac{\bar{b}+\bar{c}+\bar{d}}{3} = \frac{2}{3}(\bar{i}+\bar{j}+\bar{k})\). Thus,…
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