TS EAMCET · Maths · Three Dimensional Geometry

Let $A (1, 2, 3), B (- 1, 4, 6), C (0, - 6, 4)$ and $D (1, 1, 1)$ be the vertices of a tetrahedron, $G$ be its centroid and $G_{1}$ be the centroid of its face $B C D$ . Then $\frac{A G_{1}}{A G} =$

A $\frac{5}{3}$
B $\frac{4}{3}$
C $\frac{7}{6}$
D $\frac{5}{4}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{4}{3}$

Step-by-step Solution

Detailed explanation

Given points A1,2,3,B-1,4,6,C0,-6,4 and D1,1,1 be the vertices of a tetrahedron Now G be its centroid of tetrahedron given by 1-1+0+14,2+4-6+14,3+6+4+14≡14,14,144 And G1 be the centroid of its face BCD which is 03,-13,113 Now using distance formula we get, AG1=629 and…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Let A1,2,3,B-1,4,6,C0,-6,4 and D1,1,1 be the vertices of a tetrahedron, G be its centroid and G1 be the centroid of its face BCD. Then AG1AG=

Step-by-step Solution

Let $A (1, 2, 3), B (- 1, 4, 6), C (0, - 6, 4)$ and $D (1, 1, 1)$ be the vertices of a tetrahedron, $G$ be its centroid and $G_{1}$ be the centroid of its face $B C D$ . Then $\frac{A G_{1}}{A G} =$