Question:
If \(G(2,-1,2)\) is the centroid of tetrahedron \(OABC\) where \(O = (0,0,0)\) and \(G_1\) is the centroid of \(\Delta ABC\), then \(|OG_1| = ?\)
If \(G(2,-1,2)\) is the centroid of tetrahedron \(OABC\) where \(O = (0,0,0)\) and \(G_1\) is the centroid of \(\Delta ABC\), then \(|OG_1| = ?\)
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For a tetrahedron, relate the centroid of the tetrahedron to the centroid of a face to find distances from the origin or vertices.
Updated On: Jul 18, 2026
- 1
- \(\frac{3}{2}\)
- 4
- \(\frac{9}{2}\)
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The Correct Option is C
Solution and Explanation
Step 1: Recall formula for centroid of tetrahedron.
Centroid \(G = \frac{O + A + B + C}{4}\)
Step 2: Recall formula for centroid of triangle.
Centroid \(G_1 = \frac{A + B + C}{3}\)
Step 3: Express vector OG1 in terms of G and O.
\[ G = \frac{O + 3G_1}{4} \Rightarrow 3G_1 = 4G - O \Rightarrow G_1 = \frac{4G}{3} \]
Step 4: Compute coordinates of G1.
\[ G_1 = \frac{4}{3} (2,-1,2) = \left(\frac{8}{3}, -\frac{4}{3}, \frac{8}{3}\right) \]
Step 5: Compute distance from O.
\[ |OG_1| = \sqrt{\left(\frac{8}{3}\right)^2 + \left(-\frac{4}{3}\right)^2 + \left(\frac{8}{3}\right)^2} = \sqrt{\frac{64+16+64}{9}} = \sqrt{16} = 4 \]
Step 6: Final conclusion.
\[ \boxed{4} \]
Centroid \(G = \frac{O + A + B + C}{4}\)
Step 2: Recall formula for centroid of triangle.
Centroid \(G_1 = \frac{A + B + C}{3}\)
Step 3: Express vector OG1 in terms of G and O.
\[ G = \frac{O + 3G_1}{4} \Rightarrow 3G_1 = 4G - O \Rightarrow G_1 = \frac{4G}{3} \]
Step 4: Compute coordinates of G1.
\[ G_1 = \frac{4}{3} (2,-1,2) = \left(\frac{8}{3}, -\frac{4}{3}, \frac{8}{3}\right) \]
Step 5: Compute distance from O.
\[ |OG_1| = \sqrt{\left(\frac{8}{3}\right)^2 + \left(-\frac{4}{3}\right)^2 + \left(\frac{8}{3}\right)^2} = \sqrt{\frac{64+16+64}{9}} = \sqrt{16} = 4 \]
Step 6: Final conclusion.
\[ \boxed{4} \]
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