The position vector of the centroid of the **tetrahedron** with vertices $A(\vec a), B(\vec b), C(\vec c), D(\vec d)$ is:
A$(\vec a + \vec b + \vec c)/3$
B$(\vec a + \vec b + \vec c + \vec d)/4$
C$(\vec a + \vec b + \vec c + \vec d)/3$
D$(\vec a \cdot \vec b \cdot \vec c \cdot \vec d)$
Answer & Solution
Correct answer: B. $(\vec a + \vec b + \vec c + \vec d)/4$
For a tetrahedron, the centroid divides the line from any vertex to the centroid of the opposite face in ratio 3:1, yielding $\vec g = (\vec a + \vec b + \vec c + \vec d)/4$.
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