Let A(3, 0, -1),B(2, 10, 6),and C(1, 2, 1) be | vedclass

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(D) The midpoint $M$ of $AC$ is given by $M = \left( \frac{3+1}{2}, \frac{0+2}{2}, \frac{-1+1}{2} \right) = (2, 1, 0)$.Since $G$ divides $BM$ in the r

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$\frac{1}{\sqrt{15}}$

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(D) The midpoint $M$ of $AC$ is given by $M = \left( \frac{3+1}{2}, \frac{0+2}{2}, \frac{-1+1}{2} ight) = (2, 1, 0)$.Since $G$ divides $BM$ in the ratio $2:1$,$G$ is the centroid of $	riangle ABC$.The coordinates of the centroid $G$ are $\left( \frac{3+2+1}{3}, \frac{0+10+2}{3}, \frac{-1+6+1}{3} ight) = (2, 4, 2)$.The vector $\overrightarrow{OG} = 2\hat{i} + 4\hat{j} + 2\hat{k}$ and $\overrightarrow{OA} = 3\hat{i} + 0\hat{j} - 1\hat{k}$.The dot product $\overrightarrow{OG} \cdot \overrightarrow{OA} = (2)(3) + (4)(0) + (2)(-1) = 6 - 2 = 4$.The magnitude $|\overrightarrow{OG}| = \sqrt{2^2 + 4^2 + 2^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.The magnitude $|\overrightarrow{OA}| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.Thus,$\cos(\angle GOA) = \frac{\overrightarrow{OG} \cdot \overrightarrow{OA}}{|\overrightarrow{OG}| |\overrightarrow{OA}|} = \frac{4}{(2\sqrt{6})(\sqrt{10})} = \frac{4}{2\sqrt{60}} = \frac{2}{2\sqrt{15}} = \frac{1}{\sqrt{15}}$.

Let $A(3, 0, -1)$,$B(2, 10, 6)$,and $C(1, 2, 1)$ be the vertices of a triangle and $M$ be the midpoint of $AC$. If $G$ divides $BM$ in the ratio $2 : 1$,then $\cos(\angle GOA)$ ($O$ being the origin) is equal to

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