• # question_answer Let $A\left( 3,0,1 \right),B\left( 2,10,6 \right)$and $C\left( 1,2,1 \right)$be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then $\cos \left( \angle GOA \right)$(O being the origin) is equal to: [JEE Main 10-4-2019 Morning] A) $\frac{1}{\sqrt{30}}$                        B) $\frac{1}{6\sqrt{10}}$ C) $\frac{1}{\sqrt{15}}$            D) $\frac{1}{2\sqrt{15}}$

G is the centroid of $\Delta ABC$             $G\equiv (2,4,2)$             $\overrightarrow{OG}=2\hat{i}+4\hat{j}+2\hat{k}$             $\overrightarrow{OA}=3\hat{i}-\hat{k}$             $\cos (\angle GOA)=\frac{\overrightarrow{OG}.\overrightarrow{OA}}{|\overrightarrow{OG}|\,\,|\overrightarrow{OA}|}=\frac{1}{\sqrt{15}}$