Question:
Let $\mathrm{A}(3,0,-1), \mathrm{B}(2,10,6)$ and $\mathrm{C}(1,2,1)$ be the vertices of a triangle and $\mathrm{M}$ be the midpoint of $\mathrm{AC}$. If $\mathrm{G}$ divides $\mathrm{BM}$ in the ratio, $2: 1$, then
$\cos (\angle \mathrm{GOA})$ (O being the origin) is equal to :
Correct Option: , 3
Solution:
$G$ is the centroid of $\triangle A B C$
$G \equiv(2,4,2)$
$\overrightarrow{\mathrm{OG}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{OA}}=3 \hat{\mathrm{i}}-\hat{\mathrm{k}}$
$\cos (\angle \mathrm{GOA})=\frac{\overrightarrow{\mathrm{OG}} \cdot \overrightarrow{\mathrm{OA}}}{|\overrightarrow{\mathrm{OG}}||\overrightarrow{\mathrm{OA}}|}=\frac{1}{\sqrt{15}}$