Question

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 :

• A. $\frac{1}{\sqrt{30}}$
• B. $\frac{1}{\sqrt{10}}$
• C. $\frac{1}{\sqrt{15}}$
• D. $\frac{1}{ 2 \sqrt{15}}$

Answer 1

Answer: (C)

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