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 mid point of $AC$. If $G$ divides $BM$ in the in the ratio, $2:1,$ then $\cos \left(\angle GOA\right)$, ($O$ being the origin) is equal to :

• A. $\frac{1}{\sqrt{15}}$
• B. $\frac{1}{\sqrt{30}}$
• C. $\frac{1}{2\sqrt{15}}$
• D. $\frac{1}{6\sqrt{10}}$

Answer 1

Answer: (A)

Given:

$A(3,0,-1),B(2,10,6)$ and $C(1,2,1)$ are the vertices of a triangle $ABC$.
$G$ divided $BM$ in $2:1$ and $M$ is mid point of $AC$
$\Rightarrow G$ is centroid of the given triangle.

$\Rightarrow G=(\frac{3+2+1}{3},\frac{0+10+2}{3},\frac{-1+6+1}{3})$

$\Rightarrow G=(2,4,2)$ ---(1)

If $\theta$ be the angle $\angle GOA$ then
$\overrightarrow{OA}\cdot \overrightarrow{OG}=\left|\overrightarrow{OA}\right|\cdot \left|\overrightarrow{OG}\right|\cdot \cos \theta$ ---(2)

Here, $\overrightarrow{OA}=(3\hat{i}-0\hat{i})+(0\hat{j}-0\hat{j})+(-\hat{k}-0\hat{k})$

$\Rightarrow \overrightarrow{OA}=3\hat{i}-\hat{k}$ --(3)

$\overrightarrow{OG}=(2\hat{i}-0\hat{i})+(4\hat{j}-0\hat{j})+(2\hat{k}-0\hat{k})$

$\Rightarrow \overrightarrow{OG}=2\hat{i}+4\hat{i}+2\hat{k}$--(4)

Now, from (2)

$\overrightarrow{OA}\cdot \overrightarrow{OG}=\left|\overrightarrow{OA}\right|\cdot \left|\overrightarrow{OG}\right|\cdot \cos \theta$

$\Rightarrow (3\hat{i}-\hat{k})\cdot (2\hat{i}+5\hat{j}+2\hat{k})=\sqrt{3^2+(-1{)}^2}\cdot \sqrt{2^2+4^2+2^2}\cos \theta$

$\Rightarrow (6+0-2)=\sqrt{10}\cdot \sqrt{24}\cos \theta$

$\Rightarrow 4=\sqrt{2}.\sqrt{5}.2.\sqrt{3}.\sqrt{2}.\cos \theta$
$\Rightarrow 4=4\sqrt{15}\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{15}}$

Hence, Option A is correct.