# Let $A(3,0,-1),B(2,10,6)\: \: and\: \: C(1,2,1)$ be thevertices of a triangle and M be the midpoint of AC. If G dividesBM in the ratio , 2:1, then $cos(\angle GOA)$ ( O being the origin)is equal to : Option 1) $\frac{1}{2\sqrt{15}}$ Option 2) $\frac{1}{\sqrt{15}}$ Option 3) $\frac{1}{6\sqrt{10}}$ Option 4) $\frac{1}{\sqrt{30}}$

Given that M is mid point of AC and G divides BM in the ratio 2:1

$\therefore$ G is centroid of ABC

$G=(\frac{3+2+1}{3},\frac{0+10+2}{3},\frac{-1+6+1}{3})=(2,4,2)$

$0A=3\hat{i}+\hat{j}-\hat{k}$

$0G=2\hat{i}+4\hat{j}+2\hat{k}$

$\therefore cos(GOA)=\frac{\vec{OG}. \vec{OA}}{|\vec{OG}||\vec{OA}|}=\frac{6-2}{2\sqrt{10}\sqrt6}$

$\therefore cos(GOA)=\frac{1}{\sqrt{15}}$

So, correct option is (2)

Option 1)

$\frac{1}{2\sqrt{15}}$

Option 2)

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

Option 3)

$\frac{1}{6\sqrt{10}}$

Option 4)

$\frac{1}{\sqrt{30}}$

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