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  • Stuck here, help me understand: - Three Dimensional Geometry - JEE Main-6

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 :

  • 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}}

 

Answers (1)

best_answer

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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solutionqc

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