JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    Let \[A\left( 3,0,1 \right),B\left( 2,10,6 \right)\]and \[C\left( 1,2,1 \right)\]be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then \[\cos \left( \angle GOA \right)\](O being the origin) is equal to: [JEE Main 10-4-2019 Morning]

    A) \[\frac{1}{\sqrt{30}}\]                       

    B) \[\frac{1}{6\sqrt{10}}\]

    C) \[\frac{1}{\sqrt{15}}\]           

    D) \[\frac{1}{2\sqrt{15}}\]

    Correct Answer: C

    Solution :

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

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