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

  • question_answer If the length of the perpendicular from the point \[(\beta ,0,\beta )(\beta \ne 0)\] to the line,\[\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}\] is \[\sqrt{\frac{3}{2}},\]then \[\beta \]is equal to : [JEE Main 10-4-2019 Morning]

    A) -1

    B) 2

    C) -2                                

    D) 1

    Correct Answer: A

    Solution :

    One of the point on line is \[P\left( 0,1,1 \right)\] and given point is \[Q\left( \beta ,0,\beta  \right).\] So, \[\overrightarrow{PQ}=\beta \hat{i}-\hat{j}+(\beta +1)\hat{k}\] Hence, \[{{\beta }^{2}}+1+{{(\beta +1)}^{2}}-\frac{{{\left( \beta -\beta -1 \right)}^{2}}}{2}=\frac{3}{2}\] \[\Rightarrow 2{{\beta }^{2}}+2\beta =0\]\[\Rightarrow \beta =0,-1\] \[\Rightarrow \beta =-1\](as\[\beta \ne 0\])                             

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