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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1), having normal perpendicular to both the lines \[\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}\]and \[\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1},\]is:-        JEE Main Solved Paper-2017

    A) \[\frac{10}{\sqrt{74}}\]                 

    B)  \[\frac{20}{\sqrt{74}}\]

    C)  \[\frac{10}{\sqrt{83}}\]                

    D)  \[\frac{5}{\sqrt{83}}\]

    Correct Answer: C

    Solution :

     Normal vector                 \[\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    1 & -2 & 3  \\    2 & -1 & -1  \\ \end{matrix} \right|=5\hat{i}+7\hat{j}+3\hat{k}\]                 So plane is \[5(x-1)+7(y+1)+3(z+1)=0\]                 \[\Rightarrow \]\[5x+7y+3z+5=0\]                 Distance \[\frac{5+21-21+5}{\sqrt{25+49+9}}=\frac{10}{\sqrt{83}}\]


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