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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    Let P be the plane, which contains the line of intersection of the planes, \[x+y+z6=0\] and \[2x+3y+z+5=0\]and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :-             [JEE Main 9-4-2019 Afternoon]

    A) \[63\sqrt{5}\]              

    B) \[205\sqrt{5}\]

    C) \[17/\sqrt{5}\]             

    D) \[11/\sqrt{5}\]

    Correct Answer: D

    Solution :

    \[\lambda \left( x+y+z6 \right)+2x+3y+z+5=0\] \[(\lambda +2)x+(\lambda +3)y+(\lambda +1)z+56\lambda =0\] \[\lambda +1=0\Rightarrow \lambda =1\] \[P:x+2y+11=0\] perpendicular distance\[=\frac{11}{\sqrt{5}}\]


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