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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 26

  • question_answer
    Let P be the plane containing the line \[y+z=2,\] \[\text{x}=0\] and parallel to the line \[x-z=2,\,y=0.\] Then the distance of plane P from origin is

    A) \[\frac{1}{\sqrt{3}}\]                 

    B)        \[\frac{2}{\sqrt{3}}\]

    C) \[\sqrt{3}\]                    

    D)        \[2\]

    Correct Answer: B

    Solution :

    [b] Equation of plane through the line \[y+z=2,\] \[x=0\]is: \[(y+z-2)+\lambda x=0\]                       ...(1) This plane is parallel to the line \[x-z=2,\] \[y=0\] So, plane (1) is perpendicular to the planes \[x-z=0\]and \[y=0\] \[\therefore \,\,\lambda +0-1=0=0\] or   \[\lambda =1\] Therefore, equation of plane is: \[y+z-2-x=0\]  or   \[x-y-z+2=0\] Distance of this plane from origin \[=\frac{2}{\sqrt{3}}\].


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