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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Critical Thinking

  • question_answer
    The equation of the planes passing through the line of intersection of the planes \[3x-y-4z=0\] and \[x+3y+6=0\] whose distance from the origin is 1, are

    A) \[x-2y-2z-3=0\], \[2x+y-2z+3=0\]

    B) \[x-2y+2z-3=0\], \[2x+y+2z+3=0\]

    C) \[x+2y-2z-3=0\], \[2x-y-2z+3=0\]

    D) None of these

    Correct Answer: A

    Solution :

    • Equation of planes passing through intersecting the planes \[3x-y-4z=0\]  and \[x+3y+6=0\] is, \[(3x-y-4z)+\lambda \,(x+3y+6)=0\]       --(i)           
    • Given, distance of plane (i) from origin is 1.             
    • \[\therefore \] \[\frac{6\lambda }{\sqrt{{{(3+\lambda )}^{2}}+{{(3\lambda -1)}^{2}}+{{4}^{2}}}}=1\]           
    • or   \[36{{\lambda }^{2}}=10{{\lambda }^{2}}+26\]or \[\lambda =\pm 1\]           
    • Put the value of \[\lambda \] in (i),           
    • \[\therefore \,\,\,(3x-y-4z)\,\pm \,(x+3y+6)=0\]           
    • or   \[4x+2y-4z+6=0\] or \[2x+y-2z+3=0\]           
    • and \[2x-4y-4z-6=0\]           
    • Thus the required planes are \[x-2y-2z-3=0\] and  \[2x+y-2z+3=0\].


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