JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Plane

  • question_answer
    The distance between the planes \[x+2y+3z+7=0\] and \[2x+4y+6z+7=0\] is [MP PET 1991]

    A)            \[\frac{\sqrt{7}}{2\sqrt{2}}\]

    B)            \[\frac{7}{2}\]

    C)            \[\frac{\sqrt{7}}{2}\]

    D)            \[\frac{7}{2\sqrt{2}}\]

    Correct Answer: A

    Solution :

                Required distance = \[\frac{7-\frac{7}{2}}{\sqrt{1+4+9}}=\frac{\sqrt{7}}{2\sqrt{2}}\].


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