JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    If the points \[(1,1\lambda )\]and (-3, 0, 1) are equidistant    from    the    plane, \[3x+4y-12z+13=0,\]then \[\lambda \] satisfies the equation : JEE Main Online Paper (Held On 10 April 2015)

    A) \[3{{x}^{2}}-10x+21=0\]

    B) \[3{{x}^{2}}-10x+7=0\]

    C) \[3{{x}^{2}}+10x+7=0\]

    D) \[3{{x}^{2}}+10x-13=0\]

    Correct Answer: B

    Solution :

                     \[\left| 3+4-12\lambda +13 \right|=\left| -9+0-12+13 \right|\] \[\left| -12\lambda +20 \right|=\left| 8 \right|\] \[\left| 3\lambda -5 \right|=2\] \[9{{\lambda }^{2}}+25-30\lambda =4\] \[9{{\lambda }^{2}}-30\lambda +21=0\] \[3{{\lambda }^{2}}-10\lambda +7=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner