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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    If the points \[A(2,-9,\lambda )\] and \[B(\lambda ,-1,-3)\]lie on the opposite sides of a plane which contains the lines\[\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\] and \[\frac{x}{1}=\frac{7-y}{3}=\frac{z+7}{2},\] then the number of integral values of \[\lambda \] is

    A) \[0\]                       

    B)        \[1\]                       

    C) \[2\]         

    D)        \[3\]

    Correct Answer: C

    Solution :

    [c] We have lines: \[{{L}_{1}}:\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\] and \[{{L}_{2}}:\frac{x}{1}=\frac{7-y}{3}=\frac{z+7}{2}.\] Points on the lines \[{{L}_{1}}\] and \[{{L}_{2}}\] are:             \[P(-1-3r,\,3+2r,-2+r)\] and \[Q(t,7-3t,\,-7+2t)\] These lines are intersecting each other. So,\[-1-3r=t,\text{ }3+2r=7-3t\]and \[r-2=2t-7.\] Solving these, we get \[r=-1\]and\[t=2\]. Thus, point of intersection is \[(2,1,-3)\]. So, equation of plane is: \[\left| \begin{matrix}    x-2 & y-1 & z+3  \\    -3 & 2 & 1  \\    1 & 3 & 2  \\ \end{matrix} \right|=0\] or  \[x+y+z=0\] Points \[A\,\,(2,-9,\lambda )\]and \[B\,\,(\lambda ,-1,-3)\] lie on the opposite sides of the plane. \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2-9+\lambda )\,\,(\lambda -1-3)<0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\lambda -7)\,(\lambda -4)<0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4<\lambda <7\]             


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