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JEE Main & Advanced Sample Paper JEE Main Sample Paper-22

  • question_answer
    If the lines \[x=1\text{ }4-a,\text{ }y=-3-\lambda a,\text{ }z=1+\lambda a\] and \[x=\frac{b}{2},y=1+b,z=2-b\] are coplanar, then \[\lambda \] is equalto

    A)  - 3                              

    B)  2

    C)  1                                

    D)  - 2

    Correct Answer: D

    Solution :

    The given lines are \[\frac{x-1}{1}\,=\frac{y+3}{-\lambda }\,=\frac{z-1}{\lambda }=(a)\] and \[\frac{x-0}{1/2}\,=\frac{y-1}{1}=\frac{z-2}{-1}=(b)\] \[\therefore \] for coplanarity, we must have \[\left| \begin{matrix}    -1 & 4 & 1  \\    1 & -\lambda  & \lambda   \\    1/2 & 1 & -1  \\ \end{matrix} \right|=0\] \[\Rightarrow \,-1(\lambda -\lambda )-4\,\left( -1-\frac{\lambda }{2} \right)\,+\left( 1+\frac{\lambda }{2}\, \right)=0\] \[\Rightarrow \,4+2\lambda \,+1+\frac{\lambda }{2}=0\] \[\Rightarrow \,\,5+\frac{5\lambda }{2}\,=0\] \[\therefore \,\lambda =-2\]


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