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

  • question_answer
    If the straight lines \[x=1+s,\] \[y=-3-\lambda s,\] \[z=1+\lambda s\] and \[x=t/2,y=1+t,z=2-t\], with parameters s and \[t\] respectively, are co-planar, then \[\lambda \]equals [AIEEE 2004]

    A) 0

    B) -1

    C) -1/2

    D) -2

    Correct Answer: D

    Solution :

    • We have, \[\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=s\]                   
    • and         \[\frac{x-0}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}=t\]                   
    • Since, lines are coplanar then                   
    • \[\left| \,\begin{matrix}    {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}}  \\    {{l}_{1}} & {{m}_{1}} & {{n}_{1}}  \\    {{l}_{2}} & {{m}_{2}} & {{n}_{2}}  \\ \end{matrix}\, \right|\,=\,0\]Þ \[\left| \,\begin{matrix}    -1 & 4 & 1  \\    1 & -\lambda  & \lambda   \\    1/2 & 1 & -1  \\ \end{matrix}\, \right|\,=\,0\] 
    • On solving, \[\lambda =-2\].


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