JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    If the lines \[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\]and \[\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\]are coplanar. then k is equal to

    A)  \[\frac{3}{2}\]                                  

    B)  \[\frac{5}{2}\]

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

    D)  \[\frac{9}{2}\]

    Correct Answer: D

    Solution :

     Idea If the lines \[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}\] and \[\frac{x-{{x}_{2}}}{{{a}_{2}}}=\frac{y-{{y}_{2}}}{{{b}_{2}}}=\frac{z-{{z}_{2}}}{{{c}_{2}}}\]are coplanar, then \[\left| \begin{matrix}    {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}}  \\    {{a}_{1}} & {{b}_{1}} & {{c}_{1}}  \\    {{a}_{2}} & {{b}_{2}} & {{c}_{2}}  \\ \end{matrix} \right|=0\] Here, for the given lines if the given lines are coplanar, then\[\left| \begin{matrix}    3-1 & k-1 & 0-1  \\    2 & 3 & 4  \\    1 & 2 & 1  \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[-10+(k+1)2-1=0\]\[\Rightarrow \]\[k=9/2\] TEST Edge Equation of lines in various form related questions are asked. To solve such type of question, students are advised to understand the concept of line.


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