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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 8

  • question_answer
    A plane which passes through the point (3, 2, 0) and the line\[\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\] is

    A) \[x-y+z=1\]      

    B) \[~x+y+z=5\]

    C)  \[x+2y-z=0\]   

    D)  \[2x-y+z=5\]

    Correct Answer: A

    Solution :

    [a] : Any plane passing through (3, 2, 0) is \[a(x-3)+b(y-2)+c(z-0)=0\]                      ...(i) \[\because \]Plane is passing through the line \[\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\] \[\therefore \]\[a(3-3)+b(6-2)+c(4-0)=0\]                \[\Rightarrow \]\[0a+4b+4c=0\]                           ...(ii) Also, the plane is passing through the given line. \[\therefore \]\[a+5b+4c=0\]                             ...(iii) On solving (ii) and (iii), we get \[\frac{a}{16-20}=\frac{b}{4-0}=\frac{c}{0-4}\Rightarrow \frac{a}{-1}=\frac{b}{1}=\frac{c}{-1}\] On putting the values of a, b and c in (i), we get \[x-y+z=1\]


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