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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    The equation of a plane containing the line \[\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\] and the point (0,7, - 7) is.   JEE Main Online Paper (Held On 26-May-2012)  

    A) \[x+y+z=0\]                       

    B)                        \[x+2y+z=21\]

    C)                        \[3x-2y+5z+35=0\]

    D)                        \[3x+2y+5z+21=0\]

    Correct Answer: A

    Solution :

                    The equation of the plane containing the \[line\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\]is\[a(x+1)+b\] \[(y-3)+c(z+2)=0\]where This passes through (0, 7, - 7) \[\therefore \]\[a(0+1)+b(7-3)+c(-7+2)=0\] \[\Rightarrow \]\[a+4b-5c=0\]                   ...(B) On solving equation (A) and (B) we get \[a=1,b=1,c=1\] \[\therefore \]Required plane is\[x+1+y-3+z+2=0\] \[\Rightarrow \]\[x+y+z=0\]


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