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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    A plane passes through \[(1,\,\,-2,\,\,1)\] and is perpendicular to two planes \[2x-2y+z=0\] and\[x-y+2z=4\]. The distance of the plane from the point \[(1,\,\,2,\,\,2)\] is

    A) \[0\]                                     

    B) \[1\]

    C) \[\sqrt{2}\]                                        

    D) \[2\sqrt{2}\]

    Correct Answer: D

    Solution :

    The equation of a plane passing through\[(1,\,\,-2,\,\,1)\]is                 \[a(x-1)+b(y+2)+c(z-1)=0\]          ... (i) It is perpendicular to \[2x-2y+z=0\] and\[x-y+2z=4\] \[\therefore \]  \[2a-2b+c=0\]and\[a-b+2c=0\] \[\Rightarrow \]               \[\frac{a}{-4+1}=\frac{-b}{4-1}=\frac{c}{-2+2}\] \[\Rightarrow \]               \[\frac{a}{-3}=\frac{b}{-3}=\frac{c}{0}\] On substuting the values of \[a,\,\,b\] and \[c\] in Eq. (i), we get                 \[-3(x-1)-3(y+2)+0(z-1)=0\] \[\Rightarrow \]               \[x+y+1=0\] Distance of this plane from\[(1,\,\,2,\,\,2)\] is given by                 \[d=\frac{1+2+1}{\sqrt{1+1}}=2\sqrt{2}\]


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