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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    The equation of plane containing the line \[\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\] and 'the point \[(0,\,\,7,\,\,-7)\] is

    A) \[x+y+z=1\]

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

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

    D)  None of the above

    Correct Answer: D

    Solution :

    The equation of the plane containing the line\[\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\]is                 \[a(x+1)+b(y-3)+c(z+2)=0\]         ... (i) where\[,\]          \[-3a+2b+c=0\]                 ... (ii) This passes through\[(0,\,\,7,\,\,-7)\] \[\therefore \]  \[a+4b-5c=0\]                                    ... (iii) From Eqs. (ii) and (iii), we get                 \[\frac{a}{-14}=\frac{b}{-14}=\frac{c}{-14}\]or\[\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\] So, the required plane is\[~x+y+z=0\].


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