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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    A variable plane\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]at a unit distance from the origin cuts the coordinate axes \[A,\,\,B\] and\[C\]. Centroid \[(x,\,\,y,\,\,z)\] of \[\Delta ABC\] satisfies the equation\[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=k\]. The value of\[k\]is

    A) \[9\]                                     

    B) \[3\]

    C) \[1/9\]                                 

    D) \[1/3\]

    Correct Answer: A

    Solution :

    Given plane cuts the coordinate axes at\[A(a,\,\,0,\,\,0)\],\[B(0,\,\,b,\,\,0)\] and\[C(0,\,\,0,\,\,c)\]. It is at a unit distance from the origin. \[\therefore \]  \[\frac{1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}}}=1\] \[\Rightarrow \]               \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=1\]                  ? (i) Since, \[(x,\,\,y,\,\,z)\] is the centroid of\[\Delta ABC\]. \[\therefore \]  \[x=\frac{a}{3},\,\,y\frac{b}{3}\,\,and\,\,z=\frac{c}{3}\] \[\Rightarrow \]               \[a=3x,\,\,b=3y\,\,c=3z\] On substituting the values of a, b and c in Eq. (i), we get                 \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=9\] \[\therefore \]  \[k=9\]


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