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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Critical Thinking

  • question_answer
    A variable plane at a constant distance p from origin meets the co-ordinates axes in \[A,B,C\]. Through these points planes are drawn parallel to co-ordinate planes. Then locus of the point of intersection is

    A) \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=\frac{1}{{{p}^{2}}}\]

    B) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{p}^{2}}\]

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

    D) \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=p\]

    Correct Answer: A

    Solution :

    • Equation of plane is, \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]           
    • \[\{a,\,\,b,\,\,c\]respectively are intercepts on \[x,\,\,y,\,\,z\] axes}           
    • Then \[\frac{abc}{\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}}=p\]           
    • \[\,\Rightarrow \,\,\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=\frac{1}{{{p}^{2}}}\]           
    • Therefore locus of the point (x, y, z) is                                                              
    • \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=\frac{1}{{{p}^{2}}}\].


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