A) \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}=16{{p}^{-2}}\]
B) \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}=16{{p}^{-1}}\]
C) \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}=16\]
D) None of these
Correct Answer: A
Solution :
Plane is\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\], where \[p=\frac{1}{\sqrt{\sum\limits_{{}}^{{}}{\left( \frac{1}{{{a}^{2}}} \right)}}}\] or \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=\frac{1}{{{p}^{2}}}\] ?..(i) Now according to equation, \[x=\frac{a}{4},\ \ y=\frac{b}{4},\ \ z=\frac{c}{4}\] Put the values of x, y, z in (i), we get the locus of the centroid of the tetrahedron.
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