A) \[\left( \frac{a}{3},\frac{a}{3},\frac{a}{3} \right)\]
B) \[\left( \frac{a}{2},\frac{a}{2},\frac{a}{2} \right)\]
C) \[(a,a,a)\]
D) \[(2a,2a,2a)\]
Correct Answer: B
Solution :
[b] Let the points A(x, y, z) is equidistant from the points B(a, 0, 0), C(0, a, 0), D(0, 0, a) and E(0, 0, 0). Hence, \[{{(x-a)}^{2}}+{{y}^{2}}+{{z}^{2}}={{x}^{2}}+{{(y-a)}^{2}}+{{z}^{2}}\] \[={{x}^{2}}+{{y}^{2}}+{{(z-a)}^{2}}\] \[={{x}^{2}}+{{y}^{2}}+{{z}^{2}}\] \[\Rightarrow {{(x-a)}^{2}}+{{y}^{2}}+{{z}^{2}}={{x}^{2}}+{{(y-a)}^{2}}+{{z}^{2}}\] \[\Rightarrow {{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{z}^{2}}={{x}^{2}}+{{y}^{2}}+{{a}^{2}}\] \[-2ay+{{z}^{2}}\] \[\Rightarrow -2ax=-2ay\Rightarrow ax=ay\Rightarrow x=y\] Similarly, \[ay=az\Rightarrow y=z\Rightarrow x=y=z\] \[\therefore \,\,{{(x-a)}^{2}}+{{x}^{2}}+{{x}^{2}}={{x}^{2}}+{{x}^{2}}+{{x}^{2}}\] \[\Rightarrow {{x}^{2}}+{{a}^{2}}-2ax+{{x}^{2}}+{{x}^{2}}=3{{x}^{2}}\] \[\Rightarrow {{a}^{2}}=2ax\Rightarrow x=\frac{a}{2}\] \[\therefore \] Points is \[\left( \frac{a}{2},\frac{a}{2},\frac{a}{2} \right)\].
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