Three Dimensional Geometry Question 342
Question: What are coordinates of the point equidistant from the points (a, 0, 0), (0, a, 0), (0, 0, a) and (0, 0, 0)?
Options:
A) $ ( \frac{a}{3},\frac{a}{3},\frac{a}{3} ) $
B) $ ( \frac{a}{2},\frac{a}{2},\frac{a}{2} ) $
C) $ (a,a,a) $
D) $ (2a,2a,2a) $
Show Answer
Answer:
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}=3x^{2} $
$ \Rightarrow a^{2}=2ax\Rightarrow x=\frac{a}{2} $
$ \therefore $ Points is $ ( \frac{a}{2},\frac{a}{2},\frac{a}{2} ) $ .
BETA
