Three Dimensional Geometry Question 331
Question: What is the locus of a point which is equidistant from the points (1, 2, 3) and (3, 2, - 1)?
Options:
A) $ x+z=0 $
B) $ x-3z=0 $
C) $ x-z=0 $
D) $ x-2z=0 $
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Answer:
Correct Answer: D
Solution:
[d] Let $ (h,k,\ell ) $ be the point which is equidistant from the points (1, 2, 3) and (3, 2, -1)
$ \Rightarrow \sqrt{{{(h-1)}^{2}}+{{(k-2)}^{2}}+{{(\ell -3)}^{2}}} $
$ \Rightarrow \sqrt{{{(h-3)}^{2}}+{{(k-2)}^{2}}+{{(\ell +1)}^{2}}} $
$ \Rightarrow {{(h-1)}^{2}}+{{(\ell -3)}^{2}}={{(h-3)}^{2}}+{{(\ell +1)}^{2}} $
$ \Rightarrow h^{2}+1-2h+{{\ell }^{2}}-6\ell +9=h^{2}-6h+9+{{\ell }^{2}} $ $ +2\ell +1 $
$ \Rightarrow -2h-6\ell =-6h+2\ell \Rightarrow 6h-2h-6\ell -2\ell =0 $
$ \Rightarrow 4h-8\ell =0\Rightarrow h-2\ell =0 $ Putting h = x and $ \ell =z $ We get locus of points $ (h,k,\ell ) $ as, $ x-2z=0 $
BETA
