Three Dimensional Geometry Question 347
Question: The coordinates of point in xy-plane which is equidistant from three points A (2, 0, 3), B (0, 3, 2) and C (0, 0, 1) are
Options:
A) (3, 2, 0)
B) (3, 4, 0)
C) (0, 0, 3)
D) (2, 3, 0)
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We know that z-coordinate of every point on xy-plane is zero. So, let P(x, y, 0) be a point in xy-plane such that $ PA=PB=PC. $ Now $ PA=PB\Rightarrow PA^{2}=PB^{2} $
$ \Rightarrow {{(x-2)}^{2}}+{{(y-0)}^{2}}+{{(0-3)}^{2}} $ $ ={{(x-0)}^{2}}+{{(y-3)}^{2}}+{{(0-2)}^{2}} $
$ \Rightarrow 4x-6y=0 $ or $ 2x-3y=0…(1) $ $ PB=PC\Rightarrow PB^{2}=PC^{2} $
$ \Rightarrow {{(x-0)}^{2}}+{{(y-3)}^{2}}+{{(0-2)}^{2}} $ $ ={{(x-0)}^{2}}+{{(y-0)}^{2}}+{{(0-1)}^{2}} $
$ \Rightarrow -6y+12=0\Rightarrow y=2…(2) $ Putting y = 2 in (1), we obtain x = 3. Hence, the required point is (3, 2, 0).
BETA
