Three Dimensional Geometry Question 317
Question: Find the equation of set points P such that $ PA^{2}+PB^{2}=2K^{2}, $ where A and B are the points (3, 4, 5) and (-1, 3, -7), respectively:
Options:
A) $ K^{2}-109 $
B) $ 2K^{2}-109 $
C) $ 3K^{2}-109 $
D) $ 4K^{2}-10 $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the coordinates of point P be (x, y, z). Here, $ PA^{2}={{(x-3)}^{2}}+{{(y-4)}^{2}}+{{(z-5)}^{2}} $ $ PB^{2}={{(x+1)}^{2}}+{{(y-3)}^{2}}+{{(z+7)}^{2}} $ By the given condition $ PA^{2}+PB^{2}=2K^{2} $ We have $ {{(x-3)}^{2}}+{{(y-4)}^{2}}+{{(z-5)}^{2}}+{{(x+1)}^{2}} $ $ +{{(y-3)}^{2}}+{{(z+7)}^{2}}=2K^{2} $ i.e. $ 2x^{2}+2y^{2}+2z^{2}-4x-14y+4z=2K^{2}-109 $
BETA
