Three Dimensional Geometry Question 411
Question: A variable plane at a distance of 1 unit form the origin cuts the coordinate axes at A, B and C. if the centroid $ D(x,y,z) $ of triangle ABC satisfies the relation $ \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=k $ , then the value of k is
Options:
A) 3
B) 1
C) 1/3
D) 9
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let the equation of variable plane be $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1. $ Which meets the axes at $ A(a,0,0),B(0,b,0) $ and $ C(0,0,c) $ . The centroid of $ \Delta ABC $ is $ ( \frac{a}{3},\frac{b}{3},\frac{c}{3} ) $ and it satisfies the relation $ \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=k $ . Thus, $ \frac{9}{a^{2}}+\frac{9}{b^{2}}+\frac{9}{c^{2}}=k $ or $ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{k}{9} $ ??…(i) Also it is given that the distance of the plane $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $ from (0, 0, 0) is 1 unit. Therefore, $ \frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}}=1 $ or $ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=1 $ ?..(ii) From (i) and (ii), we get $ k/e=1, $ i.e., k=9
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