Question

A variable plane $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $ at a unit distance from the origin cuts the coordinate axes $ A,\,\,B $ and $ C $ . Centroid $ (x,\,\,y,\,\,z) $ of $ \Delta ABC $ satisfies the equation $ \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=k $ . The value of $ k $ is

• A. $ 9 $
• B. $ 3 $
• C. $ 1/9 $
• D. $ 1/3 $

Answer 1

Answer: (A)

Given plane cuts the coordinate axes at
$ A(a,\,\,0,\,\,0) $ , $ B(0,\,\,b,\,\,0) $ and $ C(0,\,\,0,\,\,c) $ .
It is at a unit distance from the origin.
$ \therefore $ $ \frac{1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}}}=1 $
$ \Rightarrow $ $ \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=1 $ .. (i)
Since, $ (x,\,\,y,\,\,z) $ is the centroid of $ \Delta ABC $ .
$ \therefore $ $ x=\frac{a}{3},\,\,y\frac{b}{3}\,\,and\,\,z=\frac{c}{3} $
$ \Rightarrow $ $ a=3x,\,\,b=3y\,\,c=3z $
On substituting the values of a, b and c in Eq. (i), we get
$ \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=9 $
$ \therefore $ $ k=9 $