Question

A variable plane at a distance of $1$ unit from the origin cuts 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 find the value of $k$

• A. $7$
• B. $10$
• C. $2$
• D. $9$

Answer 1

The correct option is D $9$

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 $C(0,0,c)$ respectively.

Centroid of $△ABC$ is $(\frac{a}{3},\frac{b}{3},\frac{c}{3})$ and it satisfies the relation $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$

$\Rightarrow \frac{9}{a^2}+\frac{9}{b^2}+\frac{9}{c^2}=k\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{k}{9}$ ...(1)

Also given that the distance of plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ from $(0,0,0)$ is $1$ unit.

$\Rightarrow \frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1$ ...(2)

From (1) and (2), we get

$\frac{k}{9}=1\Rightarrow k=9$