Question

A variable plane at a distance of $1$ unit from the origin cuts the co-ordinate 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

• A. $3$
• B. $1$
• C. $1/3$
• D. $9$

Answer 1

The correct option is D $9$

Let the Dcs be $cos\alpha ,cos\beta ,cos\gamma$ .

We have ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

The distance from origin to plane is $1$

The coordinates of $A$ is $(\frac{1}{cos\alpha },0,0)$

The coordinates of $B$ is $(0,\frac{1}{cos\beta },0)$

The coordinates of $C$ is $(0,0,\frac{1}{cos\gamma )}$

The centriod of $ABC$ is $(x,y,z)=(\frac{1}{3cos\alpha },\frac{1}{3cos\beta },\frac{1}{3cos\gamma })$

So we get $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=9({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma )=9$

Therefore option $D$ is correct