Question

If a variable plane, at a distance of $3$ units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of $\Delta$ABC is

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3$
• B. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1$
• C. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{9}$
• D. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=9$

Answer 1

The correct option is B $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1$

Let the plane equation be $ax+by+cz+d=0$

Distance of the plane from origin is 3,

$\therefore \frac{d}{\sqrt{a^2+b^2+c^2}}=3$

i.e. $d^2=9(a^2+b^2+c^2)...\left(1\right)$

Now, the plane would intersect the $x$ axis at $A(\frac{-d}{a},0,0),y$ axis at $B(0,\frac{-d}{b},0)$ and $z$ axis at $C(0,0,\frac{-d}{c})$

The centroid of this triangle would have the co-ordinates $(\frac{-d}{3a},\frac{-d}{3b},\frac{-d}{3c})$

Let $h=\frac{-d}{3a},k=\frac{-d}{3b},l=\frac{-d}{3c}$

Using equation (1), we can write $\frac{1}{9}=\frac{1}{9h^2}+\frac{1}{9k^2}+\frac{1}{9l^2}$

$\Rightarrow 1=\frac{1}{h^2}+\frac{1}{k^2}+\frac{1}{l^2}$