Question

If a plane is at a distance $3/2$ from the origin $0$, and meets the axes in $A$, $B$ and $C$, the coordinates of the centroid can be

• A. $(1,\frac{1}{\sqrt{2}},1)$
• B. $\left(\frac{1}{\sqrt{2}},1,1\right)$
• C. $(1,1,\frac{1}{\sqrt{2}})$
• D. $(1,1,1)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\left(\frac{1}{\sqrt{2}},1,1\right)$
B $(1,1,\frac{1}{\sqrt{2}})$
C $(1,\frac{1}{\sqrt{2}},1)$

Plane at a distance of $\frac{3}{2}$ from origin

So plane is $ax+by+cz+k=0$

$D=\frac{k}{\sqrt{a^2+b^2+c^2}}=\frac{3}{2}$

$D=\frac{a^2+b^2+c^2}{k^2}=\frac{4}{9}$

Now

$x(\frac{-k}{a},0,0)$

$y(0,\frac{-k}{b},0)$

$z(0,0,\frac{-k}{c})$

$Centroid(\frac{-k}{3a},\frac{-k}{3b},\frac{-k}{3c})$

Now let $(\frac{-k}{3a}=1,\frac{-k}{3c}=1)$

$\frac{-a}{k}=\frac{1}{3},\frac{-c}{k}=\frac{1}{3}$

$\frac{b^2}{k^2}=\frac{4}{9}-\frac{1}{9}-\frac{1}{9}=\frac{2}{9}$

$-\frac{k}{3b}=\frac{1}{\sqrt{2}}$

Hence the options A,B,C satisfy