Question

The plane $ax+by+cz=1$ meets the coordinates axes at A, B and C. The centroid of $∆ABC$ is

• A. $(3a,3b,3c)$
• B. $\left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)$
• C. $\left(\frac{1}{3a},\frac{1}{3b},\frac{1}{3c}\right)$
• D. $(a,b,c)$

Answer 1

The correct option is C

$\left(\frac{1}{3a},\frac{1}{3b},\frac{1}{3c}\right)$

Find the coordinates of the centroid

The plane $ax+by+cz=1$ meets the coordinate axis in A,B,C, then the coordinates will be,

A$(a,0,0)$, B$(0,b,0)$ and C$(0,0,c)$

The equation of the plane in intercept form is,

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

The intercepts that the plane makes on the axis is $\frac{1}{a},\frac{1}{b},\frac{1}{c}$

Let C denotes the centroid, then,

C$=\left(\frac{{\displaystyle \frac{1}{a}}+0+0}{3},\frac{0+{\displaystyle \frac{1}{b}}+0}{3},\frac{0+0+{\displaystyle \frac{1}{c}}}{3}\right)$

Therefore, the coordinates of the centroid will be $\frac{1}{3a},\frac{1}{3b},\frac{1}{3c}$

Hence, the correct option is (C) $\left(\frac{1}{3a},\frac{1}{3b},\frac{1}{3c}\right)$.