Question

The plane $ax+by+cz=1$ meets the coordinate axes in $A,B$ and $C$. The centroid of the triangle is:

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

Answer 1

The correct option is D $(\frac{1}{3a},\frac{1}{3b},\frac{1}{3c})$

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

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

The equation of the plane in intercept form is,

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

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

Let C denotes the centroid, then,

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

Therefore, the coordinates of the centroid will be $\left(\frac{1}{3a},\frac{1}{3b},\frac{1}{3c}\right)$.