Question

The equation of the plane which makes with the coordinate axes, a triangle with centroid $(\alpha ,\beta ,\gamma )$ is given by

• A. $\alpha x+\beta y+\gamma z=1$
• B. $\alpha x+\beta y+\gamma z=3$
• C. $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$
• D. $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$

Answer 1

The correct option is D $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$

Let the equation of the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Then,it meets the axes at
$A(a,0,0),B(0,b,0)$ and $C(0,0,c)$
$\therefore$ centroid of $\Delta ABC$ is $G(\frac{a}{3},\frac{b}{3},\frac{c}{3})$.
$\therefore \frac{a}{3}=\alpha ,\frac{b}{3}=\beta$ and $\frac{c}{3}=\gamma$
$\Rightarrow a=3\alpha ,b=3\beta$ and $c=3\gamma$
$\therefore$ The required equation of the plane is
$\frac{x}{3\alpha }+\frac{y}{3\beta }+\frac{z}{3\gamma }=1$
$\Rightarrow \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$