Question

Find the equation of plane with intercepts $3,2$ and $6$ on the $x,y$ and $z$ axes respectively.

Answer 1

The equation of a plane with intercepts a, b, c on x, y and z-axis respectively is

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

Given: Intercept on x-axis$=3$

$\therefore a=3$

Intercept on y-axis$=2$

$\therefore b=2$

Intercept on z-axis$=6$

$\therefore c=6$

Equation of plane is

$\frac{x}{3}+\frac{y}{2}+\frac{z}{6}=1$

$\frac{2x}{6}+\frac{3y}{6}+\frac{z}{6}=1$

(or) $2x+3y+z=6$ is the required equation of the plane.