Question

Find the equation of plane with the intercepts $2,3$ and $4$ on $x,y$ and $z$ axis respectively.

Answer 1

Equation of a plane in intercept form : $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

$x-intercept=a=2$

$y-intercept=b=3$

$z-intercept=c=4$

$\therefore$ Equation of plane = $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$

$\frac{6x+4y+3z}{12}=1$

$6x+4y+3z=12$

$\therefore$ Equation of plane $6x+4y+3z=12$