Question

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

Answer 1

Let the plane equation be $ax+by+cz=d$

x intercept = 2 $\Rightarrow \frac{d}{a}=2\Rightarrow a=d/2$

y intercept = 3 $\Rightarrow \frac{d}{b}=3\Rightarrow b=\frac{d}{3}$

z intercept = 4 $\Rightarrow \frac{d}{c}=4\Rightarrow c=\frac{d}{4}$

$\therefore$ equation of plane $\Rightarrow \frac{d}{2}x+\frac{d}{3}y+\frac{d}{4}z=d$

$\Rightarrow \frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$

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

equation of plane $\Rightarrow 6x+4y+3z=12$