Question

Find the intercepts cut-off by the plane 2x+y-z=5.

Answer 1

Given plane 2x+y-z=5 can be written as

$\frac{2x}{5}+\frac{y}{5}-\frac{z}{5}=1\Rightarrow \frac{x}{5/2}+\frac{y}{5}+\frac{z}{-5}=1$ (On dividing by 5 both sides)

It is known that the equation of a plane in intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where a,b,c are the intercepts cut-off by the plane at X,Y and Z-axes respectively.

Therefore, for the given equation, $a=\frac{5}{2},b=5andc=-5.$

Hence, the intercepts of the plane are $\frac{5}{2}$,5 and -5.