Question

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

Answer 1

2x+y-z=5...(1)

Dividing both sides of equation(1) by 5, we obtain

$\frac{2}{5}x+\frac{y}{5}-\frac{z}{5}=1$

$\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1$...(2)

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 planet at x, y, and z axes respectively.

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

Thus, the intercepts cut off by the plane are $\frac{5}{2}$, 5, and -5.