Question

Find the equation of the plane whose intercepts on the axes of x, y and z are respectively 3, 4 and -5.

Answer 1

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

a = 3

b = 4

c = -5

\(\frac{x}{3}+\frac{y}{4}-\frac{z}{5}=1\)

\(=\frac{20 x+15 y-12 z}{60}=1\)

\(=20 x+15 y-12 z=60\)