Find the line of intersection of the planes 𝑥+𝑦−2𝑧=2 and 2𝑥−𝑦+𝑧=1.

Answer 1

Solving for y we obtain that y=2z-x+2 and y=2x+z-1. Hence, 2z-x+2=2x+z-1 or 3x=z+3 or x=(1/3)z+1. Thus, y=(2/3)z+2+z-1 or y=(5/3)z+1. Therefore, the intersection line is given by the expression x=(1/3)t+1, y=(5/3)t+1 and z=t where t is in R.