The equation of a plane through the line of intersection of the planes x+2y=3, y-2z+1=0, a…

Question

The equation of a plane through the line of intersection of the planes $x+2y=3$, $y-2z+1=0$, and perpendicular to the first plane, is:

Accepted Answer

Family: $(x+2y-3)+\lambda(y-2z+1)=0$, i.e. $x+(2+\lambda)y-2\lambda z+(\lambda-3)=0$. Perpendicular to $x+2y=3$ (normal $(1,2,0)$):
$$1+2(2+\lambda)=0\implies\lambda=-\tfrac52.$$
Plane: $x-\tfrac{y}2+5z-\tfrac{11}2=0\implies 2x-y+10z-11=0$.

Solutions · 1