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:

Answer 1

Equation of a plane through the line of intersection of the planes
x+2y=3,y-2z+1=0 is
$(x+2y-3)+\lambda (y-2z+1)=0\Rightarrow x+(2+\lambda )y-2\lambda \left(z\right)-3+\lambda =0...\left(i\right)$
Now, plane (i) is $\perp$ to x+2y =3
$\therefore$ Their dot product is zero
i.e $1+2(2+\lambda )=0\Rightarrow \lambda =-\frac{5}{2}$
Thus, required plane is
$x+(2-\frac{5}{2})y-2x\frac{-5}{2}\left(z\right)-3-\frac{5}{2}=0\Rightarrow x-\frac{y}{2}+5z-\frac{11}{2}=0\Rightarrow 2x-y+10z-11=0$