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:

• A. $2x-y-10z=9$
• B. $2x-y+7z=11$
• C. $2x-y+10z=11$
• D. $2x-y-9z=10$

Answer 1

Answer: (C)

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