Question

The plane through the intersection of the planes $x+2y+z-1=0$ and $2x+y+3z-2=0$ is perpendicular to the planes $x+y+z-1=0$ and $x+ky+3z-1=0$. Then the value of $k$ is

• A. $-\frac{5}{2}$
• B. $-\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $\frac{5}{2}$

Equation of the plane through the intersection of planes $x+2y+z-1=0$ and $2x+y+3z-2=0$ is given by
$(x+2y+z-1)+\lambda (2x+y+3z-2)=0\Rightarrow x(1+2\lambda )+y(2+\lambda )+z(1+3\lambda )-(2\lambda +1)=0\dots \left(1\right)$

Given that plane $\left(1\right)$ is perpendicular to $x+y+z-1=0$
$\Rightarrow 1+2\lambda +2+\lambda +1+3\lambda =0$
$\Rightarrow \lambda =-\frac{2}{3}$

Also given that plane $\left(1\right)$ is perpendicular to $x+ky+3z-1=0$
$\Rightarrow 1+2\lambda +k(2+\lambda )+3(1+3\lambda )=0$
$\Rightarrow 4+11\lambda +(2+\lambda )k=0$
$\Rightarrow 4-\frac{22}{3}+\frac{4}{3}k=0(\because \lambda =-\frac{2}{3})$
$\Rightarrow k=\frac{10}{4}=\frac{5}{2}$