Question

The equation of the plane containing the line of intersection of the planes $x+y+4z-6=0$ and $2x+4y+5=0$ and making an angle $\frac{\pi }{4}$ with the plane $x-z+5=0$ is

• A. $12x+5y-16z+30=0$
• B. $5x+3y-16z+29=0$
• C. $x+5y-6z+9=0$
• D. $2x+8y-16z+39=0$

Answer 1

The correct option is D $2x+8y-16z+39=0$

The equation of a plane containing the line of intersection of the planes $x+y+4z-6=0$ and $2x+4y+5=0$ is,
$(x+y+4z-6)+\lambda (2x+4y+5)=0\cdots \left(i\right)$
$\Rightarrow (2\lambda +1)x+(4\lambda +1)y+4z+5\lambda -6=0$
Now this plane is making angle $\frac{\pi }{4}$ with the plane $x-z+5=0$
$\therefore \cos \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}}\right|$
$\Rightarrow \cos \frac{\pi }{4}=\left|\frac{2\lambda +1-4}{\sqrt{\left(\right(2\lambda +1{)}^2+(4\lambda +1{)}^2+(4{)}^2\left)\right(1^2+1^2)}}\right|$
$\Rightarrow \sqrt{20{\lambda }^2+12\lambda +18}=|2\lambda -3|$
$\Rightarrow 16{\lambda }^2+24\lambda +9=0$
$\Rightarrow (4\lambda +3{)}^2=0$
$\Rightarrow \lambda =\frac{-3}{4}$
Putting the value of $\lambda$ in equation $\left(i\right),$ we get
$4x+4y+16z-24-6x-12y-15=0$
$\Rightarrow 2x+8y-16z+39=0$