Question

Let $P_1:x+y+z+1=0,P_2:x-y+2z+1=0,P_3:3x+y+4z+7=0$ be three planes, then the distance of the line of intersection of the planes $P_1=0$ and $P_2=0$ from the plane $P_3=0$ is

• A. $\frac{2}{\sqrt{26}}$
• B. $\frac{4}{\sqrt{26}}$
• C. $\frac{1}{\sqrt{26}}$
• D. $\frac{7}{\sqrt{26}}$

Answer 1

The correct option is B $\frac{4}{\sqrt{26}}$

The line of intersection of planes $P_1$ and $P_2$ must be parallel to the plane $P_3$.
The family of planes of $P_1$ and $P_2$ $\lambda P_1+P_2=0\Rightarrow (\lambda +1)x+(\lambda -1)y+(\lambda +2)z+\lambda +1=0$
The distance between the line and the plane=the distance between the family of planes and the plane $P_3$. For this, the family of plane must parallel to the given plane.
$\Rightarrow \lambda =2$

The distance between the planes $3x+y+4z+3=0$ and $3x+y+4z+7=0$ is $\frac{4}{\sqrt{26}}$