Question

If $p_1,p_2,p_3$ denote the distance of the plane $2x-3y+4z+2=0$ from the planes $2x-3y+4z+6=0,4x-6y+8z+3=0$ and $2x-3y+4z-6=0$ respectively, then

• A. $p_1+8p_2-p_3=0$
• B. ${p_3}^2=16{p_2}^2$
• C. $8{p_2}^2={p_1}^2$
• D. $p_1+2p_2+3p_3=\sqrt{29}$

Answer 1

Answer: (A)

$p_1+8p_2-p_3=0$

Since the planes are all parallel planes,

$p_1=\frac{|2-6|}{\sqrt{2^2+3^2+4^2}}=\frac{4}{\sqrt{4+9+16}}=\frac{4}{\sqrt{29}}$

Equation of the plane $4x-6y+8z+3=0$ can be written as $2x-3y+4z+\frac{3}{2}=0$

So, $p_2=\frac{|2-\frac{3}{2}|}{\sqrt{2^2+3^2+4^2}}=\frac{1}{2\sqrt{29}}$

and $p_3=\frac{|2+6|}{\sqrt{2^2+3^2+4^2}}=\frac{8}{\sqrt{29}}$

$\Rightarrow p_1+8p_2-p_3=0$