Question

If $P_1,P_2,P_3$ denotes the perpendicular distances of the plane $2x-3y+4z+2=0$ from the parallel 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. $P_3=16P_2$
• C. $8P_2=P_1$
• D. $P_1+2P_2+3P_3=\sqrt{29}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $P_1+8P_2-P_3=0$
B $P_3=16P_2$
C $8P_2=P_1$
D $P_1+2P_2+3P_3=\sqrt{29}$

Since the planes are 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 $P_3=\frac{|2+6|}{\sqrt{2^2+3^2+4^2}}=\frac{8}{\sqrt{29}}$

So, from options:

(A) $P_1+8P_2-P_3=\frac{4}{\sqrt{29}}-8\times \frac{1}{2\sqrt{29}}-\frac{8}{\sqrt{29}}=\frac{8}{\sqrt{29}}-\frac{8}{\sqrt{29}}=0$

(B) $16P_2=\frac{16}{2\sqrt{29}}=\frac{8}{\sqrt{29}}=P_3$

(C) $8P_2=\frac{8}{2\sqrt{29}}=\frac{4}{\sqrt{29}}=P_1$

(D) $P_1+2P_2+3P_3=\frac{4}{\sqrt{29}}+\frac{2}{2\sqrt{29}}+3\times \frac{8}{\sqrt{29}}=\frac{29}{\sqrt{29}}=\sqrt{29}$