Question

If $p_1,p_2,p_3$ denote the distances 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. $p_3^2=16p_2^2$
• C. $8p_2^2=p_1^2$
• D. $p_1+2p_2+3p_3=\sqrt{29}$

Answer 1

Answer: (A), (D)

The correct options are
A $p_1+8p_2-p_3=0$
D $p_1+2p_2+3p_3=\sqrt{29}$

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 $p_3=\frac{|2+6|}{\sqrt{2^2+3^2+4^2}}=\frac{8}{\sqrt{29}}\Rightarrow p_1+8p_2-p_3=0$