Question

Distance between the parallel planes 2x - y + 3z + 4 = 0 and 6x - 3y + 9z - 3 = 0 is:

• A. $\frac{5}{\sqrt{3}}$
• B. $\frac{4}{\sqrt{6}}$
• C. $\frac{5}{\sqrt{14}}$
• D. $\frac{3}{2\sqrt{3}}$

Answer 1

Answer: (C)

Note: The perpendicular distance from the origin to the plane ax + by + cz + d = 0 is
given by $\frac{d}{\sqrt{a^2+b^2+c^2}}$
Thus the perpendicular distance from origin to the plane 2x - y + 3z + 4 = 0 is
$P_1=\frac{4}{\sqrt{14}}$
Similarly, $P_2=\frac{-3}{3\sqrt{14}}=\frac{-1}{\sqrt{14}}$
Thus the distance between the two parallel planes
$=|P_2-P_1|=\frac{5}{\sqrt{14}}$.