Question

Distance between the two planes: 2x+3y+4z = 4 and 4x+6y+8z = 12 is

• A. 2 units
• B. 4 units
• C. 8 units
• D. \(\frac {2}{\sqrt 29}\) units

Answer 1

The Correct Option is D

The equation of the planes are

2x+3y+4z = 4          ...(1)

4x+6y+8z = 12

⇒2x+3y+4z = 6       ...(2)

It can be seen that the given planes are parallel. It is known that the distance between two parallel planes, ax+by+cz=d₁ and ax+by+cz=d₂, is given by,
D = \(|\frac {d_2-d_1}{√a^2+b^2+c^2}|\)

⇒ D = \(|\frac {6-4}{√2^2+3^2+4^2}|\)

D = \(\frac {2}{\sqrt 29}\)

Thus, the distance between the lines is \(\frac {2}{\sqrt 29}\) units.
Hence, the correct answer is D.