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

Answer: (D)

$\frac {2}{\sqrt 29}$ units

Answer 2

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.