Question

The distance between the parallel planes 2x – 3y + 6z = 5 and 6x – 9y + 18z + 20 = 0, is

A. \(\frac{5}3\)units 

B. 5√3 units 

C. \(\frac{8}5\) units 

D. 8√5 units

Answer 1

Given: The equations of the parallel planes are 2x – 3y + 6z = 5 and 6x – 9y + 18z + 20 = 0 

To find: distance between the planes 

Formula Used: Distance between two parallel planes ax + by + cz + d₁ = 0 and ax₁ + by₁ + cz₁ + d₁ = 0 is

Explanation: 

The equations of the parallel planes are: 

2x – 3y + 6z – 5 = 0

2x - 3y + 6x + \(\frac{20}3\) = 0

Therefore distance between them is

Therefore distance between the planes is \(\frac{5}3\) units