Question

The distance between the planes x + 2y – 2z + 1 = 0 and 2x + 4y – 4z – 4z + 5 = 0, is 

A. 4 units 

B. 2 units 

C. \(\frac{1}2\) units 

D. \(\frac{1}2\) units

Answer 1

Given: The equations of the planes are x + 2y – 2z + 1 = 0 and 2x + 4y – 4z – 4z + 5 = 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 planes are: 

x + 2y – 2z + 1 = 0 and 2x + 4y – 4z – 4z + 5 = 0 

Multiplying the equation of first plane by 2, 

2x + 4y – 4z + 2 = 0 

Therefore distance between them is

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