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
