If the distance between planes 4x - 2y - 4z + 1 = 0 and 4x - 2y - 4z + d = 0 is 7, then d is:
44
41 or 43
41 or 43
42 or 44
Solution:
Distance between two planes 4x - 2y - 4z + 1 = 0 and 4x - 2y - 4z + d = 0 is given by D = |1 - d| / √(4)² + (-2)² + (-4)² 7 = |1 - d| / 6 1 - d = ±42 d = 43 or -41