Question

If perpendicular distance between two planes 3x – 2y + z = 1 and 6x – 4y + 2z = k is [3/{2√(14)}] then k = ____

(A) 5, – 1

(B) – 5, 1

(C) – 5, – 1

(D) 5, – 1

Answer 1

The correct option (A) 5, – 1

Explanation:

planes: 3x – 2y + z = 1 and 6x – 4y + 2z – k = 0

i.e. 6x – 4y + 2z = 2 and 6x – 4y + 2z – k = 0

perpendicular distance between two planes = [(|2 – k|)/√(36 + 16 + 4)]

∴ [3/{2√(14)}] = [(|k – 2|)/√(56)]

∴ (3/2) = [(|k – 2|)/2]

∴ k – 2 = 3 or k – 2 = – 3

∴ k = 5 or k = – 1.