Question
Find the value of k for which the planes 3 x– 6 y– 2 z = 7 and 2x + y'– k z = 5 are perpendicular to each other.
Solution
The equations of planes are
3x – 6y – 2z–7 = 0
and 2 x + y – k z – 5 = 0
Since two planes are perpendicular to each other.
∴ (3) (2) + (–6) (1) + (–2) (–k) = 0 [∵a1a2 + b1,b2 + c1c2 = 0]
∴ 6 – 6 + 2k = 0 ⇒ 2k = 0 ⇒ K = 0
