Question

Find the constant k so that the planes x-2y+kz=0 and 2x+5y-z=0 are at right ang les.Find the equation of the plane through (1 -1 -1) and perpendicular to these planes.

I got the constant k = -8
so request you to find the equation of the plane ​

Answer 1

Answer:

k = -8

Step-by-step explanation:

Given that

x - 2y + kz = 0

2x + 5y - z = 0

(a1, b1, c1) = (1, -2, k)

(a2, b2, c2) = (2, 5, -1)

⇒ cos θ = |a1a2 + b1b2 + c1c2|/√a1² + b1² + c1² √a2² + b2² + c2²

⇒ cos θ = |1.2 + (-2)5 + k(-1)|/√1² + (-2)² + k² √2² + 5² + (-1)²

Angle between these planes would be 90°.

⇒  θ = 90°

⇒ cos 90° = 0

⇒ 0 = |1.2 + (-2)5 + k(-1)|/√1² + (-2)² + k² √2² + 5² + (-1)²

⇒ 1.2 + (-2.5) + k(-1) = 0

⇒ 2 - 10 - k = 0

⇒ -8 -k = 0

⇒ k = -8

∴ Given planes are at right angle if k = -8 .

Hope it helps you!