Given planes, 2x – 4y + 3z = 5 and x + 2y + λz = 5
We know that planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are at right angles
if, a1a2 + b1b2 + c1c2 = 0 ……(a)
We have, a1 = 2, b1 = – 4, c1 = 3 and a2 = 1, b2 = 2, c2 = λ
Using (a) we have, a1a2 + b1b2 + c1c2 = (2)(1) + (– 4)(2) + (3)(λ) = 0
⇒ 2 – 8 + 3λ = 0
⇒ 6 = 3λ
⇒ 2 = λ
Hence, for λ = 2 the given planes are perpendicular.