Question

The distance between the planes x + 2y + 3z + 7 = 0 and 2x + 4y + 6z + 7 = 0 is …

(a) √7/2√2

(b) 7/2

(c) √7/2

(d) 7/2√2

Answer 1

(a) √7/2√2

x + 2y + 3z + 1 = 0; 2x + 4y + 6z + 7 = 0 

Multiplying 2 on both sides 

2x + 4y + 6z + 14 = 0. 

a = 2, b = 4, c = 6, d₁ = 14, d₂ = ?

Distance