Question

Find the equation of the plane which passes through the point (3, 4, –1) and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also, find the distance between the two planes

Answer 1

The required equation parallel to the plane

2x – 3y + 5z + 7 = 0 .........(1)

2x – 3y + 5z + λ = 0   ..........(2)

This passes through (3, 4, -1)

(2) ⇒ 2(3) – 3(4) + 5(–1) + λ = 0

6 – 12 – 5 + 1 = 0

λ = 11

(2) ⇒ The required equation is 2x – 3y + 5z + 11 =0  ........(3)

∴ Now, distance between the above parallel lines (1) and (3)

a = 2

b = – 3

c = 5

d₁ = 7

d₂ = 11

Distance = `|("d"_1 - "d"_2)/sqrt("a"^2 + "b"^2 + "c"^2)|`

= `|(7 - 11)/sqrt(2^2 + 3^2 + 5^2)|`

= `|4/sqrt(38)|`

d = `4/sqrt(38)`