Question

Equations of planes parallel to the plane x - 2y + 2z + 4 = 0 which are at a distance of one unit from the point (1, 2, 3) are _______.

Options

• x + 2y + 2z = - 6, x + 2y + 2z = 5

• x - 2y - 6 = 0, x - 2y + z = 6

• x + 2y + 2z = 6, x + 2y + 2z = 0

• x - 2y + 2z = 0, x - 2y + 2z - 6 = 0

Answer 1

Equations of planes parallel to the plane x - 2y + 2z + 4 = 0 which are at a distance of one unit from the point (1, 2, 3) are x - 2y + 2z = 0, x - 2y + 2z - 6 = 0.

Explanation:

x - 2y + 2z = 0    .....(i)

Equation of plane parallel to plane (i) is

x - 2y + 2z - 6 = 0     ....(ii)

Since, distance of plane (ii) from point (1, 2, 3) is 1 unit

`therefore (|1(1) - 2(2) + 2(3) + "k"|)/(sqrt((1)^2 + (- 2)^2 + (2)^2))` = 1

`=> (|1 - 4 + 6 + "k"|)/(sqrt(1 + 4 + 4))` = 1

`=> (|3 + "k"|)/(sqrt9)` = 1

⇒ |3 + k| = 3 ⇒ 3 + k = ± 3

⇒ k = 0, -6

∴ Required equation of plane are

x - 2y + 2z + 0 = 0 and x - 2y + 2z - 6 = 0

⇒ x - 2y + 2z = 0 and x - 2y + 2z - 6 = 0