Question

A plane passes through (1, - 2, 1) and is perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z = 4. The distance of the plane from the point (1, 2, 2) is ______.

विकल्प

• 0

• 1

• `sqrt2`

• `2sqrt2`

Answer 1

A plane passes through (1, - 2, 1) and is perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z = 4. The distance of the plane from the point (1, 2, 2) is `underline(2sqrt2)`.

Explanation:

The equation of a plane passing through (1, -2, 1) is

a(x - 1) + b(y + 2) + c(z - 1) = 0     ...(i)

Plane (i) is perpendicular to planes

2x - 2y + z = 0 and x - y + 2z = 4,

∴ 2a - 2b + c = 0, and    ....(ii)

a - b + 2c = 0    ....(iii)

Solving (ii) and (iii), we get

a = - 3, b = -3, c = 0

Substituting the values of a, b, c in equation (i), we get

x + y + 1 = 0

∴ The distance ofthis plane from (1, 2, 2) is

d = `|(1 + 2 + 1)/(sqrt(1 + 1))| = 2 sqrt2`