JEE Advanced · 2006 · 09 Apr 2006 · Three Dimensional Geometry

A plane passes through (1, -2, 1) and is perpendicular to two planes

2 x - 2 y + z = 0

and

x - y + 2 z = 4

. The distance of the plane from the point (1, 2, 2) is

Answer: (D)

Solution

The required plane passes through

(1, - 2, 1)

and is perpendicular to both given planes.

Its normal vector is perpendicular to the normals of both planes:

n_{1} = (2, - 2, 1)

and

n_{2} = (1, - 1, 2)

.

The normal to the required plane is:

n = n_{1} \times n_{2} = \hat{i} 21 \hat{j} - 2 - 1 \hat{k} 12 = (- 3, - 3, 0)

Simplifying:

n = (1, 1, 0)

.

The plane equation is

1 (x - 1) + 1 (y + 2) + 0 (z - 1) = 0

, which gives

x + y + 1 = 0

.

Distance from

(1, 2, 2)

to the plane:

d = \frac{∣1 + 2 + 1∣}{1 ^{2} + 1 ^{2}} = \frac{4}{2} = 22

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option