A plane passes through the point and is perpendicular to two planes and , then the distance of the plane from the point is
- A.0
- B.1
- C.
- D.
Show answer and solution
Correct option: D
Solution
The normal to the required plane is perpendicular to the normals of both given planes, so it equals their cross product:
n = (2,-2,1) × (1,-1,2) = |i j k; 2 -2 1; 1 -1 2|
= i((-2)(2)-(1)(-1)) - j((2)(2)-(1)(1)) + k((2)(-1)-(-2)(1))
= i(-4+1) - j(4-1) + k(-2+2)
= (-3, -3, 0), simplified to (1, 1, 0).
Plane through (1,-2,1) with normal (1,1,0): 1(x-1) + 1(y+2) + 0(z-1) = 0 → x + y + 1 = 0.
Distance from (1,2,2): |1 + 2 + 1|/√(1²+1²) = 4/√2 = 2√2.
n = (2,-2,1) × (1,-1,2) = |i j k; 2 -2 1; 1 -1 2|
= i((-2)(2)-(1)(-1)) - j((2)(2)-(1)(1)) + k((2)(-1)-(-2)(1))
= i(-4+1) - j(4-1) + k(-2+2)
= (-3, -3, 0), simplified to (1, 1, 0).
Plane through (1,-2,1) with normal (1,1,0): 1(x-1) + 1(y+2) + 0(z-1) = 0 → x + y + 1 = 0.
Distance from (1,2,2): |1 + 2 + 1|/√(1²+1²) = 4/√2 = 2√2.
[AIEEE 2006]