JEE Advanced2006MathematicsThree Dimensional GeometryActual
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 , then the distance of the plane from the point (1,2,2) is
Options
- A0
- B1
- C2
- D2 2
Correct answer
D. 2 2
Step-by-step solution
Let the equation of plane be, a(x-1)+b(y+2)+c(z-1)=0 which is perpendicular to 2 x-2 y+z=0 and x-y+2 z=4 array llrl & 2 a-2 b+c & =0 and a-b+2 c=0 or & a -2 1 & = b 2 = c -2 & -12 & 1 & -1 & a -3 & = b -3 = c 0 & a 1 = b 1 = c 0 array So, the equation of plane is, x-1+y+2=0 or x+y+1=0 , its distance from the point (1,2,2) is |1+2+1| 2 =2 2 .