AP EAMCET201923 Apr 2019Morning ShiftMathematicsThree Dimensional GeometryActual
The equation of the plane ( ) through the line of intersection of the planes ( ₁ x+3 y-6=0 ), and ( ₂ 3 x-y+4 z=0 ) is ( ₁+ ₂=0 ). If the plane ( ) is at unit distance from the origin, then an equation of the plane ( ) is
Options
- A(2 x+y+2 z-3=0 )
- B(2 x-y-2 z+3=0 )
- C(2 x+y+2 z+3=0 )
- D(x+2 y+2 z+3=0 )
Correct answer
A. (2 x+y+2 z-3=0 )
Step-by-step solution
Given equation of plane are ( aligned & ₁=x+3 y-6=0 & ₂=3 x-y+4 z=0 aligned ) Given, ( ₁+ ₂=0 ) ( aligned (x+3 y-6)+ (3 x-y+4 z) & =0 (1+3 ) x+(3- ) y+4 z-6 & =0 (i) aligned ) Perpendicular distance from ((0,0,0) ) to above plane is 1. ( aligned & |a x₁+b y₁+c z₁+d | a^2+b^2+c^2 =1 & |-6| (1+3 )^2+(3- )^2+(4 )^2 =1 & 1+9 ^2+6 +9+ ^2-6 +16 ^2=36 & & 26 ^2=26 & & = 1 aligned ) ( ) Substitute ( =1 ) in Eq. (i), we get ( aligned 4 x+2 y+4 z-6 & =0 2 x+y+2 z-3 & =0 aligned ) Hence, answer is (a).