    1. Chapter 11 Class 12 Three Dimensional Geometry
2. Serial order wise
3. Ex 11.3

Transcript

Ex 11.3, 11 Find the equation of the plane thro ugh the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x y + z = 0. Equation of a plane passing through the intersection of planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 (A1x + B1y + C1z d1) + (A2x + b2y + c2z d2) = 0 Given the planes passes through Equation of plane is ( 1 x + 1 y + 1 z = d1) + ( 2 x + 2 y + 2 z = d2) = 0 Putting values (1x + 1y + 1z 1) + (2x + 3y + 4z 5) = 0 x + y + z 1 + 2 x + 3 y + 4 z 5 = 0 (1 + 2 ) x + (1 + 3 )y + (1 + 4 ) z + ( 1 5 ) = 0 Also, the plane is perpendicular to the plane x y + z = 0 So, the normal vector to be the plane is perpendicular to the normal vector of x y + z = 0. Since, is perpendicular to , 1 2 + b1b2 + c1 c2 = 0 (1 + 2 ) 1 + (1 + 3 ) 1 + (1 + 4 ) 1 = 0 1 + 2 1 3 + 1 + 4 = 0 1 + 3 = 0 1 = 3 = Putting value of in (1), (1 + 2 ) x + (1 + 3 )y + (1 + 4 ) z + ( 1 5 ) = 0 1+2 1 3 x + 1+3 1 3 y + 1+4 1 3 z + 1 5 1 3 = 0 1 2 3 x + 1 1 y + 1 4 3 z + 1+ 5 3 = 0 1 3 x + 0y 1 3 z + 2 3 = 0 1 3 x 1 3 z + 2 3 = 0 1 3 (x z + 2) = 0 x z + 2 = 0 Therefore, the equation of the plane is x z + 2 = 0

Ex 11.3 