Ex 11.3, 11 - Plane through line of intersection of planes - Ex 11.3

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

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