Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

Misc 17 Find the equation of the plane which contains the line of intersection of the planes . ( + 2 + 3 ) 4 = 0 , . (2 + ) + 5 = 0 and which is perpendicular to the plane . (5 + 3 6 ) + 8 = 0 . Equation of a plane passing through the intersection of the places A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is (A1x + B1y + C1z d1) + (A2x + B2y + C2z d2) = 0 Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 Equation of plane is (A1x + B1y + C1z d1) + (A2x + B2y + C2z = d2) = 0 Putting values (1x + 2y + 3z 4) + ( 2x 1y + 1z 5) = 0 (1 2 ) x + (2 )y + (3 + ) z + ( 4 5 ) = 0 Now, the plane is perpendicular to the plane .(5 + 3 6 ) + 8 = 0 So, normal to plane will be perpendicular to normal of .(5 + 3 6 ) + 8 = 0 Now, .(5 + 3 6 ) + 8 = 0 .(5 + 3 6 ) = 8 .(5 + 3 6 ) = 8 .( 5 3 + 6 ) = 8 Finding direction cosines of & Since, is perpendicular to 1 2 + b1 b2 + c1 c2 = 0 (1 2 ) 5 + (2 ) 3 + (3 + ) 6 = 0 5 + 10 6 + 3 + 18 + 6 = 0 19 + 7 = 0 = Putting value of in (1), (1 2 ) x + (2 )y + (3 + ) z + ( 4 5 ) = 0 1 2 7 19 x + 2 7 19 y + 3+ 7 19 z + 4 5 7 19 = 0 1 + 14 19 x + 2 + 7 19 y + 3 7 19 z + 4 + 35 19 = 0 33 19 x + 45 19 y + 50 19 z 41 19 = 0 1 19 (33x + 45y + 50z 41) = 0 33x + 45y + 50z 41 = 0 Therefore, the equation of the plane is 33x + 45y + 50z = 41.