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

Transcript

Misc 15 Find the equation of the plane passing through the line of intersection of the planes . ( + + ) =1 and . (2 + 3 ) + 4 = 0 and parallel to x-axis. Equation of a plane passing through the intersection of two planes _1x + B1y + _1z = d1 and _2x + B2y + _2z = d2 is ( _ "x " +" B1y" + _ "z d1 " ) + ( _ "x" +"B2y" + _ "z d2 " ) = 0. Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 . ( + + ) = 1 Putting = x + y + z , (x + y + z ).( + + ) = 1 (x 1) + (Y 1) + (z 1) = 1 1x + 1y + 1z = 1 Comparing with _1 "x "+"B1y "+" " _1 "z = d1" _1 = 1 , _1 = 1 , _1 = 1 , _1 = 1 . (2 + 3 ) + 4 = 0 . (2 + 3 ) = 4 . (2 + 3 ) = 4 . ( 2 3 + ) = 4 Putting = x + y + z , (x + y + z ).(-2 3 + 1 ) = 4 (x 2) + (y 3) + (z 1) = 4 2x 3y + 1z = 4 Comparing with _2 "x "+"B2y "+" " _2 "z = d2" _2 = 2 , _2 = 3 , _2 = 1 , _2 = 4 Equation of plane is ( _1 "x " +" B1y" + _1 "z d1 " ) + ( _2 "x" +"B2y" + _2 "z d2 " ) = 0 (1x + 1y + 1z 1) + ( 2x 3y + 1z 4) = 0 (1 2 ) x + (1 3 )y + (1 + ) z + ( 1 4 ) = 0 Also, the plane is parallel to x axis. So, Normal vector to the plane is perpendicular to x axis. Finding direction ratios of normal and x-axis Theory : Two lines with direction ratios 1, b1, c1 and 2, b2, c2 are perpendicular if 1 2 + b1b2 + c1 c2 = 0 = (1 2 ) + (1 3 ) + (1 + ) Direction ratios = (1 2 ), (1 3 ), (1 + ) 1 = 1 2 , b1 = 1 3 , c1 = 1 + ( ) = 1 + 0 + 0 Direction ratios = 1, 0, 0 2 = 1, b2 = 0, c2 = 0, So, 1 2 + b1 b2 + c1 c2 = 0 (1 2 ) 1 + (1 3 ) 0 + (1 + ) 0 = 0 (1 2 ) + 0 + 0 = 0 1 = 2 = / Putting value of in (1) (1 2. 1/2)x + (1 3. 1/2) y + (1 + 1/2) + ( 1 4.1/2) = 0 (1 1)x + (1 3/2) y + (1 + 1/2) z + ( 1 2) = 0 0x 1/2 y + 3/2 z 3 = 0 0x 1/2 y + 3/2 z = 3 y + 3z = 6 0 = y 3z + 6 y 3z + 6 = 0 Therefore, the equation of the plane is y 3z + 6 = 0

Miscellaneous 