Misc 15 - Chapter 11 Class 12 Three Dimensional Geometry

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