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

Transcript

Ex 11.3, 10 Find the vector equation of the plane passing through the intersection of the planes . (2 + 2 3 ) = 7, .(2 + 5 + 3 ) = 9 and through the point (2, 1, 3). The vector equation of a plane passing through the intersection of planes . 1 = d1 and . 2 = d2 and also through the point (x1, y1, z1) is .( + ) = d1 + d2 Given, the plane passes through So, equation of the plane is . (2 +2 3 )+ (2 + 5 + 3 ) = 7 + .9 . 2 + 2 3 + 2 + 5 + 3 = 7 + 9 . (2+2 ) +(2+5 ) +( +3 ) = 9 + 7 Now, to find , put = x + y + z (x + y + z ). (2 + 2 ) + (2+5 ) +( 3 + 3 ) = 9 + 7 x(2 + 2 ) + y (2 + 5 ) + ( 3+3 ) = 9 + 7 The plane passes through (2, 1, 3) Putting (2, 1, 3) in (2), 2(2 + 2 ) + 1(2 + 5 ) + 3( 3 + 3x) = 9 + 7 4 + 4 + 2 + 5 + ( 9) + 9 = 9 + 7 18 9 = 7 + 3 9 = 10 = Putting value of in (1), . 2+ 2. 10 9 + 2+ 5. 10 9 + 3 + 3. 10 9 = 9. 10 9 + 7 . 2 + 20 9 + 2 + 50 9 + 3 + 30 9 = 10 + 7 . 38 9 + 68 9 + 3 9 = 17 1 9 . 38 + 68 + 3 = 17 . 38 +68 +3 = 17 9 . + + = 153 Therefore, the vector equation of the required plane is . 38 +68 +3 = 153

Ex 11.3 