Ex 11.3, 10 - Find vector equation of plane passing through - Equation of plane - Passing Through Intersection Of Planes

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

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

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