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Ex 11.3, 11 - Plane through line of intersection of planes - Ex 11.3

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Ex 11.3, 11 Find the equation of the plane thro ugh the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. Equation of a plane passing through the intersection of planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 (A1x + B1y + C1z − d1) + 𝜆 (A2x + b2y + c2z − d2) = 0 Given the planes passes through Equation of plane is ( 𝐴﷮1﷯x + 𝐵﷮1﷯y + 𝐶﷮1﷯z = d1) + 𝜆 ( 𝐴﷮2﷯x + 𝐵﷮2﷯y + 𝐶﷮2﷯z = d2) = 0 Putting values (1x + 1y + 1z − 1) + 𝜆 (2x + 3y + 4z − 5) = 0 x + y + z − 1 + 2𝜆 x + 3𝜆y + 4𝜆z − 5𝜆 = 0 (1 + 2𝜆) x + (1 + 3𝜆)y + (1 + 4𝜆) z + (−1 −5𝜆) = 0 Also, the plane is perpendicular to the plane x − y + z = 0 So, the normal vector 𝑁﷯ to be the plane is perpendicular to the normal vector of x − y + z = 0. Since, 𝑵﷯ is perpendicular to 𝒏﷯, 𝑎1 𝑎2 + b1b2 + c1 c2 = 0 (1 + 2𝜆) × 1 + (1 + 3𝜆) × − 1 + (1 + 4𝜆) × 1 = 0 1 + 2𝜆 − 1 − 3𝜆 + 1 + 4𝜆 = 0 1 + 3𝜆 = 0 −1 = 3𝜆 ∴ 𝜆 = −𝟏﷮𝟑﷯ Putting value of 𝜆 in (1), (1 + 2𝜆) x + (1 + 3𝜆)y + (1 + 4𝜆) z + (−1 −5𝜆) = 0 1+2 × −1﷮3﷯﷯x + 1+3× −1﷮3﷯﷯ y + 1+4 × −1﷮3﷯﷯ z + −1−5× −1﷮3﷯﷯ = 0 1− 2﷮3﷯﷯ x + 1−1﷯ y + 1− 4﷮3﷯﷯ z + − 1+ 5﷮3﷯﷯ = 0 1﷮3﷯ x + 0y − 1﷮3﷯ z + 2﷮3﷯ = 0 1﷮3﷯ x − 1﷮3﷯ z + 2﷮3﷯ = 0 1﷮3﷯ (x − z + 2) = 0 x − z + 2 = 0 Therefore, the equation of the plane is x − z + 2 = 0

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