Last updated at March 11, 2017 by Teachoo

Transcript

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 ( 𝐴1x + 𝐵1y + 𝐶1z = d1) + 𝜆 ( 𝐴2x + 𝐵2y + 𝐶2z = 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 = −𝟏𝟑 Putting value of 𝜆 in (1), (1 + 2𝜆) x + (1 + 3𝜆)y + (1 + 4𝜆) z + (−1 −5𝜆) = 0 1+2 × −13x + 1+3× −13 y + 1+4 × −13 z + −1−5× −13 = 0 1− 23 x + 1−1 y + 1− 43 z + − 1+ 53 = 0 13 x + 0y − 13 z + 23 = 0 13 x − 13 z + 23 = 0 13 (x − z + 2) = 0 x − z + 2 = 0 Therefore, the equation of the plane is x − z + 2 = 0

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .