Ex 11.3, 9 - Equation of plane through intersection of planes - Ex 11.3

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  1. Chapter 11 Class 12 Three Dimensional Geometry
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Ex 11.3, 9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1). Equation of a plane passing through the intersection of planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 and through the point (x1, y1, z1) is (A1x + B1y + C1z − d1) + 𝜆 (A2x + B2y + C2z − d2) = 0 Given, plane passes through Equation of plane is (3x − 1y + 2z − 4) + 𝜆 (1x + 1y + 1z − 2) = 0 3x − y + 2z − 4 + 𝜆x + 𝜆y + 𝜆z – 2𝜆 = 0 (3 + 𝜆) x + (−1 + 𝜆)y + (2 + 𝜆)z + (−4 − 2𝜆) = 0 We now find the value of 𝜆 The plane passes through (2, 2, 1) Putting (2, 2, 1) in (1), (3 + 𝜆) x + (−1 + 𝜆)y + (2 + 𝜆)z + (−4 − 2𝜆) = 0 (3 + 𝜆) × 2 + (−1 + 𝜆) × 2 + (2 + 𝜆) × 1 + (− 4 − 2𝜆) = 0 6 + 2𝜆 − 2 + 2𝜆 + 2 + 𝜆 − 4 − 2𝜆 = 0 3𝜆 + 2 = 0 3𝜆 = −2 ∴ 𝜆 = −𝟐﷮𝟑﷯ Putting value of 𝜆 in (1), (3 + 𝜆) x + (−1 + 𝜆)y + (2 + 𝜆)z + (−4 − 2𝜆) = 0 3+ −2﷮3﷯﷯﷯ x + −1+ −2﷮3﷯﷯﷯y + 2+ −2﷮3﷯﷯﷯z + −4−2× −2﷮3﷯﷯ = 0 3− 2﷮3﷯﷯𝑥 + −1− 2﷮3﷯﷯ y + 2− 2﷮3﷯﷯z + −4+ 4﷮3﷯﷯ = 0 7𝑥﷮3﷯ − 5𝑦﷮3﷯ + 4𝑧﷮3﷯ − 8﷮3﷯ = 0 1﷮3﷯ (7x − 5y + 4z − 8) = 0 7x − 5y + 4z − 8 = 0 ∴ The equation of plane is 7x − 5y + 4z − 8 = 0

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