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The equation of the plane passing through the line of intersection of the planes x + y + z = 1, 2x + 3y + 4z = 7, and perpendicular to the plane x −5y + 3z = 5 is given by A. x + 2y + 3z −6 = 0 B. x + 2y + 3z +6 = 0 C. 3x + 4y +5z − 8 = 0 D. 3x + 4y + 5z + 8 = 0 |
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Answer» Correct option A. x + 2y + 3z −6 = 0 Explanation: Equation of plane passing through the intersection of lines x + y + z = 1 and 2x + 3y + 4z = 7 is: x + y + z – 1 + λ (2x + 3y + 4z – 7) = 0 Or x(1+2λ) + y(1+3λ) + z(1+4λ) –(1-7λ) = 0 →(1) This plane is perpendicular to the line, x – 5y – 3z = 5 ∴ Dot product of these will be equal to zero. ∴ (1+2λ)1 - (1+3λ)5 + (1+4λ)3 = 0 So λ = -1 Putting in (1) Required equation is: x + 2y + 3z – 6 = 0 |
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