The plane 2x – y + 3z + 5 = 0 is rotated through 90 0 about its line of intersection with the plane 5x – 4y + 2z + 1 = 0. The equation of the plane in the new position is
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Equation of a plane passing through the line of intersection of the given planes is
2x – y + 3z + 5 + λ (5x – 4y – 2z + 1 ) = 0
or (2 + 5 λ )x – (1 + 4 λ )y + (3 – 2 λ )z + 5 + λ = 0
This will be perpendicular to the plane
2x – y + 3z + 5 = 0
if 2(2 + 5 λ ) + (1 + 4 λ ) + 3(3 – 2 λ ) = 0
⇒ λ = –7/4 and the required equation of the plane is
4(2x – y + 3z + 5) – 7(5x – 4y – 2z + 1) = 0
27x – 24y – 26z – 13 = 0
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