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Question

The plane 2xy+3z+5=0 is rotated through 90 about its line of intersection with the plane 5x4y2z+1=0. Let the equation of the plane in the new position be kx+my26z13=0. Find k+m? (Note: k and m are integers in shortest form.)

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Solution

Plane through the intersection of given planes is
(2xy+3z+5)+λ(5x4y2z+1)=0
(2+5λ)x(1+4λ)y+(32λ)z+(5+λ)=0 ...( 1 )
This plane makes an angle of 90 with the plane
2xy+3z+5=0
Applying the condition of perpendicularity
a1a2=0,
2(2+5λ)+1(1+4λ)+3(32λ)=0
8λ+14=0
λ=74
required equation of plane from (1) on substituting λ=74 is
27x24y26z13=0
k+m=3

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