Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.
The plane 2...
Question
The plane 2x−y+3z+5=0 is rotated through 90∘ about its line of intersection with the plane 5x−4y−2z+1=0. Let the equation of the plane in the new position be kx+my−26z−13=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 (2x−y+3z+5)+λ(5x−4y−2z+1)=0 (2+5λ)x−(1+4λ)y+(3−2λ)z+(5+λ)=0 ...( 1 ) This plane makes an angle of 90∘ with the plane 2x−y+3z+5=0 Applying the condition of perpendicularity ∑a1a2=0, 2(2+5λ)+1(1+4λ)+3(3−2λ)=0 8λ+14=0
∴λ=−74 ∴ required equation of plane from (1) on substituting λ=−74 is 27x−24y−26z−13=0 ⇒k+m=3