Given a cartesian equation of a plane , how can find the equation of a rotated plane [closed]

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I want to rotate a plane represented by the equation z = 6 , by n degrees along y axis and find the new equation of the plane. how can this be done?

Thanks

Answer 1

Base point (0,0,6) after rotation will lie in XZ plane with coordinates

(x0, y0, z0) = (-6*sin(Fi), 0, 6*cos(Fi))

normal vector

n = (A,B,C) = (-sin(Fi), 0, cos(Fi))

so new plane equation is (for explanation see beginning of the article)

A*(x-x0)+B*(y-y0)+C*(z-z0)=0
or
-sin(Fi)*x + cos(Fi)*z - 6 = 0