If the plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$, then prove that the equation of the plane in its new position is $ax + by \pm \left( {\sqrt {{a^2} + {b^2}} \tan \alpha } \right)z = 0$.

Equation of the plane is $ax + by = 0$……..(i)

$\therefore$ Equation of the plane after new position is

$\frac{{ax\cos \alpha }}{{\sqrt {{a^2} + {b^2}} }} + \frac{{by\cos \alpha }}{{\sqrt {{b^2} + {a^2}} }} \pm z\sin \alpha = 0$

$\Rightarrow$ $\frac{{ax}}{{\sqrt {{a^2} + {b^2}} }} + \frac{{by}}{{\sqrt {{b^2} + {a^2}} }} \pm z\tan \alpha = 0$

[on dividing by $\cos \alpha$]
$\Rightarrow$ $ax + by \pm z\tan \alpha \sqrt {{\alpha ^2} + {b^2}}$

$= 0$ [on multiplying with $\sqrt {{a^2} + {b^2}}$]
Alternate Method
Given, planes are $ax + by = 0$

…..(i)
and $z = 0$

……(ii)
Therefore, the equation of any plane passing through the line of intersection of planes

(i) and (ii) may be taken as $ax + by + k = 0$.

Then, direction cosines of a normal to the plane

(iii) are $\frac{a}{{\sqrt {{a^2} + {b^2} + {k^2}} }},\frac{b}{{\sqrt {{a^2} + {b^2} + {k^2}} }}$,$\frac{c}{{\sqrt {{a^2} + {b^2} + {k^2}} }}$

and direction cosines of the normal to the plane

(i) are $\frac{a}{{\sqrt {{a^2} + {b^2}} }},\frac{b}{{\sqrt {{a^2} + {b^2}} }}$,0.
Since, the angle between the planes (i) and (ii) is $\alpha$,

$\therefore \cos \alpha = \frac{{a \cdot a + b \cdot b + k \cdot 0}}{{\sqrt {{a^2} + {b^2} + {k^2}} \sqrt {{a^2} + {b^2}} }}$

$= \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2} + {b^2} + {k^2}}}}$

$\Rightarrow$ ${k^2}{\cos ^2}\alpha = {a^2}\left( {1 - {{\cos }^2}\alpha } \right) + {b^2}\left( {1 - {{\cos }^2}\alpha } \right)$

$\Rightarrow$ ${k^2} = \frac{{\left( {{a^2} + {b^2}} \right){{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}$

$k = \pm \sqrt {{a^2} + {b^2}} \tan \alpha$
On putting this value in plane (iii),

we get the equation of the plane as

$ax + by + z\sqrt {{a^2} + {b^2}} \tan \alpha = 0$