Question

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

Answer 1

Given,
Plane 1: $ax+by=0$
Plane 2: $z=0$

Equation of any plane through the line of intersection of these planes is:

Plane 3: $ax+by+kz=0;k\in R$

Angle between two planes is equal to the angle between their normals. $\alpha$ the angle between plane 3 and plane 1.

Direction ratios of normal to plane 1 are $(a,b,0)$

Direction ratios of normal to plane 3 are $(a,b,k)$

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

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

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

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

$\Rightarrow k=\pm \sqrt{a^2+b^2}\tan \alpha$

So, equation of the plane is

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

Hence, $c=0$