Question

The plane $lx+my=0$ is rotated about its line of intersection with the plane $z=0$ through an angle $\alpha$. The equation of the plane in its new position is

• A. $lx+my\pm z\sqrt{l^2+m^2}\sin \alpha =0$
• B. $lx+my\pm z\sqrt{l^2+m^2}\tan \alpha =0$
• C. $lx+my\pm z\sqrt{l^2+m^2}\cot \alpha =0$
• D. None of these

Answer 1

The correct option is B $lx+my\pm z\sqrt{l^2+m^2}\tan \alpha =0$

The plane has been rotated about its line of intersection with the plane $z=0;$ hence it will pass through the intersection of the plane $lx+my=0$ and $z=0$.

Let the equation of the plane after rotation be $lx+my+\lambda z=0;$ ...(1)

Then it makes an angle $\alpha$ with the plane

$lx+my=0.$ ...(2)

$\therefore \cos \alpha =\frac{l.l+m.m+\lambda .0}{\sqrt{(l^2+m^2+n^2)}\sqrt{(l^2+m^2)}}=\frac{\sqrt{(l^2+m^2)}}{\sqrt{l^2+m^2+{\lambda }^2}}$

$\therefore (l^2+m^2+{\lambda }^2){\cos }^2\alpha =l^2+m^2$

${\lambda }^2{\cos }^2\alpha =(l^2+m^2)(1-{\cos }^2\alpha ),$

i.e., ${\lambda }^2=(l^2+m^2){\tan }^2\alpha$

$\lambda =\pm \sqrt{l^2+m^2}\tan \alpha .$

Putting this value of $\lambda$ in (1), the required plane is given by,

$lx+my\pm z\sqrt{l^2+m^2}\tan \alpha =0.$