Question

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

• A. $lx+my\pm z\sqrt{l^2+m^2}\cot \theta =0$
• B. $lx+my\pm z\sqrt{l+m}\tan \theta =0$
• C. $lx+my\pm z\sqrt{l+m}\cot \theta =0$
• D. $lx+my\pm z\sqrt{l^2+m^2}\tan \theta =0$

Answer 1

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

Given two planes are

$P_1:lx+my=0$

$P_2:z=0$

eq of planes passing through the intersection of planes is

$P_1+\lambda P_2=0$

$lx+my+\lambda z=0$---------------(1)

now the planes $P_1$ is rotated through angle $\theta$

SO the angle between the rotated plane $lx+my+\lambda z=0$ and $P_1$ is $\theta$

to find value of $\lambda$

$\cos \theta =\frac{n_{new}\cdot n_1}{\left|n_{new}\right|\left|n_1\right|}$

$\cos \theta =\frac{l^2+m^2}{\sqrt{l^2+m^2}\sqrt{l^2+m^2+{\lambda }^2}}$

$\cos \theta =\frac{\sqrt{l^2+m^2}}{\sqrt{l^2+m^2+{\lambda }^2}}$

$\sqrt{l^2+m^2+{\lambda }^2}=\frac{\sqrt{l^2+m^2}}{\cos \theta }$

on squaring both sides

$l^2+m^2+{\lambda }^2=\frac{l^2+m^2}{{\cos }^2\theta }$

${\lambda }^2=\frac{l^2+m^2}{{\cos }^2\theta }-l^2-m^2$

${\lambda }^2=\frac{l^2+m^2-l^2{\cos }^2\theta -m^2{\cos }^2\theta }{{\cos }^2\theta }$

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

$si{n}^2\theta +{\cos }^2\theta =1$

${\lambda }^2=\frac{l^2{\sin }^2\theta +m^2{\sin }^2\theta }{{\cos }^2\theta }$

${\lambda }^2=\frac{l^2{\sin }^2\theta }{{\cos }^2\theta }+\frac{m^2{\sin }^2\theta }{{\cos }^2\theta }$

${\lambda }^2=l^2{\tan }^2\theta +m^2{\tan }^2\theta$

${\lambda }^2={\tan }^2\theta (l^2+m^2)$

$\lambda =\sqrt{{\tan }^2\theta (l^2+m^2)}$

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

putting $\lambda$ in eq (1)

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