Question

The plane $ax+by=0$ is rotated about its line of intersection with the plane $z=0$ through an angle $\theta (\theta \ne \frac{n\pi }{2},nϵI).$
The equation of the plane in the new position can be given as

• A. $ax+by+z\sqrt{a^2+b^2}\tan \theta =0$
• B. $ax+by+z\sqrt{a^2+b^2}\cot \theta =0$
• C. $ax+by-z\sqrt{a^2+b^2}\tan \theta =0$
• D. $ax+by-z\sqrt{a^2+b^2}\cot \theta =0$

Answer 1

Answer: (A), (C)

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

Given plane $P_1:ax+by=0$ and $P_2:z=0$
$P_1+\lambda P_2=0$
$ax+by+\lambda z=\mathrm{0....}\left(1\right)$
Now, the angle between $P_1$ and the line of intersection is
$\cos \theta =\frac{a^2+b^2}{\sqrt{a^2+b^2}\sqrt{a^2+b^2+{\lambda }^2}}$
$\sqrt{a^2+b^2+{\lambda }^2}=\sqrt{a^2+b^2}\sec \theta$
$a^2+b^2+{\lambda }^2=(a^2+b^2){\sec }^2\theta$
${\lambda }^2=(a^2+b^2)[{\sec }^2\theta -1]$
${\lambda }^2=(a^2+b^2){\tan }^2\theta$
$\lambda =\pm \sqrt{a^2+b^2}\tan \theta$
$\therefore ax+by\pm \sqrt{a^2+b^2}\tan \theta z=0$
Hence, options 'A' and 'C' are correct.