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)

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

Equation of any plane passing through the line of intersection of the planes

$ax+by=0$ ...$\left(1\right)$

and $z=0$ ...$\left(2\right)$

is $ax+by+\lambda z=0$ ...$\left(3\right)$

Since plane $\left(3\right)$ makes an angle $\alpha$ with the plane $\left(1\right)$, we have

$\cos \theta =\frac{|a.a+b.b+c.c|}{\sqrt{a^2+b^2}\sqrt{a^2+b^2+{\lambda }^2}}$

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

$\Rightarrow {\cos }^2\theta =\frac{a^2+b^2}{a^2+b^2+{\lambda }^2}\Rightarrow {\sec }^2\theta =\frac{a^2+b^2+{\lambda }^2}{a^2+b^2}=1+\frac{{\lambda }^2}{a^2+b^2}$

$\Rightarrow {\sec }^2\theta -1=\frac{{\lambda }^2}{a^2+b^2}\Rightarrow {\tan }^2\theta =\frac{{\lambda }^2}{a^2+b^2}$

Substituting these values of $\lambda$ in $\left(3\right)$, the equation of the plane in its new position is

$ax+by\pm \left(\sqrt{a^2+b^2}\tan \theta \right)z=0$