Question

The plane \(\ell x+m y=0\) is rotated about its line of intersection with the plane \(z=0\) through an angle \(\alpha\). The equation changes to
(A) \(\ell x+m y \pm \tan \alpha \sqrt{\ell^{2}+m^{2}}=0\)
(B) \(\ell x+m y \pm z \tan \alpha \sqrt{\ell^{2}+m^{2}+1}=0\)
(C) \(\ell x+m y \pm z \tan \alpha \sqrt{\ell^{2}+1}=0\)
(D) \(\ell x+m y \pm z \tan \alpha \sqrt{\ell^{2}+m^{2}}=0\)

Answer 1

Ans:(D)
Hint \(: P_{1}: \ell x+m y=0, \quad P_{2}=z=0\)
Plane through common line of \(P_{1}\) and \(P_{2}\)
\(P_{3}: \ell x+m y+n z=0\)
angle between \(P_{1}\) and \(P_{3}=\alpha\)
\(\therefore \cos \alpha=\hat{n}_{1} \hat{n}_{2}\)
\(=\frac{\ell^{2}+\mathrm{m}^{2}}{\sqrt{\ell^{2}+\mathrm{m}^{2}} \sqrt{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}}=\sqrt{\frac{\ell^{2}+\mathrm{m}^{2}}{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}}\)
$$
\begin{aligned}
&\Rightarrow \cos ^{2} \alpha=\frac{\ell^{2}+\mathrm{m}^{2}}{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}} \\
&\Rightarrow \mathrm{n}^{2}=\left(\ell^{2}+\mathrm{m}^{2}\right) \tan ^{2} \alpha \\
&\Rightarrow \mathrm{n}=\pm \sqrt{\ell^{2}+\mathrm{m}^{2}} \tan \alpha \\
&\mathrm{P}_{3}: \ell \mathrm{x}+\mathrm{my} \pm \sqrt{\ell^{2}+\mathrm{m}^{2}} \tan \alpha=0
\end{aligned}
$$

Answer 2

cialis drug <a href="https://ordergnonline.com/">cialis 20mg usa</a> buy ed pills cheap

Answer 3

tadalafil 10mg us <a href="https://ordergnonline.com/">cialis prices</a> best ed drugs