Let PM be perpendicular from the point $\mathrm{P}(1,2,3)$ to $\mathrm{x}-\mathrm{y}$ plane. If $\ov

Question

Let PM be perpendicular from the point $\mathrm{P}(1,2,3)$ to $\mathrm{x}-\mathrm{y}$ plane. If $\overrightarrow{O P}$ makes an angle $	heta$ with the positive direction of z-axis and $\overrightarrow{O M}$ makes an angle $\phi$ with the positive direction of x-axis, where O is the origin and $	heta$ and $\phi$ are acute angles, then:

Professor PC Thomas Classes · Answer

If $\mathrm{P} \mathrm{be}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ then from the figure, $x=r \sin 	heta \cos \phi, y=r \sin 	heta \sin \phi, z=r \cos 	heta$
$
\begin{aligned}
& \Rightarrow \quad 1=r \sin 	heta \cos \phi, 2=r \sin 	heta \sin \phi, 3=r \cos 	heta \
& \Rightarrow \quad 1^2+2^2+3^2=r^2= \pm \sqrt{14} \
& 	herefore \quad \sin 	heta \cos \phi=\frac{1}{\sqrt{14}}, \sin 	heta \sin \phi=\frac{1}{\sqrt{14}} \
& \sin 	heta \sin \phi=\frac{2}{\sqrt{14}}, \cos 	heta=\frac{3}{\sqrt{14}}
\end{aligned}
$
(neglecting negative sign as $	heta$ and $\phi$ are acute)
$
	herefore \frac{\sin 	heta \sin \phi}{\sin 	heta \cos \phi}=\frac{2}{1} \Rightarrow 	an \phi=2
$
Also, $	an 	heta=\frac{\sqrt{5}}{3}$