Question

Let $PM$ be the perpendicular from the point $P(1,2,3)$ to the $xy$ plane. If $\overrightarrow{OP}$ makes an angle $\theta$ with the positive direction of the $z$-axis and $\overrightarrow{OM}$ makes an angle $\varphi$ with the positive direction of $x$-axis, where $O$ is the origin and $\theta$ and $\varphi$ are acute angles, then

• A. $\cos \theta \cos \varphi =\frac{1}{\sqrt{14}}$
  
• B. $\sin \theta \sin \varphi =\frac{2}{\sqrt{14}}$
  
• C. $\tan \varphi =2$
  
• D. $\tan \theta =\frac{\sqrt{5}}{3}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\sin \theta \sin \varphi =\frac{2}{\sqrt{14}}$

C $\tan \varphi =2$

D $\tan \theta =\frac{\sqrt{5}}{3}$

If $P$ be $(x,y,z),$ then from the figure,
$x=r\sin \theta \cos \varphi ,y=r\sin \theta \sin \varphi$ and $z=r\cos \theta$
$1=r\sin \theta \cos \varphi ,2=r\sin \theta \sin \varphi$ and $3=r\cos \theta$
$\Rightarrow 1^2+2^2+3^2=r^2$
$\Rightarrow r=\pm \sqrt{14}$
$\therefore \sin \theta \cos \varphi =\frac{1}{\sqrt{14}},$
$\sin \theta \sin \varphi =\frac{2}{\sqrt{14}}$ and $\cos \theta =\frac{3}{\sqrt{14}}$
(neglecting negative sign as $\theta$ and $\varphi$ are acute)

$\frac{\sin \theta \sin \varphi }{\sin \theta \cos \varphi }=\frac{2}{1}$
$\Rightarrow \tan \varphi =2$
Also, $\tan \theta =\frac{\sqrt{5}}{3}$