Question

Let $PM$ be the perpendicular from the point $P(1,2,3)$ to the $x-y$ 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. $\cos \theta \cos \varphi =1/\sqrt{14}$
• B. $\sin \theta \sin \varphi =2/\sqrt{14}$
• C. $\tan \varphi =2$
• D. $\tan \theta =\sqrt{5}/3$

Answer 1

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

The correct options are
B $\sin \theta \sin \varphi =2/\sqrt{14}$
C $\tan \varphi =2$
D $\tan \theta =\sqrt{5}/3$

If P has the coordinates $(x,y,z)$
Then, $x=r\sin \theta \cos \varphi ,y=r\sin \theta \sin \varphi ,z=r\cos \theta$
So, $1=r\sin \theta \cos \varphi$ ....(i)
$2=r\sin \theta \sin \varphi$ .....(ii)
$3=r\cos \theta$ .....(iii)
Squaring and adding $\left(i\right)$ and $\left(ii\right)$, we get
$r^2{\sin }^2\theta ({\sin }^2\varphi +{\cos }^2\varphi )=5$
$\Rightarrow r^2{\sin }^2\theta =5$
$\Rightarrow r\sin \theta =\sqrt{5}$ ....(iv)
Dividing $\left(iv\right)$ by $\left(iii\right)$, we get
$\tan \theta =\frac{\sqrt{5}}{3}$
Dividing $\left(ii\right)$ by $\left(i\right)$, we get
$\Rightarrow \tan \varphi =2$
Since, $r^2=x^2+y^2+z^2$
$r^2=14$
$\Rightarrow r=\sqrt{14}$
Hence, by $\left(ii\right)$
$\sin \theta \sin \varphi =\frac{2}{\sqrt{14}}$
Hence, options 'B', 'C' and 'D' are correct.