Question

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

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

Answer 1

Answer: (A), (B), (C)

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

P be $(x,y,z)$

$x=r\sin \theta \cdot \cos \varphi ;y=r\sin \theta \cdot \sin \varphi ;z=r\cos \theta$

$1=r\sin \theta \cdot \cos \varphi ;2=r\sin \theta \cdot \sin \varphi ;3=r\cos \theta$

$1^2+2^2+3^2=r^2{\sin }^2\theta {\cos }^2\varphi +r^2{\sin }^2\theta {\sin }^2\varphi +r^2{\cos }^2\theta$

$=r^2{\sin }^2\theta ({\cos }^2\varphi +{\sin }^2\varphi )+r^2{\cos }^2\theta$

$=r^2{\sin }^2\theta +r^2{\cos }^2\theta =r^2$

$r^2=14$

$r=\pm \sqrt{14}$

$\sin \theta \cos \varphi =\frac{1}{\sqrt{14}};\sin \theta \sin \varphi =\frac{2}{\sqrt{14}},$

$\cos \theta =\frac{3}{\sqrt{14}}$ (neglecting. (-)ve sign)



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

$\tan \varphi =2\tan \theta =\frac{\sqrt{5}}{3}$