Question

If $x=r\cos \alpha \cos \beta \cos \gamma ,y=r\cos \alpha \cos \beta \sin \gamma ,z=r\sin \alpha \cos \beta ,\mu =r\sin \beta$, then $x^2+y^2+z^2+{\mu }^2=$

• A. $1$
• B. $0$
• C. $r$
• D. $r^2$

Answer 1

The correct option is D $r^2$

(D) $x^2+y^2$
$=r^{2}co{s}^2\alpha co{s}^2\beta co{s}^2\gamma +r^{2}co{s}^2\alpha co{s}^2\beta si{n}^2\gamma$
$=r^{2}co{s}^2\alpha co{s}^2\beta (co{s}^2\gamma +si{n}^2\gamma )$
$=r^{2}co{s}^2\alpha co{s}^2\beta$
$(x^2+y^2)+z^2=r^{2}co{s}^2\alpha co{s}^2\beta +r^{2}si{n}^2\alpha co{s}^2\beta$
$=r^{2}co{s}^2\beta (co{s}^2\alpha +si{n}^2\alpha )$
$\therefore x^2+y^2+z^2+{\mu }^2=r^{2}co{s}^2\beta +r^{2}si{n}^2\beta =r^2$