Question

If $x=r\sin \alpha \cos \beta ,y=r\sin \alpha \sin \beta \text{and}z=r\cos \alpha ,$ then

• A. $x^2+y^2+z^2=r^2$
• B. $x^2-y^2+z^2=r^2$
• C. $x^2+y^2-z^2=r^2$
• D. $z^2+y^2-x^2=r^2$

Answer 1

The correct option is A $x^2+y^2+z^2=r^2$

Given,

$x=r\sin \alpha \cos \beta$ ...(i)

$y=r\sin \alpha \sin \beta$ ...(ii)

$z=r\cos \alpha$ ...(iii)

Squaring and adding (i) and (ii), we get,

$x^2+y^2=r^2{\sin }^2\alpha {\cos }^2\beta +r^2{\sin }^2\alpha {\sin }^2\beta =r^2{\sin }^2\alpha ({\cos }^2\beta +{\sin }^2\beta )=r^2{\sin }^2\alpha$ ...(iv)

Now, squaring (iv) and adding it to (iii), we get,

$x^2+y^2+z^2=r^2{\sin }^2\alpha +r^2{\cos }^2\alpha =r^2({\sin }^2\alpha +{\cos }^2\alpha )=r^2$

Hence, correct option is A.