A) \[\frac{{{r}^{2}}}{2}\]
B) \[{{r}^{2}}\]
C) \[{{r}^{2}}-1\]
D) \[{{r}^{2}}+1\]
Correct Answer: B
Solution :
Given,\[x\text{ }=\text{ }r\text{ }sin\text{ }A\text{ }cos\text{ }B\] ... (i) \[y\text{ }=\text{ }r\text{ }sin\text{ }A\,sin\text{ B}\] \[\Leftrightarrow \,\,{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A\,.\,{{\sin }^{2}}B\] ... (ii) \[z=r\cos A\] \[\Leftrightarrow \,{{z}^{2}}={{r}^{2}}{{\cos }^{2}}A\] (iii) On adding Eqs. (i), (ii) and (iii), we get \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}\] \[={{r}^{2}}{{\sin }^{2}}A{{\cos }^{2}}B+{{r}^{2}}\,{{\sin }^{2}}A\,.\,{{\sin }^{2}}B\] \[+{{r}^{2}}{{\cos }^{2}}A\] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}{{\sin }^{2}}A\left( {{\cos }^{2}}B+\sin 2 \right)B\] \[+{{r}^{2}}{{\cos }^{2}}A\] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}{{\sin }^{2}}A+{{r}^{2}}{{\cos }^{2}}A\] \[={{r}^{2}}\left[ {{\sin }^{2}}A+{{\cos }^{2}}A \right]\] \[={{r}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
