A) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+xyz=0\]
B) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=0\]
C) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+xyz=1\]
D) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1\]
Correct Answer: D
Solution :
Given that \[{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \] \[\Rightarrow \,\,{{\cos }^{-1}}(x)+{{\cos }^{-1}}(y)+{{\cos }^{-1}}(z)={{\cos }^{-1}}(-1)\] \[\Rightarrow \,\,{{\cos }^{-1}}(x)+{{\cos }^{-1}}(y)={{\cos }^{-1}}(-1)-{{\cos }^{-1}}(z)\] \[\Rightarrow \,\,{{\cos }^{-1}}(xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}})}={{\cos }^{-1}}\,\left\{ (-1)\,\,(z) \right\}\] \[\Rightarrow \,\,xy-\sqrt{(1-{{x}^{2}})\,\,(1-{{y}^{2}})}=-z\] \[\Rightarrow \,\,(xy+z)=\sqrt{(1-{{x}^{2}})\,\,(1-{{y}^{2}})}\] Squaring both sides we get\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1\]. Trick: Put \[x=y=z=\frac{1}{2},\] so that \[{{\cos }^{-1}}\frac{1}{2}+{{\cos }^{-1}}\frac{1}{2}+{{\cos }^{-1}}\frac{1}{2}=\pi \] Obviously (d) holds for these values of x, y, z.
You need to login to perform this action.
You will be redirected in
3 sec
