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Obtain the vector and cartesian equation of thesphere whose centre is $(1 , -1 , 1 )$ and radius is the same as that of the sphere $|\overrightarrow{r}-(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{2k})|=5.$

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  • Vector equation of a sphere with centre $ \overrightarrow c$ and radius $a$ is $|\overrightarrow r-\overrightarrow c|=a$ When the centre is at the origin, the vector equation is $|\overrightarrow r|=a$ Cartesian equation is $ (x-c_1)^2+(y-c_2)^2+(z-c_3)^2=a^2$ where $(c_1, c_2, c_3 )$ is the centre.
The sphere $|\overrightarrow r-(\overrightarrow i+\overrightarrow j+2\overrightarrow k)|=5$ is 5 units.
The required sphere has its centre at $(1, -1, 1)(\overrightarrow c=\overrightarrow i-\overrightarrow j+\overrightarrow k)$ and has a radius of 5 units.
The vector equation of the sphere is
$ |\overrightarrow r-\overrightarrow c|=5$ i.e., $|\overrightarrow r-(\overrightarrow i-\overrightarrow j+\overrightarrow k)|=5$
The cartesian equation is
$ (x-c_1)^2+(y-c_2)^2+(z-c_3)^2=a^2$
$ (x-1)^2+(y+1)^2+(z-1)^2=25$
$ x^2+y^2+z^2-2x+2y-2z+1+1+1-25=0$
$ \therefore x^2+y^2+z^2-2x+2y-2z-22=0$
answered Jun 18, 2013 by thanvigandhi_1

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