# Prove that the center of the sphere circumscribing the tetrahedron is given by position vector

If $OABC$ is a tetrahedron where $O$ is the origin and $A,B$ and $C$ are the other three vertices with position vectors $\vec{a},\vec{b}$ and $\vec{c},$ respectively,then prove that the center of the sphere circumscribing the tetrahedron is given by position vector $$\vec{r}=\frac{a^2(\vec{b}\times\vec{c})+b^2(\vec{c}\times\vec{a})+c^2(\vec{a}\times\vec{b})}{2[\vec{a}\hspace{0.25cm}\vec{b}\hspace{0.25cm}\vec{c}]}$$,where $[\vec{a}\hspace{0.25cm}\vec{b}\hspace{0.25cm}\vec{c}]$ denotes the scalar triple product of $\vec{a},\vec{b},\vec{c}$.

My Attempt:
Since $O$ be the origin and the position vectors of $A,B,C$ are $\vec{a},\vec{b},\vec{c}$.Let the center of the circumsphere be $P$ whose positin vector is $\vec{r}$
$\vec{PA}=\vec{a}-\vec{r},\vec{PB}=\vec{b}-\vec{r},\vec{PC}=\vec{c}-\vec{r}$,where $|\vec{PA}|=|\vec{PB}|=|\vec{PC}|=|\vec{OP}|=|\vec{r}|$
Since $|\vec{a}-\vec{r}|=|\vec{r}|\Rightarrow (\vec{a}-\vec{r}).(\vec{a}-\vec{r})=\vec{r}.\vec{r}$
$\Rightarrow a^2-2\vec{a}.\vec{r}+r^2=r^2$
$a^2=2\vec{a}.\vec{r}$
Similarly,$b^2=2\vec{b}.\vec{r},c^2=2\vec{c}.\vec{r}$

Since the points $A,B,C$ and $O$ form a tetrahedron we know that the vectors $\vec{a},\vec{b}$ and $\vec{c}$ are not coplanar and hence the vectors $\vec{a}\times\vec{b},\vec{b}\times\vec{c}$ and $\vec{c}\times\vec{a}$ are also not coplanar. This means that we can express the position vector of the circumcentre as a linear combination of these vectors.
So let $\vec{r}=\lambda (\vec{a}\times\vec{b})+\mu(\vec{b}\times\vec{c})+\gamma (\vec{c}\times\vec{a})$
Then $\vec{r}\cdot\vec{a}=\mu[\vec{a} \space\space\vec{b}\space\space \vec{c}]$ , $\vec{r}\cdot\vec{b}=\gamma[\vec{a} \space\space\vec{b}\space\space \vec{c}]$ , $\vec{r}\cdot\vec{c}=\lambda[\vec{a} \space\space\vec{b}\space\space \vec{c}]$
Now just use the values of $\vec{r}\cdot\vec{a} , \vec{r}\cdot\vec{b}$ and $\vec{r}\cdot\vec{c}$ that you obtained in your attempt.