Given a sphere $S$ whose center is at O and whose radius is 1 , take three points A , B and C on $S$ such that $$\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } } = \overrightarrow { \mathrm { OB } } \cdot \overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { OC } } \cdot \overrightarrow { \mathrm { OA } } = 0 .$$ Note that $\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } }$, etc., refers to the inner product of the two vectors. (1) It follows that $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { AC } } = \square \mathbf { A } , | \overrightarrow { \mathrm { AB } } | = \sqrt { \mathbf { B } } , \cos \angle \mathrm { BAC } = \frac { \square \mathbf { C } } { \square }$ and the area of the triangle ABC is $\frac { \sqrt { \mathbf { E } } } { \mathbf { F } }$. (2) Let G be the center of gravity of triangle ABC and P be the intersection point of the ray (half line) OG and sphere $S$. Since $\overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { G } } ( \overrightarrow { \mathrm { OA } } + \overrightarrow { \mathrm { OB } } + \overrightarrow { \mathrm { OC } } )$, we have $$\begin{gathered}
| \overrightarrow { \mathrm { OG } } | = \frac { \sqrt { \mathbf { I } } } { \sqrt { \mathbf { J } } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \square \mathbf { K } - \sqrt { \square \mathbf { L } } } { \square \mathbf { M } } \\
\overrightarrow { \mathrm { AG } } \cdot \overrightarrow { \mathrm { PG } } = \square \mathbf { N } .
\end{gathered}$$ Hence the volume of the tetrahedron PABC is $\frac { \sqrt { \mathbf { Q } } - \mathbf { P } } { \mathbf { Q } }$.
Given a sphere $S$ whose center is at O and whose radius is 1 , take three points A , B and C on $S$ such that
$$\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } } = \overrightarrow { \mathrm { OB } } \cdot \overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { OC } } \cdot \overrightarrow { \mathrm { OA } } = 0 .$$
Note that $\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } }$, etc., refers to the inner product of the two vectors.
(1) It follows that $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { AC } } = \square \mathbf { A } , | \overrightarrow { \mathrm { AB } } | = \sqrt { \mathbf { B } } , \cos \angle \mathrm { BAC } = \frac { \square \mathbf { C } } { \square }$\\
and the area of the triangle ABC is $\frac { \sqrt { \mathbf { E } } } { \mathbf { F } }$.
(2) Let G be the center of gravity of triangle ABC and P be the intersection point of the ray (half line) OG and sphere $S$.
Since $\overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { G } } ( \overrightarrow { \mathrm { OA } } + \overrightarrow { \mathrm { OB } } + \overrightarrow { \mathrm { OC } } )$, we have
$$\begin{gathered}
| \overrightarrow { \mathrm { OG } } | = \frac { \sqrt { \mathbf { I } } } { \sqrt { \mathbf { J } } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \square \mathbf { K } - \sqrt { \square \mathbf { L } } } { \square \mathbf { M } } \\
\overrightarrow { \mathrm { AG } } \cdot \overrightarrow { \mathrm { PG } } = \square \mathbf { N } .
\end{gathered}$$
Hence the volume of the tetrahedron PABC is $\frac { \sqrt { \mathbf { Q } } - \mathbf { P } } { \mathbf { Q } }$.