Volume of Tetrahedron
Tetrahedron is a pyramid having a triangular base. Therefore,
Volume = ⅓ [(Height) x (Area of base)]
\(=\frac{1}{3}h\times \frac{1}{2}\left| \overrightarrow{AB}\times \overrightarrow{AC} \right|\),
\(=\frac{1}{6}h\left| \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a} \right|\),
Now h is projection of on vector which is normal to the plane.
Vector normal to the plane ABC is \(\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\). Therefore,
\(h=\frac{\overrightarrow{a}.\left( \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a} \right)}{\left| \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a} \right|}\),
\(=\frac{\left[ \overrightarrow{a}\ \ \overrightarrow{b}\ \ \overrightarrow{c} \right]}{\left| \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a} \right|}\),
Volume \(=\frac{1}{6}\frac{\left[ \overrightarrow{a}\ \ \overrightarrow{b}\ \ \overrightarrow{c} \right]}{\left| \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a} \right|}\times \left| \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a} \right|\),
\(=\frac{1}{6}\left[ \overrightarrow{a}\ \ \overrightarrow{b}\ \ \overrightarrow{c} \right]\).