If the vectors $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ form the sides $$BC,$$ $$CA$$ and $$AB$$ respectively of a triangle $$ABC,$$ then

Let the vectors $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $$\overrightarrow d $$ be such that
$$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow c \times \overrightarrow d } \right) = 0.$$ Let $${P_1}$$ and $${P_2}$$ be planes determined
by the pairs of vectors $$\overrightarrow a .\overrightarrow b $$ and $$\overrightarrow c .\overrightarrow d $$ respectively. Then the angle between $${P_1}$$ and $${P_2}$$ is

If $$\overrightarrow a \,,\,\overrightarrow b $$ and $$\overrightarrow c $$ are unit coplanar vectors, then the scalar triple product $$\left[ {2\overrightarrow a - \overrightarrow b ,2\overrightarrow b - \overrightarrow c ,2\overrightarrow c - \overrightarrow a } \right] = $$

The dimension of $$\left( {{1 \over 2}} \right){\varepsilon _0}{E^2}$$
( $${\varepsilon _0}$$ : permittivity of free space, E electric field )