Question

For any two vectors \[\vec{a}\] and \[\vec{b}\] , find \[\vec{a} . \left( \vec{b} \times \vec{a} \right) .\]

 

 

 

 

Answer 1

\[\text{ Let } :\]
\[ \vec{a} = a_1 \hat{ i }  + a_2 \hat{ j }  + a_3 \hat{ k}  \]
\[ \vec{b} = b_1 \hat{ i  }+ b_2 \hat{ j }  + b_3 \hat{ k }  \]
\[ \vec{b} \times \vec{a} = \begin{vmatrix}\hat{ i }  & \hat{ j }  & \hat{ k }  \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3\end{vmatrix}\]
\[ = \hat{ i } \left( b_2 a_3 - b_3 a_2 \right) - \hat{ j }  \left( b_1 a_3 - b_3 a_1 \right) + \hat{ k }\left( b_1 a_2 - b_2 a_1 \right)\]
\[\text{ Now } ,\]
\[ \vec{a} . \left( \vec{b} \times \vec{a} \right)\]
\[ = \left( a_1 \hat{ i }  + a_2 \hat{ j }  + a_3 \hat{ k} \right) . \left[ \hat{ i }  \left( b_2 a_3 - b_3 a_2 \right) - \hat{ j } \left( b_1 a_3 - b_3 a_1 \right) + \hat{k}  \left( b_1 a_2 - b_2 a_1 \right) \right]\]
\[ = a_1 \left( b_2 a_3 - b_3 a_2 \right) - a_2 \left( b_1 a_3 - b_3 a_1 \right) + a_3 \left( b_1 a_2 - b_2 a_1 \right)\]
\[ = a_1 b_2 a_3 - a_1 b_3 a_2 - a_2 b_1 a_3 + a_2 b_3 a_1 + a_3 b_1 a_2 - a_3 b_2 a_1 \]
\[ = 0\]