Question

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

 

Answer 1

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