If either \[\vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0} , \text{ then } \vec{a} \times \vec{b} = \vec{0} .\] Is the converse true? Justify your answer with an example.
Solution
\[\text{ If } \vec{a} = \vec{0} \text{ or } \vec{b} =0, \text{ then } \left| \vec{a} \right| \left| \vec{b} \right| \sin \theta \hat{ n } = \vec{0 .} \]
\[ \Rightarrow \vec{a} \times \vec{b} = \vec{0} \]
\[\text{ But the converse is not true as whenever } \vec{a} \times \vec{b} = \vec{0} , \text{ we cannot be sure that either } \vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0} .\]
\[\text{ For example } :\]
\[ \vec{a} = \hat{ i } + 2 \hat{ j } + 3 \hat{ k } \]
\[ \vec{b} = \hat{ i } + 2 \hat{ j } + 3 \hat{ k } \]
\[\text{ Here } ,\]
\[ \vec{a} \neq0\]
\[ \vec{b} \neq0\]
\[\text{ But } \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 2 & 3 \\ 1 & 2 & 3\end{vmatrix}\]
\[ = 0 \hat{ i } + 0 \hat{ j } + 0 \hat{ k } \]
\[ = \vec{0}\]
