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Question
If either \[\vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0}\] then \[\vec{a} \cdot \vec{b} = 0 .\] But the converse need not be true. Justify your answer with an example.
Solution
\[\text{ Let us assume that either }\left| \vec{a} \right|=0 \text{ or } \left| \vec{b} \right| = 0\]
\[Then, \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta = 0....................... (\theta \text{ is the angle between } \vec{a} \text{ and } \vec{b} )\]
\[\text{ Now, let us assume that } \vec{a} . \vec{b} = 0\]
\[ \Rightarrow \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta = 0\]
\[\text{ But here we cannot say that either }\left| \vec{a} \right|=0 \text{ or }\left| \vec{b} \right| = 0 ............. \text{ (Because even cos } \theta \text{ can be zero) }\]
\[\text{ For example, let} \]
\[ \vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k} \text{ and } \vec{b} = - 3 \hat{i} + 2 \hat{k} \]
\[Here,\left| \vec{a} \right|=\sqrt{4 + 1 + 9}=\sqrt{14}\neq0\]
\[\left| \vec{b} \right| = \sqrt{9 + 4} = \sqrt{13}\neq0\]
\[\text{But } \vec{a} . \vec{b} = \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) . \left( - 3 \hat{i} + 2 \hat(k) \right) = - 6 + 0 + 6 = 0\]
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