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प्रश्न
Define \[\vec{a} \times \vec{b}\] and prove that \[\left| \vec{a} \times \vec{b} \right| = \left( \vec{a} . \vec{b} \right)\] tan θ, where θ is the angle between \[\vec{a} \text{ and } \vec{b}\] .
उत्तर
\[\text{ If } \vec{a} \text{ and } \vec{b} \text{ are two non-zero non-parallel vectors, then the vector product denoted by } \vec{a} \times \vec{b} \text{ is defined as } \vec{a} \times \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \sin \theta \hat{ η } . \]
\[\text{ Here } ,\theta \text{ is the angle between } \vec{a} \text{ and } \vec{b} \text{ and } \hat{η } \text{ is the unit vector perpendicular to the plane of } \vec{a} \text{ and } \vec{b} \text{ such that } \vec{a} , b \text{ and } \hat{ η } \text{ form a right handed system. } \]
\[LHS=\left| \vec{a} \times \vec{b} \right|\]
\[ =\left| \vec{a} \right| \left| \vec{b} \right| \sin \theta \]
\[ = \left| \vec{a} \right| \left| \vec{b} \right| \sin \theta \times \frac{\cos \theta}{\cos \theta}\]
\[ = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta \frac{\sin \theta}{\cos \theta}\]
\[ = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta \tan \theta\]
\[ = \left( \vec{a} . \vec{b} \right) \tan \theta\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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