Question

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 \[\vec{a} \text{ and }  \vec{b}\] .

 

 

 

Answer 1

\[\text{ If }  \vec{a} \text{ and }  \vec{b} \text{ are two non-zero non-parallel vectors, then the vector product denoted by }  \vec{a} \times \vec{b} \text{ is defined as }  \vec{a} \times \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \sin \theta   \hat{ η } . \] 

\[\text{ Here } ,\theta \text{ is the angle between } \vec{a} \text{ and } \vec{b} \text{ and }  \hat{η } \text{  is the unit vector perpendicular to the plane of } \vec{a} \text{ and }  \vec{b} \text{ such that } \vec{a} , b \text{ and } \hat{ η } \text{  form a right handed system. } \]

\[LHS=\left| \vec{a} \times \vec{b} \right|\]

\[ =\left| \vec{a} \right| \left| \vec{b} \right| \sin \theta \]

\[ = \left| \vec{a} \right| \left| \vec{b} \right| \sin \theta \times \frac{\cos \theta}{\cos \theta}\]

\[ = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta \frac{\sin \theta}{\cos \theta}\]

\[ = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta \tan \theta\]

\[ = \left( \vec{a} . \vec{b} \right) \tan \theta\]

\[ = RHS\]

\[\text{ Hence proved } .\]