question_answer

If \[\vec{a}\] and \[\vec{b}\] are two vectors, then show that \[{{(\vec{a}\times \vec{b})}^{2}}+{{(\vec{a}\cdot \vec{b})}^{2}}={{a}^{2}}{{b}^{2}}.\]

Answer:

\[{{(\vec{a}\times \vec{b})}^{2}}=|\vec{a}\times \vec{b}{{|}^{2}}\]      \[[\because \,\,\,{{(\vec{a})}^{2}}=\vec{a}\cdot \vec{a}=|\vec{a}{{|}^{2}}]\] \[=\,\,|\vec{a}{{|}^{2}}|\vec{b}{{|}^{2}}{{\sin }^{2}}\theta ={{a}^{2}}{{b}^{2}}(1-{{\cos }^{2}}\theta )\] \[={{a}^{2}}{{b}^{2}}-{{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta \] \[={{a}^{2}}{{b}^{2}}-|\vec{a}{{|}^{2}}|\vec{b}{{|}^{2}}{{\cos }^{2}}\theta \] \[={{a}^{2}}{{b}^{2}}-{{(\vec{a}\cdot \vec{b})}^{2}}\] \[[\because \vec{a}\cdot \vec{b}=\,\,|\vec{a}||\vec{b}|\cos \theta ]\] \[\therefore {{(\vec{a}\times \vec{b})}^{2}}+{{(\vec{a}\cdot \vec{b})}^{2}}={{a}^{2}}{{b}^{2}}\]           Hence proved.