For any two vectors \vec{a}\; and\; \vec{b} , prove that  \left ( \vec{a}\times \vec{b} \right )^{2}= \vec{a}^{2}\; \vec{b}^{2}-\left ( \vec{a}\cdot \vec{b} \right )^{2}
 

 

 

 

 
 
 
 
 

Answers (1)

consider \left ( \bar{a}\times \bar{b} \right )^2= \left \{ \left | \vec{a} \right | \left | \vec{b} \right |\sin \theta \right \}^{2}
\Rightarrow \left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}\sin ^{2}\theta
\Rightarrow \left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}\left ( 1-\cos ^{2}\theta \right )
\Rightarrow \left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}-\left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}\cos ^{2}\theta
\Rightarrow\left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}-\left \{ \left | \vec{a} \right |\left | \vec{b} \right | \cos \theta \right \}^{2}
\therefore \left ( \vec{a}\times \vec{b} \right )^{2}= \vec{a}^{2}\vec{b}^{2}-\left ( \vec{a}\cdot \vec{b} \right )^{2}

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