|vec{a} times vec{b}|^2 + (vec{a} cdot vec{b} | vedclass

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(B) We know that for any two vectors $\vec{a}$ and $\vec{b}$,the magnitude of the cross product is given by $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\ve

Vedclass · Accepted Answer

$(\vec{a} \cdot \vec{a}) (\vec{b} \cdot \vec{b})$

Vedclass · Answer

(B) We know that for any two vectors $\vec{a}$ and $\vec{b}$,the magnitude of the cross product is given by $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta$. Squaring this,we get $|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta$. The dot product is given by $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$. Squaring this,we get $(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 	heta$. Adding these two expressions: $|\vec{a} 	imes \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 (\sin^2 	heta + \cos^2 	heta)$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have: $|\vec{a} 	imes \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 = (\vec{a} \cdot \vec{a}) (\vec{b} \cdot \vec{b})$.

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = \dots$

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