If vec{u} = vec{a} - vec{b} and vec{v} = vec{ | vedclass

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(A) Given $\vec{u} = \vec{a} - \vec{b}$ and $\vec{v} = \vec{a} + \vec{b}$.Calculating the cross product: $\vec{u} 	imes \vec{v} = (\vec{a} - \vec{b})

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$2 \sqrt{16 - (\vec{a} \cdot \vec{b})^2}$

Vedclass · Answer

(A) Given $\vec{u} = \vec{a} - \vec{b}$ and $\vec{v} = \vec{a} + \vec{b}$.Calculating the cross product: $\vec{u} 	imes \vec{v} = (\vec{a} - \vec{b}) 	imes (\vec{a} + \vec{b})$.Using the distributive property: $\vec{u} 	imes \vec{v} = \vec{a} 	imes \vec{a} + \vec{a} 	imes \vec{b} - \vec{b} 	imes \vec{a} - \vec{b} 	imes \vec{b}$.Since $\vec{a} 	imes \vec{a} = 0$ and $\vec{b} 	imes \vec{b} = 0$,and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$,we get $\vec{u} 	imes \vec{v} = 0 + \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b} - 0 = 2(\vec{a} 	imes \vec{b})$.Taking the magnitude: $|\vec{u} 	imes \vec{v}| = 2|\vec{a} 	imes \vec{b}|$.Using the identity $|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$.Given $|\vec{a}| = 2$ and $|\vec{b}| = 2$,then $|\vec{a}|^2 = 4$ and $|\vec{b}|^2 = 4$.$|\vec{u} 	imes \vec{v}| = 2 \sqrt{|\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2} = 2 \sqrt{4 	imes 4 - (\vec{a} \cdot \vec{b})^2} = 2 \sqrt{16 - (\vec{a} \cdot \vec{b})^2}$.

If $\vec{u} = \vec{a} - \vec{b}$ and $\vec{v} = \vec{a} + \vec{b}$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\vec{u} \times \vec{v}| = ......$

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