If overrightarrow{u}=overrightarrow{a}-overri | vedclass

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Question

(A) Given $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$.$\overrightarrow{u

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$.$\overrightarrow{u} 	imes \overrightarrow{v} = (\overrightarrow{a}-\overrightarrow{b}) 	imes (\overrightarrow{a}+\overrightarrow{b})$$= \overrightarrow{a} 	imes \overrightarrow{a} + \overrightarrow{a} 	imes \overrightarrow{b} - \overrightarrow{b} 	imes \overrightarrow{a} - \overrightarrow{b} 	imes \overrightarrow{b}$Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$ and $\overrightarrow{b} 	imes \overrightarrow{b} = 0$,and $\overrightarrow{b} 	imes \overrightarrow{a} = -(\overrightarrow{a} 	imes \overrightarrow{b})$:$= 0 + \overrightarrow{a} 	imes \overrightarrow{b} + \overrightarrow{a} 	imes \overrightarrow{b} - 0 = 2(\overrightarrow{a} 	imes \overrightarrow{b})$Taking the magnitude: $|\overrightarrow{u} 	imes \overrightarrow{v}| = 2|\overrightarrow{a} 	imes \overrightarrow{b}|$Using the identity $|\overrightarrow{a} 	imes \overrightarrow{b}|^2 = |\overrightarrow{a}|^2|\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2$:$|\overrightarrow{u} 	imes \overrightarrow{v}| = 2 \sqrt{|\overrightarrow{a}|^2|\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$Given $|\overrightarrow{a}| = 2$ and $|\overrightarrow{b}| = 2$,so $|\overrightarrow{a}|^2 = 4$ and $|\overrightarrow{b}|^2 = 4$:$|\overrightarrow{u} 	imes \overrightarrow{v}| = 2 \sqrt{(4)(4) - (\overrightarrow{a} \cdot \overrightarrow{b})^2} = 2 \sqrt{16 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$

If $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$,$\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$,$|\overrightarrow{a}|=|\overrightarrow{b}|=2$,then $|\overrightarrow{u} \times \overrightarrow{v}|$ is equal to

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