For any two vectors vec{a} and vec{b},|vec{a} | vedclass

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

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

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(C) We know that the magnitude of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta$,where $	heta$ is the angle between the vectors.Squaring both sides,we get $|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta$.Using the trigonometric identity $\sin^2 	heta = 1 - \cos^2 	heta$,we have:$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 (1 - \cos^2 	heta)$$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2 	heta$Since the dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$,it follows that $(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 	heta$.Substituting this into the equation,we get:$|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$.

For any two vectors $\vec{a}$ and $\vec{b}$,$|\vec{a} \times \vec{b}|^2$ is equal to

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