The value of frac{(vec{a} times vec{b})^2+(ve | vedclass

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

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$\frac{1}{2}$

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(C) We know that for any two vectors $\vec{a}$ and $\vec{b}$,the magnitude of the cross product is $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta$,where $	heta$ is the angle between them. Squaring both sides,we get $(\vec{a} 	imes \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta$. Similarly,the dot product is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$. Squaring both sides,we get $(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 	heta$. Substituting these into the given expression: $\frac{(\vec{a} 	imes \vec{b})^2 + (\vec{a} \cdot \vec{b})^2}{2|\vec{a}|^2 |\vec{b}|^2} = \frac{|\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 	heta}{2|\vec{a}|^2 |\vec{b}|^2}$ $= \frac{|\vec{a}|^2 |\vec{b}|^2 (\sin^2 	heta + \cos^2 	heta)}{2|\vec{a}|^2 |\vec{b}|^2}$ Since $\sin^2 	heta + \cos^2 	heta = 1$,the expression simplifies to: $= \frac{|\vec{a}|^2 |\vec{b}|^2 (1)}{2|\vec{a}|^2 |\vec{b}|^2} = \frac{1}{2}$.

The value of $\frac{(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2}{2|\vec{a}|^2|\vec{b}|^2}$ is

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