The value of |vec{a} cdot vec{b}|^2 + |vec{a} | vedclass

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(B) We know that the dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$.Thus, $|\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^

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$|\vec{a}|^2 |\vec{b}|^2$

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(B) We know that the dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$.Thus, $|\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 	heta$.The magnitude of the cross product is defined as $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta$.Thus, $|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta$.Adding these two expressions:$|\vec{a} \cdot \vec{b}|^2 + |\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 	heta + |\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta$.Factor out $|\vec{a}|^2 |\vec{b}|^2$:$= |\vec{a}|^2 |\vec{b}|^2 (\cos^2 	heta + \sin^2 	heta)$.Since $\cos^2 	heta + \sin^2 	heta = 1$, the result is $|\vec{a}|^2 |\vec{b}|^2$.

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

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