If |vec{A} times vec{B}| = sqrt{3} vec{A} cdo | vedclass

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(D) Given that $|\vec{A} 	imes \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$.Using the definitions of cross product and dot product,we have $AB \sin 	h

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$(A^2 + B^2 + AB)^{1/2}$

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(D) Given that $|\vec{A} 	imes \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$.Using the definitions of cross product and dot product,we have $AB \sin 	heta = \sqrt{3} AB \cos 	heta$.Dividing both sides by $AB \cos 	heta$,we get $	an 	heta = \sqrt{3}$,which implies $	heta = 60^\circ$.The magnitude of the resultant vector is given by $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Substituting $	heta = 60^\circ$ and $\cos 60^\circ = 1/2$,we get $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB(1/2)} = (A^2 + B^2 + AB)^{1/2}$.

If $|\vec{A} \times \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$,then what is the value of $|\vec{A} + \vec{B}|$?

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