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

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(D) Given: $|\vec{A} 	imes \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$Using the definitions of cross and dot products: $AB \sin 	heta = \sqrt{3} AB \

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

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(D) Given: $|\vec{A} 	imes \vec{B}| = \sqrt{3} \vec{A} \cdot \vec{B}$Using the definitions of cross and dot products: $AB \sin 	heta = \sqrt{3} AB \cos 	heta$Dividing both sides by $AB \cos 	heta$: $	an 	heta = \sqrt{3}$Therefore,the angle between the vectors is $	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}$: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 60^{\circ}}$Since $\cos 60^{\circ} = 1/2$: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB(1/2)} = \sqrt{A^2 + B^2 + AB}$Thus,$|\vec{A} + \vec{B}| = (A^2 + B^2 + AB)^{1/2}$

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

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